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Guido
04-13-2010, 01:18 AM
Howdy--------- Local fabricator wanting ideas/hardware for measuring torque which is being transmitted by a rotating shaft. Max rpm is 3,000, max torque may be as high as 800 lb-ft and max HP is limited to 600.

Strain guages come to mind, with attendant problems of getting a measurement signal off the spinning shaft. Any ideas or derivations of a localized signal transmitting/receiving device/system?

All thoughts appreciated.

--G

macona
04-13-2010, 01:31 AM
There are inline load cells with slip rings that do what you want. Omega should have what you need:

http://www.omega.com/toc_asp/subsectionSC.asp?subsection=F16&book=Pressure

Edit, maybe, maybe not. Biggest they have is rated for 10000 in/lb or 833ft/lb

Part #TQ501-10K $4075

becksmachine
04-13-2010, 01:34 AM
Can you put instruments/strain gages on the driving motor mounts? Might save the problems associated with slip rings etc.

Dave

macona
04-13-2010, 01:43 AM
I think you would be asking for problems doing something like that.

Heres a new in the box sensor on ebay:

http://cgi.ebay.com/Himmelstein-MCRT-Torquemeter-9-02TS-1-4-NEW-10-000-lb_W0QQitemZ380153794946QQcmdZViewItemQQptZLH_Defa ultDomain_0?hash=item5882ee9182#ht_500wt_1182

barts
04-13-2010, 01:47 AM
http://www.sensotec.com/torque.asp

In particular the clamp-on torque cells look handy....

- Bart

darryl
04-13-2010, 02:22 AM
I imagine you would want a digital readout of some kind. In that case, some electronics will be required. It's not too much of a stretch to say that a shaft-mounted device could be battery powered and send a signal to a receiving and processing circuit. I don't know of any such device, but I'm sure it exists somewhere.

macona
04-13-2010, 03:01 AM
Omega has a line of readouts that connect to them as well as the others.

The Artful Bodger
04-13-2010, 04:13 AM
Use an ampmeter to compare current between loaded and unloaded conditions?

Evan
04-13-2010, 06:40 AM
Measure the twist in the shaft by using optical sensors at each end that pick up an index mark once per rev. I could throw something together like that in a few hours. Cheap and easy. A commercial frequency counter with a pulse duration mode will give an accurate readout of the time difference which can be referenced to a converson table developed during calibration. Or, you can design and build a direct reading unit like this one. With only a slight modification it would do the job. It counts the duration from pulse to pulse by preloading the counters to maximum count and counting down until the next pulse latches the count to the displays. By adjusting the clock generator it can be made to read in whatever units are convenient.

http://ixian.ca/pics7/pulse.jpg

J Tiers
04-13-2010, 09:11 AM
The smartest idea may be to measure the reaction at the drive point. That requires virtually no calibration aside from physical measurements.

Measuring the twist of the shaft requires knowing things about the shaft, which is of course not impossible....... but.

Another approach, depending on precision and real-time requirements might be to measure the motor current... that can be one-time calibrated to the motor and will be good until the motor is re-wound or is damaged.

Evan
04-13-2010, 09:17 AM
Meauring the twist doesn't require knowing anything about the shaft, only the load.

vincemulhollon
04-13-2010, 02:32 PM
[QUOTE=GuidoMax rpm is 3,000, max torque may be as high as 800 lb-ft and max HP is limited to 600.[/QUOTE]

Bodger has the right answer, motor current with maybe a slight correction for RPM (as in, higher frictional/cooling fan losses at 3000 rpm than 30 rpm)

Real question is, (800*3000)/5252 = 456 horsepower... So your peak torque and peak rpm seem to limit you well under your 600 limit. But I'm reading it as a 600 HP, 3000 rpm motor is bolted to a shaft or gearbox that'll shatter above 800 lb-ft, which is inadequate shaft for a 600 HP motor.

In my opinion you should replace the shaft/gearbox/whatever with one that'll survive at least 1600 lb-ft (and thats a factor of safety of only 50%). Alternately, program the VFD, or install a restrictor plate, or swap out the motor to maybe a 400 HP one.

Easy way to remember the HP equation, BTW, is to just memorize 5280 aka feet in a mile. Its close enough.

Guido
04-13-2010, 03:16 PM
Been brainstorming this problem and a solution for years. Should have been more detailed with the problem of wanting to measure/display HP and torque on the dashboard of a Peterbuilt moving down the highway at speed. Idea popped up when we saw how turboprop engines display torque in real time. A nonlinear pressure guage is mounted on the panel, with pressure from engine compressor stage number so-and-so measured and displayed. The pressure guage displays in ft-lb of torque.

There's gotta be a market for a device to be added to an over the highway diesel truck tractor, which would reliably display torque/HP. Instant operator visual of motor performance.

Everyone's thinking and input is appreciated.

--G

kf2qd
04-13-2010, 03:37 PM
Put some type of transducer under the comporession side motor mount and figuure the distance to the opposite side motor mount. You then have a force and a distance which can calulate to torque. 800Lbs, and 24" spacing = 400 ft lbs.

ADGO_Racing
04-13-2010, 04:46 PM
Put some type of transducer under the comporession side motor mount and figuure the distance to the opposite side motor mount. You then have a force and a distance which can calulate to torque. 800Lbs, and 24" spacing = 400 ft lbs.


Since the engine is trying to spin around the crankshaft, I think you would only need the distance from the center line of the crank to the load cell. Not the mount to mount distance. The cell could be placed on either side, I would say the side in tension, as it would be much easier to attach a cell to the top of the mount. Rather than having to lift an engine of that size out of its mounts. Less to disassemble that way too.

macona
04-13-2010, 04:47 PM
Now you are going to have to have some sort of filter to clean out all the noise in the system.

ADGO_Racing
04-13-2010, 04:47 PM
Meauring the twist doesn't require knowing anything about the shaft, only the load.


You do need to know the diameter of the shaft, and the length between measuring points to calculate the twist.

philbur
04-13-2010, 05:25 PM
No but turning the twist into a torque value does require that you know something about the shaft. If he knew the load he could calculate the torque directly by measuring the rpm, but that is not the case.

Phil:)


Meauring the twist doesn't require knowing anything about the shaft, only the load.

J Tiers
04-13-2010, 07:53 PM
Meauring the twist doesn't require knowing anything about the shaft, only the load.

I see that others have already dealt with this.........

One DOES assume that a numeric measurement with some basis in normal units and reality is wanted......

Evan
04-13-2010, 08:46 PM
You do need to know the diameter of the shaft, and the length between measuring points to calculate the twist.



No but turning the twist into a torque value does require that you know something about the shaft. If he knew the load he could calculate the torque directly by measuring the rpm, but that is not the case.


Nope. All you need to know is how much twist is produced by a particular amount of torque on the output. That can be a purely relative value that is determined by an experienced operator who can tell when the system is close to it's limit. All the other possible methods face exactly the same issue if an absolute value of torque is required to be known and that is somehow measuring the actual torque on the output.

darryl
04-13-2010, 10:31 PM
I wonder how much twist a suitable shaft can withstand and still return to its original 'alignment'- for what it's worth, I just had a thought, so I'll contribute it-

Have the piece of shaft, the drive couplings on both ends, and two pieces of reasonably close fitting tubes placed end to end between the couplings. Clamp the ends of the tubes to the shaft at the ends where the couplings are. While at rest, scribe a mark across both tubes where they meet. If you twist the shaft, those marks will go out of alignment. How far out is determined by how much you twist the shaft.

You can calibrate that by putting a known force on the end of a known length of bar that's attached to one end of the shaft. Scratch the mating marks across for each torque value you apply. In this case, you might want a range up to maybe 1200 ft/lbs if you'll be routinely measuring roughly 800 ft/lb.

In operation, you read it using a strobe light to freeze the markings, or you could trigger the strobe from the rotating shaft.

J Tiers
04-13-2010, 10:34 PM
No but turning the twist into a torque value does require that you know something about the shaft.



Nope. All you need to know is how much twist is produced by a particular amount of torque on the output.

That would be "knowing something about the shaft"........;)

Evan
04-13-2010, 10:46 PM
Nope. All it is is knowing the relative alignment of each end of the shaft to the other. That does not reveal the properties of the shaft nor does it matter. That isn't what is being measured. It is a value that is directly coupled to the torque load and can be calibrated in absolute terms without even knowing what size or shape the shaft is.

J Tiers
04-14-2010, 12:49 AM
Nope. All it is is knowing the relative alignment of each end of the shaft to the other. That does not reveal the properties of the shaft nor does it matter. That isn't what is being measured. It is a value that is directly coupled to the torque load and can be calibrated in absolute terms without even knowing what size or shape the shaft is.

So, a shaft on a system has 0.7mm relative movement on the circumference between two marks 3 meters apart. End of data.

What is the torque on the shaft in NM?

Evan
04-14-2010, 12:56 AM
In case you missed it here it is again.

All you need to know is how much twist is produced by a particular amount of torque on the output. That can be a purely relative value that is determined by an experienced operator who can tell when the system is close to it's limit. All the other possible methods face exactly the same issue if an absolute value of torque is required to be known and that is somehow measuring the actual torque on the output.


It still doesn't require knowing anything about the properties of the shaft.


So, a shaft on a system has 0.7mm relative movement on the circumference between two marks 3 meters apart. End of data

By the way, we don't even need to know that much. All we need to know is that there is a timing difference between when the marks pass the sensors and that also doesn't need to be quantified. The most we need to know is that X timing difference corresponds to N torque.

J Tiers
04-14-2010, 01:01 AM
Evan, you said you don't need any data on the shaft, and I'm holding you to it...

What's the torque? Come on, give...... inquiring minds want to know.


Oh,, you mean you need MORE DATA? Like the relation of torque to deflection in that particular shaft?

Ah, so you DO want to "know something about the shaft", even though you SAID you didn't need to know that....




By the way, we don't even need to know that much. All we need to know is that there is a timing difference between when the marks pass the sensors and that also doesn't need to be quantified. The most we need to know is that X timing difference corresponds to N torque.

Which would be "knowing something about the shaft".....................

We will now return to step one and repeat......... in different words, but coming back to the same thing.......Since a timing difference on a particular size shaft at X rpm, i.e some actual physical system, corresponds precisely to a distance......

All you can get without SOME form of absolute calibration is an "assumed" relative relation between an "assumed" maximum output, and some distance ot timing difference, whichever you prefer to call it. "we are at about 76% of what we think is maximum, but we don't know the actual maximum number"........

Evan
04-14-2010, 01:06 AM
You can measure the torque on the output end of the shaft with a dyno. That is all you need to know to calibrate the system. If you don't understand that I can't make it any simpler. That doesn't give you information about the properties of the shaft since it doesn't even tell you the amount of deflection that takes place. You don't need to know that.

The only information that you obtain is a phase difference between input and output. That is enough.

J Tiers
04-14-2010, 01:11 AM
You can measure the torque on the output end of the shaft with a dyno. That is all you need to know to calibrate the system. If you don't understand that I can't make it any simpler.

Ah, but you only NOW want this dyno, which anyone would realize right away is one possible way to do it.

before you said you didn't NEED it.......

of course a dyno is not the only way......

If you know dimensions and material of the shaft, you can come reasonably close to reality with a calculated value for teh torque to deflection figure..

I understand this is how ships power was figured at various speeds, I don't seriously believe that the whole system was hooked to an underwater dyno system with 200,000 HP capacity while the ship was floating in the harbor.

Evan
04-14-2010, 01:17 AM
before you said you didn't NEED it.......



I said no such thing. Go back and read the postings Jerry. The dyno is for calibration which is precisely what I said in two different posts.

For the third time:

All the other possible methods face exactly the same issue if an absolute value of torque is required to be known and that is somehow measuring the actual torque on the output.


I don't seriously believe that the whole system was hooked to an underwater dyno system with 200,000 HP capacity while the ship was floating in the harbor.


Yes it was. It's called a propellor. That is exactly how some dynomometers work.

gearedloco
04-14-2010, 02:08 AM
Been brainstorming this problem and a solution for years. Should have been more detailed with the problem of wanting to measure/display HP and torque on the dashboard of a Peterbuilt moving down the highway at speed. Idea popped up when we saw how turboprop engines display torque in real time. A nonlinear pressure guage is mounted on the panel, with pressure from engine compressor stage number so-and-so measured and displayed. The pressure guage displays in ft-lb of torque.

There's gotta be a market for a device to be added to an over the highway diesel truck tractor, which would reliably display torque/HP. Instant operator visual of motor performance.

Everyone's thinking and input is appreciated.--G

Way back in, oh maybe 1958, my dad and I were driving west on the PA Turnpike on the section that climbs towards the eastern-most tunnel. We had a similar conversation, just from the standpoint that it would be interesting to watch the changes due to throttle changes. That section of road was kind of twisty, and the grade wasn't particularly constant.

We came to the conclusion that some sort of way to measure the movement of the motor mounts would be a possibility, but we had no way to calibrate it.
Our conversation drifted off to other things of similar importance.:cool:

By the by, I don't know if big trucks have the engine/transmission set up the same way that cars did then, but the torque measured that way would have to have the torque multiplication of the torque converter (if one is used) and the transmission factored in if one really wanted to know the actual torque at the crankshaft. I'd suspect the torque converter factor to be a tad-bit on the un-constant side.;)

I haven't been over that stretch of highway in quite a few years, but I'm sure the road no longer follows the path it did then. They might have even bypassed the tunnels.

-bill

winchman
04-14-2010, 03:27 AM
Way back in the late '60s when I was working for Avco-Lycoming, we measured the torque on the engines under test. A water brake was mounted to a ring supported by four short beams. Each beam had strain gages bonded to two sides, and their values changed as the beams flexed from the torque absorbed by the brake. The beams and their mounting rings were machined from one piece of steel tubing. The systems were very durable and reasonably accurate.

You could probably substitute a dial indicator for the strain gages for a less complicated system.

The Artful Bodger
04-14-2010, 04:59 AM
Did someone mention this already? If measuring the twist is what is required that is easily indicated by triggering a strobe at one end of the shaft and aiming the strobe at a scale stuck on the other end of the shaft.

Weston Bye
04-14-2010, 06:54 AM
I wonder how much twist a suitable shaft can withstand and still return to its original 'alignment'- for what it's worth, I just had a thought, so I'll contribute it-

Have the piece of shaft, the drive couplings on both ends, and two pieces of reasonably close fitting tubes placed end to end between the couplings. Clamp the ends of the tubes to the shaft at the ends where the couplings are. While at rest, scribe a mark across both tubes where they meet. If you twist the shaft, those marks will go out of alignment. How far out is determined by how much you twist the shaft.

You can calibrate that by putting a known force on the end of a known length of bar that's attached to one end of the shaft. Scratch the mating marks across for each torque value you apply. In this case, you might want a range up to maybe 1200 ft/lbs if you'll be routinely measuring roughly 800 ft/lb.

In operation, you read it using a strobe light to freeze the markings, or you could trigger the strobe from the rotating shaft.

Well described! Such a mechanical arrangement is often used to measure torque. However, rather than the strobe and index marks, (not a bad idea for a manual setup) a pair of gears are fitted on the ends of each tube in place of the index marks. A pair of gear tooth sensors are placed adjacent to the gears and the phase difference of the sensor outputs is measured with a circuit such as Evan suggested, or more simply with an op-amp circuit (RPM needs to be factored in either case) feeding an analog meter movement - perhaps more appropriate in the Peterbuilt application that Guido has in mind.

J Tiers
04-14-2010, 09:20 AM
Yes it was. It's called a propellor. That is exactly how some dynomometers work.

OK. the shaft is turning a propellor... NOW what's the torque?

Didn't think that helped much.......

Aside from that being entirely encompassed and made totally *obvious* by my prior statement about ship power measurement.... you are just transferring the calibration problem to a different point on the shaft.....

if you REALLY want a measurement, as I and a couple others have pointed out, you may get an absolute measurement by measuring the reaction at the engine mounts..... At that point you don't need any shaft dingamusses.... but it may be a lot less convenient in certain cases.

BTW, according to my shipbuilding text, it was not unknown to calibrate the shaft, or a section of it, at the works, so that later measurements could be accurately made...

A force and lever are a means of absolute calibration. Any known force and known lever will do if it provides a measurable deflection in same conditions as in use.

Evan
04-14-2010, 09:28 AM
OK. the shaft is turning a propellor... NOW what's the torque?



You sure are going to great lengths to try and cover up your mistake. Reminds me of a cat trying to scratch dirt over some *hit on a rug.

The torque is whatever the drive system is rated to produce. I am sure somebody knows that. The propellor is the load. Duh.

rowbare
04-14-2010, 11:34 AM
You can measure the torque on the output end of the shaft with a dyno. That is all you need to know to calibrate the system. If you don't understand that I can't make it any simpler. That doesn't give you information about the properties of the shaft since it doesn't even tell you the amount of deflection that takes place. You don't need to know that.

The only information that you obtain is a phase difference between input and output. That is enough.

If you have a dyno, what is the point in dicking around with reference marks on a shaft. Read the torque off of the dyno.

If the OP had a dyno, he wouldn't have needed to start this thread.

I am puzzled, isn't the phase difference a function of the deflection? In that case using a dyno to determine what phase difference corresponds to what amount of torque IS telling you something about the shaft...

bob

Weston Bye
04-14-2010, 01:58 PM
Been brainstorming this problem and a solution for years. Should have been more detailed with the problem of wanting to measure/display HP and torque on the dashboard of a Peterbuilt moving down the highway at speed. Idea popped up when we saw how turboprop engines display torque in real time. A nonlinear pressure guage is mounted on the panel, with pressure from engine compressor stage number so-and-so measured and displayed. The pressure guage displays in ft-lb of torque.

There's gotta be a market for a device to be added to an over the highway diesel truck tractor, which would reliably display torque/HP. Instant operator visual of motor performance.

Everyone's thinking and input is appreciated.

--G

The most accessable place on the Peterbuilt would be the driveshaft between the transmission and the differential. Anyplace else would require considerable reengineering of the transmission and/or engine. Torque reaction motor mounts would have to contend with gravity and inertia factors. The driveshaft location suggests a bolt-on aftermarket solution.

Some engineering would be necessary to size components properly and choose the appropriate materials. The torque element would have to be sized to twist a little bit, but not yield. Perhaps some torque limiting positive stops. A balancing act of engineering - I've seen stock drive shafts twisted in two.

Evan
04-14-2010, 02:18 PM
In that case using a dyno to determine what phase difference corresponds to what amount of torque IS telling you something about the shaft...


The dyno is for calibration only. You don't leave it attached to the system. Tell us then what knowing the phase difference tells you about the shaft? It doesn't tell you how much it twists since that hasn't been measured and we don't need to know that. All we need to know is what degree of twist corresponds to the established torque limit.

Rpm differences can be automatically factored into the phase difference measurement by just using a tachometer output from the engine as the phase clock. There is no need to develop any sort of compensation table or other complication.

RPease
04-14-2010, 02:40 PM
Easy way to remember the HP equation, BTW, is to just memorize 5280 aka feet in a mile. Its close enough.


And I thought the "easy way" was to count the number of feet and divide by 4.........Of course, that assumes each horse has 4 feet.........:D

I thought this thread needed a little humor........;)

Rod

PaulT
04-14-2010, 02:48 PM
Measuring the shaft rotational deflection might work, but you should keep in mind that there will be a lot of vibrational and harmonical related components in that measured output. Those likely would have be well filtered to get a meaningful measurement, I don't think just a simple digital display of the difference in rotation would be useful, it would just be flying all over the place.

Its possible that if the overall speed of the shaft is pretty constant that a simple weighted or sliding averaging method would work.

I've worked on developing instrumentation for dyno's, and handling measurement noise and vibrational components in the measurements is a significant issue.

If I had a choice and its practical to implement I'd go for the load cell/beam approach, I think there would be less measurement noise to deal with there versus measuring the shaft deflection, particularly if the shaft has a reasonable safety factor.

If you do decide to try the shaft deflection method you may not have to use encoders to do that. If there is anything with a reasonable radius on either end of the shaft (or if you can add "measurement" wheels) just glue magnets on the outside radius of the wheels and set up proximity switches to catch the magnets going by. Then you just need to time the difference between the switch transitions. This could be implemented pretty cheaply using a microcontroller with an internal timer/counter.

I still think this output would need to be filtered or at least averaged to get a useful result.

Paul T.

RWO
04-14-2010, 03:05 PM
The problem has been solved for many years. See http://www.sensotec.com/torque.asp?category=wireless for many different designs to fit nearly any mechanical configuration.

RWO

Swamp Donkey
04-14-2010, 03:26 PM
Drive shaft torque gauge.

http://www.cosworth.com/Default.aspx?id=1095593

The Artful Bodger
04-14-2010, 03:37 PM
In the case of the truck there will be some movement in the suspension in response to torque and that may be a more convenient place to mount a strain or movement sensor.

philbur
04-14-2010, 06:11 PM
Which means you need to know something about the shaft, ie its torsional stiffness.

Phil:)


Nope. All you need to know is how much twist is produced by a particular amount of torque on the output.

Evan
04-14-2010, 06:17 PM
No you don't. You only need to calibrate whatever phase difference occurs with a certain applied load. You don't even need to know what material the shaft is made from. It could be metal or wood or fiberglass or day old french bread. Why is this so difficult to understand?

X load produces Y twist. Period.

philbur
04-14-2010, 06:27 PM
Which is effectively the torsional stiffness, you just haven't bother to give it it's name.

Your point seems to be that if you don't call a black sheep "a black sheep" then its not a black sheep.

Phil:)


X load produces Y twist. Period.

ADGO_Racing
04-14-2010, 07:04 PM
I think what Evan is after is a poor way of measuring the twist in the shaft, based on Bubbas calibrated ear...When Bubba says "If it grunts anymore it'll f^&%in fly to bits"(Add your favorite regional hill billy accent:p ). Then measure the twist. This being the "Maximum Twist". That is probably ok for a backwoods arrangement. But the reality is you DO NOT know anything about the HP or the shaft. The addition of a Dyno, will give you a maximum Bubba applied torque, which still tells you NOTHING ABOUT the shaft other than: this much Bubba=this much twist. What is needed is the diameter, the wall thickness (If any), the length, and material type. Will all allow you to calculate the maximum allowable stress for a given torque.

Evan's method will not tell you if you are operating in a safe range. As Bubbas calibrated ear, may well be (And usually is) pushing things to the MAX. So you may be right on the verge , but repeatedly pushing to Bubbas max, will cyclically weaken the shaft and one day operating within the Bubba calibrated system will fly to pieces anyway. Bubbas calibration does not take into account other factors, such as shock loads, a whole new problem.

Plus the point of the original post was/is to do something which doesn't so much worry about the shaft as it is already a reliable shaft, but to give the driver some feedback information. This is easily accomplished with a load cell under an engine mount and knowing the distance from the Crank center line to the load cell. A small electronics package can condition the signal and display the information.

Weston Bye
04-14-2010, 07:10 PM
No you don't. You only need to calibrate whatever phase difference occurs with a certain applied load. You don't even need to know what material the shaft is made from. It could be metal or wood or fiberglass or day old french bread. Why is this so difficult to understand?

X load produces Y twist. Period.

Evan is correct - if the shaft already exists. However, everybody else insists on knowing the properties of the shaft as if they had to engineer it into existence for the application. I would perfer to know the material properties and do the calculations - instead of empirically starting with day old french bread and working my way up to 4140, heat treated.

Given a mix of superior intellegence, knowledge, experience, intuition and luck, the TLAR method (That Looks About Right) sometimes works for new applications. I've used the method myself, but I perfer to do the calculations - not real easy as I tend to be math challenged.

Now, if I had an existing application, apply a known load, measure the deflection and call it good. In such case it would be good to know the material characteristics to know what the limits are.

J. R. Williams
04-14-2010, 07:19 PM
Contact a company by the name of "Himmelstein" as they offer a wide range of torque measuring equipment. (bring money) A big diesel engine would be a problem mounting torque measuring load cells directly on the engine mounts. Many years ago I installed a big Mack diesel engine on a test stand it rough on the drive line and couplings.

JRW

Evan
04-14-2010, 08:33 PM
What is needed is the diameter, the wall thickness (If any), the length, and material type. Will all allow you to calculate the maximum allowable stress for a given torque.



The poster didn't ask for that information, only a way to quantify the torque in an existing system.

J Tiers
04-14-2010, 10:39 PM
You sure are going to great lengths to try and cover up your mistake. Reminds me of a cat trying to scratch dirt over some *hit on a rug.

The torque is whatever the drive system is rated to produce. I am sure somebody knows that. The propellor is the load. Duh.


Sorry, no mistakes here......Unfortunately for your cute efforts to transfer the cat crap to me, you do NOT KNOW the torque the system is "rated" to produce IS ACTUALLY BEING PRODUCED..... That would be part of the reason for the testing.....to prove that the contract has been fulfilled.

For a ship, the owners contractually specified a penalty if the custom designed propulsion system did not produce the specified power output AS MOUNTED IN THE SHIP. They may or may not have allowed a bonus for overage.

Therefore, the legal system is involved, and "Mr Bow Tie" may have calculated a certain power from the turbine and boiler system (a few years ago of course), but that cuts no ice.... the owners want to know that it works in the final "product"....*

So, you may use the propellor as a load, which is rather obvious if you take measurements while under way, as is normal, but you haven't got it calibrated.... So either the shaft has been calibrated in the shop, or the shaft has been calculated, whichever the owners accept...

But you MUST know something about the shaft in order to get a numeric, quantified, traceable measure of the output by measuring what happens to the shaft.....

Since you want to be cute, Evan, I will point out that others made this same point, and your reply was that "no you don't need to know anything about the shaft......" Which would only be true if you don't care about the numbers.

You seem to have , as John S says, "backpedaled fast" on that statement.

I don't get any particular pleasure from pointing out your mistakes and re-directions of discussions, but I do very much resent someone turning everything upside down and trying to spin their own mistakes and mis-statements in a discussion so as to attempt to fasten your own mistakes onto others.

Frankly, Evan, I have NEVER ONCE seen you admit to being "wrong". The rest of us do so whenever it is appropriate.....

Now, your MODIFIED statements that you don't NEED to know anything about teh shaft IF you calibrate it, are almost correct..... EXCEPT THAT KNOWING WHAT YOU SUGGESTED


No you don't. You only need to calibrate whatever phase difference occurs with a certain applied load.

STILL CONSTITUTES KNOWING AT LEAST ONE PROPERTY OF THE SHAFT....... I am not sure why this is so difficult to understand......

You are using the shaft as a comparison meter, and the one essential thing about a meter is that you want the scale factor...... otherwise the only information you know for sure is that the conditions are now 'different" than before... possibly "more different" possibly "less different".... But how MUCH different, that you know nothing usefully numeric about unless you find the scale factor, i.e. calibrate.

Personally, given the obvious issues dealing with the shaft, I think it could be a lot more useful and direct to use a reaction of the shaft driving device against its mounts to get the torque..... That can often be easily calibrated by measuring physical distances.

And, for devices such as the prony brake, is exactly how it is done... measure reaction force on a lever pivoted at the shaft, and you have the torque. All weights and measures stuff, quite easy to calibrate. You might need to compensate for bearing friction, depending on how fussy you want to be.

For hundreds of thousands of HP in a ship, it is not practical, and the prop as a load, with the shaft as a meter makes sense... it is worth the efforts of calibration etc, to get that measurement via the shaft, as it is the easiest point to attack.

*These days, with diesel engines instead of turbines and boilers, it is credible to get a test-bed power number........ and the requirement for in-place testing is less, but still that measurement would, I am reasonably sure, be part of the test conditions, as a power check. If you look through the "Doxford" thread over at PM, you will see some rather large water brakes coupled to big marine diesels on the test bed...

Evan
04-14-2010, 11:27 PM
Frankly, Evan, I have NEVER ONCE seen you admit to being "wrong". The rest of us do so whenever it is appropriate.....


That is because you either don't bother to read all the posts in a thread (which you have admitted) or you have poor reading comprehension or both. I have on various occasions admitted to being mistaken.


I don't give a shi* about your ship Jerry. You, as usual are flailing madly about trying to distract the audience and wandering further and further from the question. The simple fact is that in this instance I am correct and you are not. Weston understands my point, you do not.


STILL CONSTITUTES KNOWING AT LEAST ONE PROPERTY OF THE SHAFT

Oh? What property? Keep in mind we don't know the actual amount of twist.

ADGO_Racing
04-14-2010, 11:52 PM
And still, as I said before Bubbas calibrated ear isn't a "Quantity". Knowing nothing about the shaft, and using this sort of test would only show you how much the shaft deflects (Twists) under Bubbas "Umm, this is enough throttle" calibration system. You cannot derive any torque from that, you cannot derive HP from that, you cannot derive a factor of safety, or safe limits from that information, unless you know something about the shaft. The twist of the shaft is useless, except in the instance that the engineering department specifies a certain deflection between two points at a given distance. Then this is ok for operators with little or no understanding of what they are measuring, and could care less. It still does not tell the operator anything about HP or torque, without a table to explain it or the proper information to calculate it.

Evan
04-15-2010, 12:08 AM
The twist of the shaft is useless, except in the instance that the engineering department specifies a certain deflection between two points at a given distance. Then this is ok for operators with little or no understanding of what they are measuring, and could care less. It still does not tell the operator anything about HP or torque, without a table to explain it or the proper information to calculate it.

This is one of the most important instruments on a helicopter. Notice anything special about it?

http://ixian.ca/pics7/torque1.jpg

Evan
04-15-2010, 12:21 AM
Gee, waddaya know? Looks like somebody else likes my idea too.



News Release Number: EJK201004091 09-Apr-10

FRCSE DET Mayport upgrades engine test cell
FRCSE DET Mayport upgrades test cell
“improves helicopter engine efficiency”

By FRCSE Public Affairs

Fleet Readiness Center Southeast (FRCSE) Detachment (DET) Mayport upgraded the Engine Test Cell propulsion support equipment in February to improve GE T700-401C helicopter engine efficiency and accuracy for the squadrons the detachment supports.

The equipment was upgraded from a reaction-based torque system to a phased-shift torque system that measures the twist of a shaft mounted between the engine and a dynamometer, an instrument used for measuring mechanical power.
Mechanical Engineer Brian Zarko from Naval Air Systems Command, Lakehurst, N.J., assisted with the upgrades. He said the engine produces the torque (turning force) and the torque meter measures the torque transmitted by a rotating shaft.

“Mayport FRCSE is the only activity currently using the new torque measurement system,” said Zarko. “The new torque meter uses two pick-ups to measure the twist of a toothed shaft for improved accuracy.”

Zarko said the system reduces the need for expensive periodic testing, which requires an engine with known performance to verify the correction factors are still accurate on the engine being tested. It requires an oil supply and return, which prompted the modification. The upgrades also replace dated equipment that can prove problematic when finding replacement parts.

Test cell operator Aviation Electrician’s Mate Terry Lamb said the new system makes calculating the torque values easier and faster for the technician at the beginning of the performance run.

“You no longer have to allow for a correction factor for the torque to each engine,” he said. “You still have to compute the torque value, but there is one less step.”

The technicians ensure the engine is outputting the correct horsepower using a dynamometer to simulate the load from a turning rotor head. They conduct a performance calibration on all required instruments and make sure all gauges are reading correctly.



http://www.navair.navy.mil/press_releases/index.cfm?fuseaction=press_release_view&press_release_id=4301&site_id=7

J Tiers
04-15-2010, 01:17 AM
T

Oh? What property? Keep in mind we don't know the actual amount of twist.

You said it yourself......... did you forget?

You want to know the reaction of the shaft in some measureable property, to a given load...... In this case you suggest the twist, as expressed by your phase difference.

Squeal and squirm and kick as much as you like...... that is "knowing something" about the shaft..... how is this difficult to understand?

You are calibrating the shaft in twist, or measured phase shift, or whatever. Nitpicky points about the method of measurement make no difference, you come out with "SHAFT UNITS" of some sort which calibration translates to internationally accepted "units". And equally nitpicky points about not calculating the shaft etc are invalid, because that is merely arguing about the MEANS of calibration, and not the FACT of it.

ANY measuring element gives you a varying output..... but you need to scale that output relative to known units in order to use it to get "real", relateable-to-real-world numbers..... Otherwise it is, as Mr Bye said, simply "Bubba units".....

When people said previously that you needed to calibrate the shaft in some way, you flat denied it, and then you turned right around and agreed both then and now that it needs to be calibrated.....as if it was your own new and novel idea. :rolleyes:

In the case of a known system, where you already know everything, can know maximum power, and have a way to calibrate the shaft, I can easily see that a twist-based (call it phase if you want) measurement could be lighter than other types, and so more desirable for a device (helicopter) with a slow-moving shaft.

In that case, you don't need to calibrate the same way as when everything is unknown, you set the system to known max, and label your output 100%, zero is obviously at rest, or with some minimum condition if at rest isn't appropriate.

Assuming the shaft is linear in response, which it should be, you now can mark the dial between those points, and a reasonable amount past 100%.

In the specific case in your quote, an engine test cell, the shaft is part of the equipment, calibrated once, and any engine connected to it can be measured by it, because they "KNOW" the shaft response to torque, and can use that to determine the engine input. That would be an "absolute" calibration, not a relative one like the instrument panel dial.

They didn't make up numbers..... I'd wager that 100% or 120% on that dial has a real meaning in terms of output to the rotor.... which means they did calibrate the system at least as I describe, and probably more thoroughly.

That PRESUMES the ability to provide known inputs to the system..... But you are STILL "knowing something" about the shaft...... see above.

You may WISH I didn't understand the point as well as Weston bye, but in fact I do quite well understand it.

of course if it suits you to deny it..... so be it, we know what's up.

Evan
04-15-2010, 02:12 AM
Squeal and squirm and kick as much as you like...... that is "knowing something" about the shaft..... how is this difficult to understand?


A phase difference isn't a property of anything. It isn't even an entity. It is no more than a concept. I thought you claim to be an engineer. :rolleyes:


You are calibrating the shaft in twist, or measured phase shift, or whatever. Nitpicky points about the method of measurement make no difference, you come out with "SHAFT UNITS" of some sort which calibration translates to internationally accepted "units". And equally nitpicky points about not calculating the shaft etc are invalid, because that is merely arguing about the MEANS of calibration, and not the FACT of it.



Nope. Notice the helicopter torque gauge? What is it telling you about the shaft? Nothing at all but that is where the value is derived. You can measure a phase difference because of torque reaction without even knowing how the torque is transmitted. There might be a shaft inside the black box, or a torque convertor, or a pair electric motors acting as a servo pair, or a magnetic drive or an eddy current drive. Measuring the phase difference tells you about the LOAD since that is the only reference you have when you calibrate it. It is also the only thing that you need to know in a helicopter and in the truck that the OP asked about.

philbur
04-15-2010, 04:36 AM
What a ridiculous statement, a phase difference is a property of a shaft under load. You are using this very property of phase difference to produce your torque gauge. You are spouting nonsense in order to justify an insult, clearly you are not an engineer.

The fact that the user of a torque gauge doesn't know anything about the shaft doesn't mean the torque gauge doesn't. When you measure (calibrate) the shaft twist for a given load you are measuring its torsion rigidity, this can equally be calculated. This value, measured or calculated, has to be input to the torque gauge by the designer or builder at some point. Clearly you understand this but equally clearly you seem unable to concede the point.

My last word on the matter.
Phil:)



A phase difference isn't a property of anything. It isn't even an entity. It is no more than a concept. I thought you claim to be an engineer. :rolleyes:

Evan
04-15-2010, 05:23 AM
What a ridiculous statement, a phase difference is a property of a shaft under load. You are using this very property of phase difference to produce your torque gauge. You are spouting nonsense in order to justify an insult, clearly you are not an engineer.



Wow! Another victim of hoof in mouth disease. It boggles the mind. The difference (a derivative) in position in time of something is not a property of anything. It is a purely mathematical concept. It has no realizable existence.

Even the values required to determine a phase difference are not properties of anything. Location is not a physical or mechanical property. Location can only be described as relative to something else as there is no such thing as absolute location in time and space.

Tell me, from what property is a phase difference derived, in your mind? RPM, Torque? You will need to show that both of those are also mechanical or physical properties. That will prove very problematic since a property of something cannot be defined by when it happens. That is called an "Event".

You would do well to avoid pursuing this line of argument further as it can only reflect badly on your knowledge and judgement.

electrosteam
04-15-2010, 09:05 AM
Hi,
As an observer from afar, Sydney Australia, I find the somewhat heated discussion on this subject interesting.

About 27 years ago (1983) I was responsible for fitting torque measurement equipment on the drive shafts to 2 x 1200 kW drives to the upper and lower rollers on an aluminium (spelling is correct for us) rolling mill.

The instrument used drive-shaft cemented strain gauges connected to a battery operated FM (frequency modulated) transmitter with a receiver and display to the mill operator, and to recording instruments.

Calibration was done at standstill with known performance current/torque values for the DC motors, followed by (as best we could with 20T blocks of aluminium at speed) steady-state runs with records of DC motor power ( volts and amps) and speed.

The calibration procedure removed the need to know anything about the shafts and we did not know what the actual torsional twist (phase shift) in the shaft was.

Torque applied to the shaft provides a torsional stress that results in a twist, or phase shift. The amount of twist is dependent on the elastic properties of the shaft, Young's Modulus of the material and the physical size.

All we knew was that a certain amount of kNm (1000s of ftlb) torque produced a predictable and repeatable display on the meter.

The important thing is that we went through a "calibration procedure" to get there.

The measurement equipment was installed on the mill to improve the knowledge data base on the statistics of block impact dynamic torque peaks to provide some predictions on the expected fatigue life of the shafts, gears and couplings.
We had just had our first major failure after 25 years of mill use ( mill commissioned in 1958 ), a shaft coupling, and the block size and mill throughput rate were increasing each year by management decree.

Hope this description of an actual installation helps,
Happy machining,
John.

J Tiers
04-15-2010, 09:41 AM
Wow! Another victim of hoof in mouth disease. It boggles the mind. The difference (a derivative) in position in time of something is not a property of anything. It is a purely mathematical concept. It has no realizable existence.

Even the values required to determine a phase difference are not properties of anything. Location is not a physical or mechanical property. Location can only be described as relative to something else as there is no such thing as absolute location in time and space.

Tell me, from what property is a phase difference derived, in your mind? RPM, Torque? You will need to show that both of those are also mechanical or physical properties. That will prove very problematic since a property of something cannot be defined by when it happens. That is called an "Event".

You would do well to avoid pursuing this line of argument further as it can only reflect badly on your knowledge and judgement.

This borders on the insane. There is NO other way to describe it..... Evan presumably is not insane, but his argument here is basically insane.

The most obvious and central point of the issue is that while the label you place on the measurement may NOT refer to something you can pick up and handle........ the EFFECT IS PHYSICAL

The shaft material may be pressed dog$hit, but the torque on it produces a PHYSICAL effect on the shaft, which YOU CHOOSE to measure/quantify by something we agree to call phase.

The effect you are measuring, whether measuring absolutely, or relatively, is physical, and each different shaft sample will be slightly different, requiring a calibration to produce results which are meaningful when related to something outside that section of shaft..... You have to know that phase results from full torque, half torque, or whatever. In fact, you really need to know two points and make an assumption of linearity.

Since you insist, the phase difference *in this case* (you must be measuring relative to something, by your own admission) derives at base from the twisting of the shaft under load. That means the "phase" derives from a PHYSICAL relative change of position at one end of a section of shaft vs the other.

Evan, let the philosophical implications go, and concentrate your mind on the fact that physical things are happening... no matter in what airy realm of "concept and 'event" you want to describe them.

The "event" here is that you are just measuring the twist of the shaft, in relative units...... Which are then, in some way related to actual physical torque so that the meter dial has meaning....

Evan
04-15-2010, 12:40 PM
The most obvious and central point of the issue is that while the label you place on the measurement may NOT refer to something you can pick up and handle........ the EFFECT IS PHYSICAL


Why yes, it is. So what? You just nailed down my argument and you don't realize it. It is an EFFECT. That does not constitute a PROPERTY of anything.

Thank you.

PaulT
04-15-2010, 01:30 PM
Yikes, its a battle of the Irresistable Force against the Immoveable Object.

This makes it a battle of a PROPERTY versus an EFFECT, with so much at stake!

Evan
04-15-2010, 04:12 PM
There is more than that. There is also the difference between Physical properties and mechanical properties.

Then there is the entire concept of phase, phase differences and phase shift.

Phase has a number of definitions depending on the subject. Materials may undergo a "phase change" which is a change in a physical property of the material. However, that has no applicability to this situation.

In this case the applicable definition of Phase is: A particular point in the time of a cycle; measured from some arbitrary zero and expressed as an angle. That is not in any way a property, only the description of the state of a system in motion. Going further, we may measure that state at more than one place and compare the results. Disregarding the fact that it is impossible to measure two separated events at the same time we can subtract one value of phase from the other and the result is the difference in phase.

A difference in phase is a derivative of the measured state of a system in motion when measured at more than one location. It is derived from measurements that do not represent properties, either physical or mechanical. Note that the measurement of phase requires the establishment of an ARBITRARY reference. Arbitrary means: Random or without a specific bias, cause or reason.

One cannot describe a physical or mechanical property by using arbitrary values. Since phase is based entirely on an arbitrary starting point for the measurement that starting point can be assigned to any point in time and any state of the system. That makes it impossible for the concept of phase difference to be used to define a mechanical property.

Weston Bye
04-15-2010, 06:03 PM
There is more than that. There is also the difference between Physical properties and mechanical properties.

Then there is the entire concept of phase, phase differences and phase shift.

Phase has a number of definitions depending on the subject. Materials may undergo a "phase change" which is a change in a physical property of the material. However, that has no applicability to this situation.

In this case the applicable definition of Phase is: A particular point in the time of a cycle; measured from some arbitrary zero and expressed as an angle. That is not in any way a property, only the description of the state of a system in motion. Going further, we may measure that state at more than one place and compare the results. Disregarding the fact that it is impossible to measure two separated events at the same time we can subtract one value of phase from the other and the result is the difference in phase.

A difference in phase is a derivative of the measured state of a system in motion when measured at more than one location. It is derived from measurements that do not represent properties, either physical or mechanical. Note that the measurement of phase requires the establishment of an ARBITRARY reference. Arbitrary means: Random or without a specific bias, cause or reason.

One cannot describe a physical or mechanical property by using arbitrary values. Since phase is based entirely on an arbitrary starting point for the measurement that starting point can be assigned to any point in time and any state of the system. That makes it impossible for the concept of phase difference to be used to define a mechanical property.

I, in my ignorance, fail to see the revelance of this to the original post. Evan (and others?) have gone off on a tangent. While I can't dispute the above treatise, I can say that I don't need the information to build a practical torque sensing device, because I have already done it.

I hearken back to Darryl:


from darryl post #21:
Have the piece of shaft, the drive couplings on both ends, and two pieces of reasonably close fitting tubes placed end to end between the couplings. Clamp the ends of the tubes to the shaft at the ends where the couplings are. While at rest, scribe a mark across both tubes where they meet. If you twist the shaft, those marks will go out of alignment. How far out is determined by how much you twist the shaft.

You can calibrate that by putting a known force on the end of a known length of bar that's attached to one end of the shaft. Scratch the mating marks across for each torque value you apply. In this case, you might want a range up to maybe 1200 ft/lbs if you'll be routinely measuring roughly 800 ft/lb.

In operation, you read it using a strobe light to freeze the markings, or you could trigger the strobe from the rotating shaft.

The strobe represents a point in time where the shaft is flexed. The fact that the whole shaft is rotating is inconsequential, and also inconvenient. We just want to know how much it is twisted at that moment, as the twist represents the torque applied. The strobe is only a method of gathering data, just as gear teeth and sensors, or strain gauges and RF links, or even slip rings are methods of getting the data off the rotating shaft. THey all convey the same information: the shaft is twisting to some degree or another. That is all that needs to be known.

Perhaps Evan's assertions are true - if you neet to be splitting infinite fine hairs.

Evan
04-15-2010, 07:43 PM
I, in my ignorance, fail to see the revelance of this to the original post.

It isn't relevant. I simply mentioned that the properties of the shaft were irrelevant to the method that I proposed to measure the torque. Jerry immediately siezed what he thought was an opportunity to try and show that I was mistaken. He then took this discussion to where it is now. I have better things to do than engage in this sort of argument but I refuse to let an alleged error in fact stand as if it were true, especially if it relates to information that I have provided.


Perhaps Evan's assertions are true - if you neet to be splitting infinite fine hairs.


It isn't splitting hairs. Knowing the difference between the physical properties, the mechanical properties and the effects that materials are subject to is the difference between structures that fall down or don't.

J Tiers
04-15-2010, 09:15 PM
[quote=Evan]That does not constitute a PROPERTY of anything.[/b]

SO WHAT.....? Your phase results from a property of the shaft, which you MUST USE to get the measurement.....

basically, its a "load"...... of crap to argue differently, so why do you?

the facts are, the shaft twists, you measure it. Call it "phase" and say it is an "event" and not related to a "property" if you want, that's just wrong and a red herring to boot.

Go there if you want to try to "faze" the audience, but that's all a crock...... That BS about "phase as an event" has no relation to engineering reality......

Squeal and kick as much as you want, the answer is the same.....You are measuring twist, and you need to relate 'amount of twist" to 'amount of torque".

Evan, people are LAUGHING AT YOU for this...... just let it drift away and preserve some shreds of dignity here.......



BTW, for once there is a real relevance to the HSM in the argument, since lot of folks do build engines, and some of them are interested in measuring properties of those engines, including torque, power etc. it's best to not obscure the subject with the "philosphy of phase".........

Evan
04-15-2010, 10:46 PM
Evan, people are LAUGHING AT YOU for this...... just let it drift away and preserve some shreds of dignity here.......


Why should I preserve your dignity? You have done this to yourself. I said originally that you didn't need to know anything about the shaft and you have nearly turned yourself inside out to justify your mistaken assertion otherwise. In the process you have revealed an astounding and troubling lack of knowledge about basic matters of physics. This has not been a good day for you Jerry.

J Tiers
04-16-2010, 12:04 AM
Why should I preserve your dignity? You have done this to yourself. I said originally that you didn't need to know anything about the shaft and you have nearly turned yourself inside out to justify your mistaken assertion otherwise. In the process you have revealed an astounding and troubling lack of knowledge about basic matters of physics. This has not been a good day for you Jerry.


LOL, way to try to turn it inside out.... :D :D

You are truly incorrigible. I am actually a bit worried about you... certain observable trends are not so good.

ADGO_Racing
04-16-2010, 05:21 AM
How many people have told Evan he is mistaken?????:eek:

This whole phase voodoo is beginning to be a ramble. A repetition of the same information. I believe the accepted definition of crazy is: doing the same thing over and over again and expecting different results.:D

Evan
04-16-2010, 12:43 PM
How many people have told Evan he is mistaken

How many people with actual experience agreed with me? Saying the same thing over and over is the only option when you are right. Just because you don't understand it doesn't make me wrong.

I will make it very simple for your benefit.

We have a box. We have no idea what is inside the box and are given absolutely no information about it. But, there is an external coupling at each end of the box. One end is locked in place and not free to turn. The other end is mounted on a bearing and free to rotate with a stub shaft to which we may apply torque.

There is a small slot at each end of the box into which we may insert a small rod so that it protrudes very slightly from the slot to act as an indicator. We apply a torque to the output with a torque wrench. We then are able to observe how much the indicator rod at each end moves. We find that at a torque of 100 ft lbs the indicator at the end with the torque wrench moves twice as far as the indicator at the other end.

That tells us how much relative deflection of the pointers occurs when a torque of 100 ft lbs is applied to the one end of the stub shaft sticking out of the box.

Now, what does it tell us about what is in the box? No assumptions are permitted. You must justify your answers based on the available evidence.

philbur
04-16-2010, 01:05 PM
It tells you the torsional rigidity of the item in the box. So simple what's the problem.

Phil:)




I will make it very simple for your benefit.

We have a box. We have no idea what is inside the box and are given absolutely no information about it. But, there is an external coupling at each end of the box. One end is locked in place and not free to turn. The other end is mounted on a bearing and free to rotate with a stub shaft to which we may apply torque.

There is a small slot at each end of the box into which we may insert a small rod so that it protrudes very slightly from the slot to act as an indicator. We apply a torque to the output with a torque wrench. We then are able to observe how much the indicator rod at each end moves. We find that at a torque of 100 ft lbs the indicator at the end with the torque wrench moves twice as far as the indicator at the other end.

That tells us how much relative deflection of the pointers occurs when a torque of 100 ft lbs is applied to the one end of the stub shaft sticking out of the box.

Now, what does it tell us about what is in the box? No assumptions are permitted. You must justify your answers based on the available evidence.

Evan
04-16-2010, 01:35 PM
It tells you the torsional rigidity of the item in the box. So simple what's the problem.


No it doesn't. You don't know the radius of gyration. There is no reason to assume that the radius is the same as the couplings.

If it isn't then you are also dealing with bending forces.

That also means that you don't know how much it has twisted or even if it has twisted. You also don't know what the mechanism of force transfer is.

To be more specific it doesn't inform you in any way as to the contents of the box. All you can deduce is that the box is able to transmit torque from one end to the other with a phase difference.

philbur
04-16-2010, 02:12 PM
Definition of torsional rigidity.

(mechanics) The ratio of the torque applied about the centroidal axis of a bar at one end of the bar to the resulting torsional angle, when the other end is held fixed.

No radius of gyration involved in torsional rigidity. If you study the equation it may help you with your misunderstanding.

Phil:)



No it doesn't. You don't know the radius of gyration. There is no reason to assume that the radius is the same as the couplings.

If it isn't then you are also dealing with bending forces.

That also means that you don't know how much it has twisted or even if it has twisted. You also don't know what the mechanism of force transfer is.

To be more specific it doesn't inform you in any way as to the contents of the box. All you can deduce is that the box is able to transmit torque from one end to the other with a phase difference.

RKW
04-16-2010, 03:41 PM
Are we having fun yet?

Regardless of who is right or wrong these banterings get pretty tiring. Isn't this stuff supposed to be fun or at least enjoyable?

We all know or work with someone we get tired of listening to so you just learn to ignore them after a while. Fortunately here you can just update your "Buddy / Ignore Lists" and the rest is magic ;-)

Weston Bye
04-16-2010, 03:58 PM
Fortunately here you can just update your "Buddy / Ignore Lists" and the rest is magic ;-)

Yabut, if you do you will eventually miss something good. All those guys do come up with some good stuff.

RKW
04-16-2010, 04:03 PM
Agreed. It was just a suggestion for folks who just can't take it anymore. My list is actually empty on this forum.


Yabut, if you do you will eventually miss something good. All those guys do come up with some good stuff.

Evan
04-16-2010, 05:13 PM
Definition of torsional rigidity.

(mechanics) The ratio of the torque applied about the centroidal axis of a bar at one end of the bar to the resulting torsional angle, when the other end is held fixed.

No radius of gyration involved in torsional rigidity. If you study the equation it may help you with your misunderstanding.


The radius of gyration also defines the torque moment of inertia.

That doesn't matter because you cannot determine the rigidity of what may be in the box. You have no reference point and the phase shift measurement doesn't give one. The most you can say is that the entire system isn't rigid, but we already knew that since nothing is rigid.

So, the question is still unaswered. What can we discover about what is in the box?

philbur
04-16-2010, 05:54 PM
One last time:

torsional rigidity = applied torque divided by the angular deflection.

This is a mathematical definition, no amount of verbal diarrhoea can obscure the fact.

Phil:)



The radius of gyration also defines the torque moment of inertia.

That doesn't matter because you cannot determine the rigidity of what may be in the box. You have no reference point and the phase shift measurement doesn't give one. The most you can say is that the entire system isn't rigid, but we already knew that since nothing is rigid.

So, the question is still unaswered. What can we discover about what is in the box?

aostling
04-16-2010, 05:56 PM
So, the question is still unaswered. What can we discover about what is in the box?

Can you supply a sketch of what you described in words? The configuration is not clear to me.

Evan
04-16-2010, 06:47 PM
One last time:

torsional rigidity = applied torque divided by the angular deflection.

This is a mathematical definition, no amount of verbal diarrhoea can obscure the fact.



You don't know the angular deflection ( of what is in the box), only the ratio.

--------------------------------------------------------

Allan, I will later but right now I don't have time as my wife is due home from work and I have a few chores to do.

aostling
04-16-2010, 07:09 PM
Allan, I will later but right now I don't have time as my wife is due home from work and I have a few chores to do.

Where's your sense of priority?

I'm hung up on the coupling. Presumably not this:

http://i168.photobucket.com/albums/u183/aostling/coupling.jpg

philbur
04-16-2010, 07:46 PM
The following is an extract from your post 71 regarding the "box" experiment, the section in bold is you measuring the angular deflection over a fixed length (between the two points on the shaft). So how can you now claim you don't know the angular deflection. The whole point of the procedure is to measure the deflection. You could more easily measure the deflection at the point you applied the torque relative to the fixed end, this would not then require any penetration of the box itself.

OK, one very last time: the torsional rigidity equals the applied torque divided by the relative angular displacement of the two pointers.

Phil:)
.....................................
There is a small slot at each end of the box into which we may insert a small rod so that it protrudes very slightly from the slot to act as an indicator. We apply a torque to the output with a torque wrench. We then are able to observe how much the indicator rod at each end moves. We find that at a torque of 100 ft lbs the indicator at the end with the torque wrench moves twice as far as the indicator at the other end.

That tells us how much relative deflection of the pointers occurs when a torque of 100 ft lbs is applied to the one end of the stub shaft sticking out of the box.
......................................

Phil:)


You don't know the angular deflection ( of what is in the box), only the ratio.

--------------------------------------------------------

Allan, I will later but right now I don't have time as my wife is due home from work and I have a few chores to do.

interrupted_cut
04-16-2010, 07:57 PM
I use a torque meter that operates as Evan describes almost every day. It is a Torquetronics device manufactured by Torquemeters Limited in the UK (www.torquemeters.com). It has a toothed wheel on each end of the measuring section, monitors the relationship of the ends by means of magnetic pickups, and the signal conditoner reads out torque, speed and power. It must first be calibrated by the use of a calibration arm and dead weights, but you don't need to know anything about the physical properties of the shaft. There is one exception: If you wanted to temperature correct the data without deadweight calibration at multiple temperatures, you would need to know the change in modulus with temperature. Luckily, this function is built in to the signal conditioner. One of the drawbacks of this type device is it is a dynamic torquemeter only, in that you cannot measure torque without significant relative motion between the pickup and the toothed wheel. In order to allow static calibration, the manufacturer has developed a clever device that interposes a motorized symmetrically toothed section in the magnetic path at both ends of the measuring section. When this is spinning, you can calibrate the non-rotating shaft. It is locked in place during normal use. The one I use is rated for 0-100 lb-ft of torque at 60,000+ rpm. We run it at 60krpm routinely, and it is very sensitive. I can consistently resolve performance changes of less than .05 lb-ft.

John Stevenson
04-16-2010, 08:33 PM
We all know or work with someone we get tired of listening to so you just learn to ignore them after a while. Fortunately here you can just update your "Buddy / Ignore Lists" and the rest is magic ;-)


Yabut, if you do you will eventually miss something good. All those guys do come up with some good stuff.

True,
I have two people in my list but they are people who, when they post never make sense to me.

Evan isn't one as I find him far more entertaining than Channel 4 either when he's right on a subject or when he's wrong which is usually the same thing :D

I tend to think of him as The Champ, named after the Austin Champ. Here's the Tiffiepedia link. http://en.wikipedia.org/wiki/Austin_Champ

This was interesting in that the reverse gear was in the rear diff giving the vehicle 5 syncromesh forward gears and 5 syncromesh reverse gears which means Evan can backpedal out of an argument just as fast as he got into it :D

.

Evan
04-16-2010, 09:04 PM
OK, one very last time: the torsional rigidity equals the applied torque divided by the relative angular displacement of the two pointers.


You are making unwarranted assumptions. The item in the box may not be under torsional load. It could be bending, or under compression, tension or shear.

You are not able to measure the angular displacement of the pointers. They are inserted in slots in the box into a receiver that you cannot see and you have no idea what the radius of motion is. There is no way to determine it and even if there was it may not represent the average state of the machine.

All you can discover is the torsional rigidity of the entire system but you can't even determine that using the measuring system I have described. However, you can calibrate the measuring system in terms of absolute torque transmitted.

Try again.

No need John.

J Tiers
04-16-2010, 09:10 PM
I use a torque meter that operates as Evan describes almost every day. It is a Torquetronics device manufactured by Torquemeters Limited in the UK (www.torquemeters.com). It has a toothed wheel on each end of the measuring section, monitors the relationship of the ends by means of magnetic pickups, and the signal conditoner reads out torque, speed and power. It must first be calibrated by the use of a calibration arm and dead weights, but you don't need to know anything about the physical properties of the shaft.

And. yet another person apparently thinks it's magic........

You do, of course know "a physical property of the shaft"...... you absolutely MUST or it cannot work.

What you know is that it deflects a certain amount per newton-meter or whatever as reflected in the change of signal......

You don't know how much that is in "real" units, and it doesn't matter........ But you DO FOR SURE know how many "magic torque meter units" it deflects per unit of torque applied......

And, the only reason you don't know the actual real deflection is that you were too lazy to correlate "magic torquemeter units" to "real" units, since after all you don't have to. But that information is available, you know just about all the things you need already.

There just is no escaping it...... you gotta know "how much" it changes per unit torque or your meter reads only "Bubba twisty units".

It's called calibration..... and it isn't just an incidental detail...

Evan
04-16-2010, 10:07 PM
Sorry Jerry. You don't even know it the box contains a shaft.


What you know is that it deflects a certain amount per newton-meter or whatever as reflected in the change of signal......

No. All you know is that it produces a particular signal when loaded to a certain torque. You have no idea how much it deflects or what the torsional rigidity of the shaft is. All you know is that it isn't rigid just like every other bit of matter in the universe.

The torque meter does not inform you about the shaft.

Arcane
04-16-2010, 10:11 PM
I'm pretty sure the box contains a cat....:D

Evan
04-16-2010, 10:33 PM
And it's name is Schrödinger...

aostling
04-16-2010, 10:49 PM
I'm pretty sure the box contains a cat....:D

I need to know what's outside the box. I've gone over Evan's verbal description repeatedly, but don't have a clue what mental picture I should adopt.

The promised sketch should make this clear.

J Tiers
04-16-2010, 11:36 PM
The torque meter does not inform you about the shaft.

Your box is NOT the same as the phase torque meter.... it is a red herring argument, which is not applicable to the original point of contention.

You have carefully structured the box to be a simple torque meter, and NOT a useful pass-thru type. So all it gives is a static reading, and we can throw it out of the argument.

The phase meter DOES provide displacement info, since the physical construction OF THE METER ELEMENTS ONLY, which is known, provides enough information that a shaft twist, or displacement can be back-calculated.

Therefore the meter gives you enough information that you can actually calculate the shaft torsional stiffness.

Of course you don't HAVE to. But CHOOSING NOT to is by no means the same thing as BEING UNABLE TO.... You just CHOOSE not to use the data you have.

If you propose as a counter a case where the meter elements are actually UNKNOWN, you just provide one added level of abstraction...... Now you STILL know the torsional stiffness, but you don't know the relation between "magic meter units" and physical displacement. So you only know the number of "magic meter units" per newton-meter after calibration.

There really isn't any device that you can stick on a shaft of totally unknown and unknowable properties and have it instantly read out torque in newton-meters or whatever.

You must calibrate.

And calibration automatically provides information about the shaft.... that is the entire reason you DO it..... to FIND that information in order to relate the readings to reality.

While you may know nothing before you calibrate, , AFTER you calibrate, you DO know the torsional stiffness. All the fol-de-rol you have brought in simply means that you may not know that property in the type of units that you might prefer.

aostling
04-16-2010, 11:54 PM
This sketch shows my lack of understanding of what the configuration looks like. It shows shafts protruding from flanged bearings on opposite sides of the box. But I expect this is not what Evan has in mind.



http://i168.photobucket.com/albums/u183/aostling/Screenshot2010-04-16at75011PM.png

Evan
04-17-2010, 12:55 AM
Total BS Jerry. There is no essential difference.

Any measurement take of the "live" system is a snapshot in time the exact same as a static system. Talk about red herrings...:rolleyes:


Here is the box.

edit: to simplify further we shall assume that the input side stick remains stationary and the output side stick moves with applied load. The result is the same either way.

http://ixian.ca/pics7/thebox.jpg


There really isn't any device that you can stick on a shaft of totally unknown and unknowable properties and have it instantly read out torque in newton-meters or whatever.


Yes, of course. Didn't you read what I wrote? Apparently not. Why do you think I mentioned a dyno way back at the beginning? Your arguments are meaningless because you aren't addressing the points. That is because you don't even know what the points are, not having read all the posts.


While you may know nothing before you calibrate, , AFTER you calibrate, you DO know the torsional stiffness.

No you don't. Who said there is a shaft? Even if there is you don't know the torsional stiffness of it. Careful how you reply because it is dead simple to demonstrate why.

J Tiers
04-17-2010, 01:24 AM
Well, assuming I HAD not read all the posts, which you have no way of knowing, that's better than not understanding the principles, and careening off on a wierd trajectory into the philosophy of "phase heaven"......

I read your comments all too well........... Including where you denied that you needed ANY information about the shaft, and then described calibrating the device, as if that were magic and not actually 'getting information about the shaft".

That's a pretty basic misconception which you ran afoul of. And one you appear to persist in.

Don't associate my comments with your magic box... it isn't relevant to what I said...... it was crafted to avoid what I said, and as such there is no association between what I said and your box as I understand your post about it.

However, this is losing its amusement quotient..... One can only read so many of your posts which essentially say "I know you are but what am I" (your post #68, I believe)...... That was probably very amusing at about 6 years of age, but has lost its appeal long ago.

Evan
04-17-2010, 01:40 AM
I don't care what you think Jerry. I let you have the last word but Philbur ressurected the thread. I am waiting to see what he has to say. You still haven't answered the question either though.

philbur
04-17-2010, 06:36 AM
You can measure the circumferential displacement of the two pointers one relative to the other with and without the applied torque. You can measure the radial distance from the input shaft axis to the tip of the pointers. So you can calculate the relative angular displacement of the two pointers due to the applied torque. Finally you can measure the applied torque required to produce the measured displacement of the two pointers.

So one very, very last time: Torsional rigidity of the item in the box is equal to the applied torque divided by the angular displacement of the two pointers.

The item in the box can be any shape you care to imagine, it still has torsional rigidity and you have just measured it. Also it can have other forces applied to it, you have measured only the impact of the applied torque.

Phil:)

http://ixian.ca/pics7/thebox.jpg

John Stevenson
04-17-2010, 06:43 AM
And it's name is Schrödinger...

I thought it was Trex.

.

Evan
04-17-2010, 09:05 AM
You can measure the circumferential displacement of the two pointers one relative to the other with and without the applied torque. You can measure the radial distance from the input shaft axis to the tip of the pointers. So you can calculate the relative angular displacement of the two pointers due to the applied torque. Finally you can measure the applied torque required to produce the measured displacement of the two pointers.


You are assuming that the pointers intersect the axis of the input shaft. They may not. You cannot determine that. If they do not your calculation will be in error.

Further, the system may be very non linear. That can be determined by calibrating the system under static load. However, non-linearity isn't a property of materials and you will only be able to guess why the system is non linear.

In fact, it may turn out that you aren't measuring torsional rigidity at all. You may only be measuring system compliance. Compliance isn't a material property either and has no units.

Here is one possible system that is in the box. It is also a realistic possibility and not some sort of trick. It turns out that you will be measuring the compressive strength of the black rubber hockey pucks that are present to absorb shock loads.

http://ixian.ca/pics7/thebox2.jpg

You measurement produces a value of phase difference beween the input and output. It does not quantify any mechanical property that can be assigned to the contents of the box. You cannot describe a hockey puck as exhibiting a 3 degree phase difference under load. The sentence is meaningless.

The measurement however is not meaningless since it can be calibrated and used to calculate an absolute value of tranmitted torque without knowing why the phase difference occurs or what is in the box.

philbur
04-17-2010, 09:34 AM
Absolutely amazing!

Firstly I am not assuming the two indicators intersect the axis. The calculation of angular displacement uses an imaginary line through the end of the pointer and the axis of the applied torque. The pointers can be wrapped 10 times around the object before they are attached to it, the external tips will still have the same relative angular displacement when you apply the torque.

When you apply torque to the object you have drawn inside the box the pointers will experience an angular displacement relative to each other. This gives you the torsional rigidity of the object. The fact that the object has other properties is immaterial.

The rest of your explanation is just irrelevant waffle, seemingly included to confuse the issue.

Phil:)



You are assuming that the pointers intersect the axis of the input shaft. They may not. You cannot determine that. If they do not your calculation will be in error.

Evan
04-17-2010, 11:06 AM
The calculation of angular displacement uses an imaginary line through the end of the pointer and the axis of the applied torque. The pointers can be wrapped 10 times around the object before they are attached to it, the external tips will still have the same relative angular displacement when you apply the torque.



You are only talking about the entire system whether you realize it or not. Depending on what is in the box the indicators may not even be measuring angular displacement. It is trivial to construct a configuration where the indicators will display movement that is a combination of forces only a portion of which is due to torque reaction. In that case you cannot determine what proportion is due to torsion but you can still calibrate it and use the result to measure torque to a load. You cannot however use the result to determine a value of torsional rigidity.


The rest of your explanation is just irrelevant waffle, seemingly included to confuse the issue.


It seems to have confused you. Of course you will attempt to dismiss it since it shows how you cannot tell anything about the properties of the method of torque transmission. You continue to evade the relevant issues.

There is an entire book on this subject. It points out clearly that in non-linear systems you need empirical data to be able to characterize the behaviour of the load transmitting element(s). In the absence of such data you will not be able to calculate the compliance curve for the compliant components so the best you can do is to treat the box as a closed system about which you know nothing except the torque transfer curve. In other words, the measurement of the phase shift tells you nothing about the shaft, only the system as a whole since the load cannot be attached directly to the shaft. It is so difficult to characterize that compliance curves are developed for different shapes of shafts and the types of connection made to the shaft as well as the shaft material and the machining techniques and material treatments that may have been applied.

A simple phase difference measurement will not allow you to characterize any of these quantities individually and you cannot determine what role each may play or even if they are present.

http://ixian.ca/pics7/thebox3.jpg

philbur
04-17-2010, 03:48 PM
The pointers measure nothing, you measured their angular displacement, so by definition you have MEASURED their angular displacmement. This angular displacement was produced by the torque.

So one very, very, very last time: Torsional rigidity is equal to the applied torque divided by the agular displacement.

Of course the torsional rigidity may not be linear, but that's just a red herring isn't it. You may need to repeat the test (calibration) over the full range of deflections to determine the linearity. You can plot a graph of the results to give you the torsional rigidity at different deflection values.

So Tr = Tc/dc

where;

Tc is the applied torque,
dc is the corresponding angular deflection,
Tr is the torsional rigidity


So now when I want to now the torque being transmitted by the shaft I measure the actual deflection and multiply it by the torsional rigidity, read of the graph I produced earlier.

or Ta = da * Tr

where:

da is the angular displacement produced by the unknown torque,
Ta is the unkown torque that you want to determine.

You may choose not to call it torsional rigidity but the rest of the world does.

Simple, no smoke, no mirrors.

Phil:)


You are only talking about the entire system whether you realize it or not. Depending on what is in the box the indicators may not even be measuring angular displacement. It is trivial to construct a configuration where the indicators will display movement that is a combination of forces only a portion of which is due to torque reaction. In that case you cannot determine what proportion is due to torsion but you can still calibrate it and use the result to measure torque to a load. You cannot however use the result to determine a value of torsional rigidity.

Evan
04-17-2010, 08:21 PM
You may choose not to call it torsional rigidity but the rest of the world does.


No, they don't. They call it Torsional Compliance.


Automating Torsional Compliance Testing of Steering Column Linkages


The Torrington Co., a subsidiary of Ingersoll-Rand, is a leading producer of precision bearings and motion control products. Torrington’s automotive components facility in Watertown, CT, generates performance curves to check the torsional compliance characteristics of steering column linkage assemblies. Their original test system used discrete torque and quadrature encoder readouts with an X-Y pen plotter to graph torque versus deflection and measure linkage performance characteristics. This setup required an engineer to graphically interpret the plotted data and perform manual calculations. Torrington recognized that the elimination of manual calculations would quickly yield improvements.



http://sine.ni.com/cs/app/doc/p/id/cs-421

If we instrument the shaft as I originally suggested we must apply an external load using some sort of equipment such as a dynamometer. There will be of necessity couplings between the shaft and the dyno. The couplings will contribute non-linearity to the torque curve applied to the shaft while the shaft itself will also present a curve that depends on the actual material from which it is made and even such characteristics as taper, wall thickness, etc.

We can indeed develop a torque curve that results in an accurate calibration of the system but there is no way to determine what contribution the shaft makes to that curve versus the rest of the calibration system. We measure only a portion of the resulting deflection which is sufficient to measure the relative LOAD on the shaft but it is not sufficient to characterise the actual response of the shaft since the torque data points are provided externally. We measure the compliance of the shaft but we cannot separate the overall system compliance from the compliance of the shaft for a particular value of torque. To do that we must test only the shaft itself in isolation from the system and even then we shall have to refer to empirically derived values to properly characterise the response of the shaft to torque.

edited to clarify

philbur
04-18-2010, 07:11 AM
More red herrings from Evan,

If you read and understand the link you posted you will see that the reference refers to measuring torsional compliance characteristics of steering column linkage assemblies. This includes reversal of the applied torque to include the free play in spline shafts and universal coupling etc and also to include the hysteresis of the system. You can see this from the graph in your link (see below). The y-axis includes both positive and negative torque values and you can see the free play at zero torque, where a certain amount of deflection occurs without any change in torque. However the slope of the curve on either side of the zero torque value gives you the torsional rigidity, because torque divided by deflection is defined (by the world) as torsional rigidity. It’s interesting to note that the torsional rigidity can be rotation direction dependent which, depending on application, may require you to perform the measurements in both directions.

http://i186.photobucket.com/albums/x36/philbur/hystgraphfp.jpg

With your black box experiment we do not have to (and should not) include data points from either side of the zero torque value when calculating the torsional rigidity. So what you are measuring in the black box (and also when calibrating the system) is torsional rigidity. You could also measure torsional compliance, although this is of no interest in the intended application.

Another red herring bites the dust.

It is kinda self evident that you can only determine the torsional rigidity of the system between the two measuring points. However as your black box was intended as an analogy of the actual meter then the user is allowed to determined where on the unknown object he has attached the pointers. Presumably he puts them accross the section of interest.

So we have now determined that we can know the torsional rigidity and also the torsional compliance of the system in the box. The list is growing.

Phil:)

Evan
04-18-2010, 08:33 AM
This includes reversal of the applied torque to include the free play in spline shafts and universal coupling etc and also to include the hysteresis of the system

Last time I checked trucks have a reverse gear.


It is kinda self evident that you can only determine the torsional rigidity of the system between the two measuring points. However as your black box was intended as an analogy of the actual meter then the user is allowed to determined where on the unknown object he has attached the pointers. Presumably he puts them accross the section of interest.



That was not specified in the orginal description. You are making an assumption. Not being able to see the contents of the box how do you know what the section of interest is? The question is what can the instrumentation reveal about the contents, not what can the operator of the test deduce by observing it.

If you havent yet realized it the box represents a transmission. Inside that box can be an unlimited variety of torque modifying elements. You can do any manner of static tests that will generate torque curves that can be used to quantify the torque transmitted.

You will end up with a calibration curve that serves to measure transmitted torque but you will have no idea what it is you calibrated.

You may not be measuring rigidity but a combination of torque multiplication/demultiplication and rigidity. You have no way to deduce what deflection contributes to the pointer movement compared to the movement produced by gearing.

philbur
04-18-2010, 09:24 AM
Last time I checked trucks have a reverse gear..
They most certainly do. What's your point.


That was not specified in the orginal description. You are making an assumption. Not being able to see the contents of the box how do you know what the section of interest is? .
Because in the context of the original question it is only the shaft you know nothing about, not the point of measurement. Once again it is self evident that you can only know the trosional rigidity of the object between the two measuring points.


The question is what can the instrumentation reveal about the contents, not what can the operator of the test deduce by observing it..
This seems to be spliting hairs with a razor blade.


If you havent yet realized it the box represents a transmission. Inside that box can be an unlimited variety of torque modifying elements. You can do any manner of static tests that will generate torque curves that can be used to quantify the torque transmitted.
You will end up with a calibration curve that serves to measure transmitted torque but you will have no idea what it is you calibrated..
This was not the question. The question was do you know anything about the object you have calibrated.


You may not be measuring rigidity but a combination of torque multiplication/demultiplication and rigidity. You have no way to deduce what deflection contributes to the pointer movement compared to the movement produced by gearing.
But you are measuring rigidity. The gearing is part of the rigidity of the system so it is correct to included it in the measurement. I once again draw you attention to:

Torsional rigidity is equal to the applied torque divided by the angular deflection.

This equation has no qualifiers attached as to what can and cannot be included in the system. Gears, torque multipliers etc all contribute to the magnitude of the deflection and therefore the torsional rigidity. They just change the slope of the curve.

There is a lot of ducking and diving going on here but not much progess. What happen to torsional compliance.

Phil:)

Evan
04-18-2010, 10:22 AM
This seems to be spliting hairs with a razor blade.


Only if you happen to have xray vision.


But you are measuring rigidity. The gearing is part of the rigidity of the system so it is correct to included it in the measurement. I once again draw you attention to:

Torsional rigidity is equal to the applied torque divided by the angular deflection.


You think you are measuring rigidity. In actual fact you may not be. The reason for the deflection of the pointer may have almost nothing to do with the rigidity of the material transmitting the torque. The rigidity of the gearing and the torque multiplication it provides are completely different properties. You cannot distinguish between the two. The object is to characterize the torque transmitting element, not the box and it's contents as a system.

Nice try Phil.

philbur
04-18-2010, 10:57 AM
In actual fact you are, always.

Torsional rigidity equals the applied torque divided by the angular displacement caused by the applied torque.

You know the torque because you applied it. You know the angular displacement because you measured it.

What mechanisms contributed to the magnitude of the angular deflection is irrelevant.

Phil:)




You think you are measuring rigidity. In actual fact you may not be.

Evan
04-18-2010, 03:07 PM
What mechanisms contributed to the magnitude of the angular deflection is irrelevant.



Only if you don't care about characterizing the properties of those mechanisms, which is the point of the exercise.

In other words, you still don't know about the shaft.

philbur
04-18-2010, 04:44 PM
But you have characterised those mechanisms, you have determined the system torsional rigidity, which includes those items. which is the point of the exercise. The only thing that interests you is the rigidity of the system between the two measuring points, not the individual rigidity of each component in the system.

Phil:)


Only if you don't care about characterizing the properties of those mechanisms, which is the point of the exercise.

In other words, you still don't know about the shaft.

ADGO_Racing
04-18-2010, 10:02 PM
Phil is right. You ARE measuring a property of the shaft, thus you are knowing something about the system. But unless you calibrate the system you have no valuable information.

I also disagree that you cannot calculate the rigidity of a shaft with varying section, or through couplings. A proper FEA will give you very close to real life deflections. The minor variations between actual and calculated, will be due to minor real life defects. Exact: NO never can be, close enough for all practical purposes: YES.

J Tiers
04-18-2010, 10:36 PM
Phil is right. You ARE measuring a property of the shaft, thus you are knowing something about the system. But unless you calibrate the system you have no valuable information.



Be prepared for squalls and other bad weather... I have been trying to get Evan to admit that for a while in this abominable thread.....

Evan is not a bad sort, but he sure is stubborn. I think he isn't Danish, must be a Swede.

Weston Bye
04-19-2010, 07:24 AM
What astonishes me is Evan's capacity for creativity along with the ability to keep up such an intense line of reasoning in arcane minutia about an inconsequential subject. Quite an investment in energy, fybromyalgia notwithstanding. Wears me out just reading the thread.

We've been tag teaming Evan for a few days now, to little effect. He will not change his position and we remain unconvinced. I hope we are not damaging his health. I'd rather see him creating and reporting on his efforts.

edit to add: This thread has run its course. There is little of value to add. Go back to machining.

Evan
04-19-2010, 10:29 AM
thus you are knowing something about the system

That I have granted many times in this thread. Knowing something about the system is what the calibration provides. It does not indentify why the shaft produces the measurements that you obtain and so does not characterize the shaft.

FEA can characterize the shaft if you ALREADY know how it is constructed and the properties of the materials from which it is made.

It is pretty damned obvious that if you can't even tell that there may be a planetary transmission in the middle of the shaft you don't have the shaft characterized.

When you measure the deflection in absence of knowledge of the system you are measuring system compliance. Compliance is a property of systems, not of materials, so knowing the compliance of the system does not inform you as to the torsional rigidity of the shaft since you do not know if it even is a shaft. The test cannot identify the factors that are responsible for the measured compliance. It is possible to arrange the box problem so that there is nothing in the box and the box itself is the transmission element.

ADGO_Racing
04-19-2010, 02:12 PM
That I have granted many times in this thread. Knowing something about the system is what the calibration provides. It does not indentify why the shaft produces the measurements that you obtain and so does not characterize the shaft.

FEA can characterize the shaft if you ALREADY know how it is constructed and the properties of the materials from which it is made.

It is pretty damned obvious that if you can't even tell that there may be a planetary transmission in the middle of the shaft you don't have the shaft characterized.

When you measure the deflection in absence of knowledge of the system you are measuring system compliance. Compliance is a property of systems, not of materials, so knowing the compliance of the system does not inform you as to the torsional rigidity of the shaft since you do not know if it even is a shaft. The test cannot identify the factors that are responsible for the measured compliance. It is possible to arrange the box problem so that there is nothing in the box and the box itself is the transmission element.


I admit that under this scenario that the you are correct. Using a "KNOWN" Torque to "CALIBRATE" a known "DISPLACEMENT". Will be sufficient for an operator to safely operate a piece of machinery.

However, your initial statement (I'm not quoting) was that you did not need to know anything about the shaft, and that angular displacement (Phase Shift, Twist, etc) was all that is necessary. While this is true, you still need to calibrate it to a KNOWN quantity. Not just an "experienced operator" saying "this is about all she'll take". I have people that buy bags of aluminum jam nuts from me. I have a torque spec for each size we make, I developed it through empirical testing. They come back constantly because they have the mentality that you tighten them until they squeak and add a quarter turn. Not true of any fastener, more especially an aluminum fastener. Same applies to many of our threaded aluminum parts. Many of these guys are what you would call "Experienced Operators". They are not who I would rely on to determine an acceptable load on a shaft. Especially one loaded as in the beginning of this thread.

philbur
04-19-2010, 03:28 PM
Compliance is a property of systems, not of materials, so knowing the compliance of the system does not inform you as to the torsional rigidity of the shaft
Who said anything about property of materials, we are discussing a property of the object in the box. Torsional rigidity is also a property of a system, or in this case an object, not of a material. You may be getting confused with Modulus of Rigidity (also known as Modulus of Shear), which is a property of materials.


So knowing the compliance of the system does not inform you as to the torsional rigidity of the shaft
The slope of the curve at any point in the compliance diagram/graph you supplied is the torsional rigidity at that torque value. So if you know the compliance of the object then you know the torsional rigidity of the object. Once again, the compliance is of no interest in the context of a torque meter, only the torsional rigidity.


since you do not know if it even is a shaft.
In the context of this discussion a shaft is any object that is capable of transmitting torque, so by your definition of the experiment the object in the box is a shaft.


It is possible to arrange the box problem so that there is nothing in the box and the box itself is the transmission element
If there is nothing in the box then the pointers cannot be attached to anything in the box and the test is meaningless. However if you measure the twist of the box when subjected to a known torque you can determine the torsional rigidity of the box, which is now acting as a shaft.

Phil:)

John Stevenson
04-19-2010, 04:26 PM
I think that's game set and match to Phil, unless Evan has more than 5 reverse gears...........




Crrrrrrrrrruuuuunnnnchhhhh.............

.

Evan
04-19-2010, 10:10 PM
Quote:


Who said anything about property of materials, we are discussing a property of the object in the box. Torsional rigidity is also a property of a system, or in this case an object, not of a material. You may be getting confused with Modulus of Rigidity (also known as Modulus of Shear), which is a property of materials.


Upon further research it turns out that your claim that the torsonal rigidity is the product of load and deflection is quite simply wrong.

For many cross sections there is no closed solution so any curve that you develop by means of the oversimplified expression you have given bears no relationship to the actual torsional rigidity of the shaft.


From the instructions for the program Mathematica
http://documents.wolfram.com/applications/structural/TorsionalAnalysis.html


For almost all these cross sections, functions for stress conditions, displacement components, stress components, twist per unit length, torsional rigidity, and warping functions are provided. The Torsion Analysis package does not include the warping function and displacement components for the cross sector Sector due to the lack of closed-form expressions.

For shafts, only the following torsional rigidities are provided:

bar of narrow rectangular cross section

hollow concentric circular section

thin circular open tube of constant thickness

hollow (similar) elliptical cross section



In other words, you cannot characterize the shaft by measuring the torsional rigidity of the system.

John, you seem to have scored an own goal.

aostling
04-19-2010, 11:41 PM
I'm quite fascinated by Evan's black box, intended to hide the coupling connecting an input shaft to an output shaft. It reminded me of something, but what? Then I remembered this page in my patent notebook, from 1977.

It's a torque wrench, designed to be not much bigger than a spark plug wrench so you could just keep it in a drawer with your other sockets. The innards, which you would never see without destroying the tool, consists of a bundle of small-diameter rods, pressed into circular plates B and D. Spacers G keep the bundle from getting too kinked during the quarter-turn (max) rotation.

I never patented this, nor built it. My notebook pages are full of stuff like this. I'll bet yours are too.

http://i168.photobucket.com/albums/u183/aostling/torquewrench.jpg

philbur
04-20-2010, 09:45 AM
Quote:
Upon further research it turns out that your claim that the torsonal rigidity is the product of load and deflection is quite simply wrong. .
I have not claimed that torsional rigidity is the product of the load and the deflection. To obtain the produce of two numbers you have to multiply them. I have stated many times, and I will state one very, very, very, very last time:

Torsional rigidity is equal to the applied torque divided by the angular displacement.

This is not a claim this is the world-wide accepted definition of torsional rigidity. It is not derived from some other function, it is a definition. If you have a reliable source that shows this definition to be "quite simply wrong" then please share.


Quote:
For many cross sections there is no closed solution so any curve that you develop by means of the oversimplified expression you have given bears no relationship to the actual torsional rigidity of the shaft. .
Evan you do not develop the curve by means of the “oversimplified expression”. You read off the torsional rigidity from the slope of the curve that you plotted from your black box experiment. The compliance curve you showed previously is plotted from measured data not produced from a calculation. Although it could be should you choose to get sufficiently sophisticated.


Quote:
From the instructions for the program Mathematica
http://documents.wolfram.com/applications/structural/TorsionalAnalysis.html
Evan this is another of your red herrings. What the link is saying is that the referenced analysis tool can only handle problems with a closed solution. So consequently it can only provide torsional rigidities where a closed solution exists. It is nonsense to use the limitations of a particular analysis tool as proof that it is not possible to calculate a particular property. In any case our whole discussion is related to measurement not calculation.


Quote: In other words, you cannot characterize the shaft by measuring the torsional rigidity of the system.
In practice you obviously measure the torsional rigidity across the part of the shaft, system or object for which you wish to know the torsional rigidity. Why would you do otherwise. So differentiating between “the shaft” and “the system” has no meaning in the context of this discussion.


Quote:
John, you seem to have scored an own goal.
I think you forgot that we changed ends at half time.:D

Phil:)

John Stevenson
04-20-2010, 10:59 AM
I think you forgot that we changed ends at half time.:D

Phil:)

ROTFLMAO :D :D

.

Evan
04-20-2010, 11:33 AM
Who said anything about property of materials, we are discussing a property of the object in the box. Torsional rigidity is also a property of a system, or in this case an object, not of a material. You may be getting confused with Modulus of Rigidity (also known as Modulus of Shear), which is a property of materials.


Sorry Phil, but is seems you are the one that is confused. Nobody switched ends.

You are now spouting obviously incorrect nonsense. Torsional Rigidity, Modulus of rigidity, Shear modulus, Shear modulus of rigidity and Modulus of Shear are all part of the same property as they are all aspects of Modulus of Elasticity.



Modulus of Elasticity


Rate of change of strain as a function of stress. The slope of the straight line portion of a stress-strain diagram. Tangent modulus of elasticity is the slope of the stress-strain diagram at any point. Secant modulus of elasticity is stress divided by strainat any given value of stress or strain. It also is called stress-strain ratio.

Tangent and secant modulus of elasticity are equal, up to the proportional limit of a material. Depending on the type of loading represented by the stress-strain diagram, modulus of elasticity may be reported as: compressive modulus of elasticity (or modulus of elasticity in compression); flexural modulus of elasticity (or modulus of elasticity in flexure); shear modulus of elasticity (or modulus of elasticity in shear); tensile modulus of elasticity (or modulus of elasticity in tension); or torsional modulus of elasticity (or modulus of elasticity in torsion). Modulus of elasticity may be determined by dynamic testing, where it can be derived from complex modulus. Modulus used alone generally refers to tensile modulus of elasticity. Shear modulus is almost always equal to torsional modulus and both are called modulus of rigidity. Moduli of elasticity in tension and compression are approximately equal and are known as Young's Modulus. Modulus of rigidity is related to Young's Modulus by the equation:

where E is Young's Modulus (psi), G is modulus of rigidity (psi) and r is Poisson's ratio. Modulus of elasticity also is called elastic modulus and coefficient of elasticity.


Torsion Test


Method for determining behavior of materials subjected to twisting loads. Data from torsion test is used to construct a stress-strain diagram and to determine elastic limit torsional modulus of elasticity, modulus of rupture in torsion, and Torsional Strength. Shear properties are often determined in a torsion test. (ASTM E-143)

Torsional Strength

Measure of the ability of a material to withstand a twisting load. It is the Ultimate Strength of a material subjected to torsional loading, and is the maximum torsional stress that a material sustains before rupture. Alternate terms are modulus of rupture and shear strength.

http://www.instron.us/wa/resourcecenter/glossaryterm.aspx

http://ixian.ca/pics7/torsiontest.jpg

Seems to be a marked similarity to my proposal. There is also a bit more to the formula than you describe.

http://ixian.ca/pics7/torsiontest2.jpg

You clearly either do not know what you are talking about or are trying to mislead the readers.

End of discussion.

philbur
04-20-2010, 03:57 PM
Sorry Phil, but is seems you are the one that is confused. Nobody switched ends.

You are now spouting obviously incorrect nonsense. Torsional Rigidity, Modulus of rigidity, Shear modulus, Shear modulus of rigidity and Modulus of Shear are all part of the same property as they are all aspects of Modulus of Elasticity. .
Which particular part is nonsense.

Evan if you look at equation (1) and (2) below you will see they are not part of the same property. The Torsional Rigidity (a property of the shaft) is actually used to determine the Modulus of Rigidity (a property of the material of the shaft). To a layman they may appear similar but the mathematics does not bear this out.

Torsional Rigidity = torque/angular displacement …………………..(1)

Modulus of Rigidity = Torsional Rigidity x length/polar moment of inertia. …………………(2)

(Polar moment of inertia is a factor related to the cross-sectional dimensions of the object)

More to the point, you can calculate the Torsional Rigidity without knowing any dimensional information about the object. In contrast you can only calculate the Modulus of Rigidity if you have dimensional information about the object. This can be seen from equations (1) and (2).

So you see that, in the context of your black box experiment, you can know the Torsional Rigidity (which is a property of the shaft) but not the Modulus of Rigidity (which is a property of the material).

Reproducing pages of technical publications with mark-ups looks impressive but contributes nothing to your reasoning. Especially when you clearly haven't bother to understand the information contained in them. I should point out that your first quote, with high-lights, doesn't even contain the words "Torsional Rigidity", so it is not clear what the specific purpose of this quote is. It looks like a lot of snow to me.

Making a general claim of nonsense, supported only by pages irrelevant quotes and mark-ups, followed by further general claims of "you don't know what you are talking about" is not a very scientific approach to proving your point.

Anyway, Good luck.
Phil:)

PS: Evan I am not happy continuing this discussion but while you continue to blow smoke up my rear-end I will continue to blow it back.

John Stevenson
04-20-2010, 06:26 PM
http://ixian.ca/pics7/torsiontest.jpg



Is that the same plank of wood you made the lathe bed out of ?

.

MrSleepy
04-20-2010, 07:48 PM
So......... do we have a winner yet ?..

The Artful Bodger
04-20-2010, 07:52 PM
I think you forgot that we changed ends at half time.:D

Phil:)

I hope that really was half time and not just the end of the first over.;)

J Tiers
04-20-2010, 11:02 PM
So......... do we have a winner yet ?..

A winner suggests an actual question as to the result..... WAAAAAYYYYYY back on page 1 of this horse-crap the same exact factual points that are being made on page 4 were already established...

But one of the participants didn't agree....... hence 3 more pages of flailing to get to exactly the same place.....

We certainly have a "Red Queen" in this discussion..... In several ways.....

Not only having to run as fast as possible to just stay in one place, but also being able to "believe as many as a dozen impossible things before breakfast" (I can't be bothered to look up the actual quote, you get the point)

Evan
04-21-2010, 03:06 AM
We have one participant that would have you believe that Torsional Rigidity is a simple calculation unrelated to the physical properties of the member transmitting torque. That is entirely incorrect.

It further seems that he is depending on other members not understanding the reason why it is incorrect and is using the "baffle them with bull" approach in dismissing the the evidence that I have presented to support my contention.

Here are some more incontrovertible illustrations that Torsional Rigidity cannot be quantified by a simple measurment of angular deflection. The reason is that, as I maintain and the science shows, Torsional Rigidity is not an extrinsic property but an intrinsic property that requires knowledge of the material and cross section of the torque element and cannot be calculated accurately at all for most shapes.



Torsional rigidity of non-circular bars in mechanisms
and machines

ABSTRACT

Determining the exact value of the torsional rigidity for cylindrical bars is limited to a few cross-sections. Exact solutions can be found in Mathematical Theory of Elasticity [Sokonlnikoff, Mathematical Theory of Elasticity, McGraw-Hill, New York, pp. 120] for the circle, ellipse and equilateral triangle. Good approximations for rectangular cross-sections are available in Theory of Elasticity, [Timoshenko, Goodier, Theory of Elasticity, McGraw-Hill, New York, p. 309] and approximations for various cross-sections such as trapezoids, isosceles triangle, semicircles, circles with a keyway, etc. are found in handbooks such as the one written by Roark [p. 309].

A method to approximate the torsional rigidity of any cylindrical solid cross-section is presented on this paper. Cross sections are classified by similarity to an ellipse, rectangle and/or a triangle. An aspect ratio of length to width is used as the second criteria to enter a graph where the torsional rigidity of the cross-section is given as a ratio to that of a circle of equal area.
......
http://ixian.ca/pics7/tr1.jpg

http://ixian.ca/pics7/tr2.jpg

http://ixian.ca/pics7/tr3.jpg

http://www.ewp.rpi.edu/hartford/~obriee/EP/Shaft_comparison/Najera_2005_Mechanism-and-Machine-Theory.pdf


Well, so much for the simple derivative approach. I wonder why they don't use it?

Any calculation of Torsional Rigidity taken as a simple derivative term of deflection and torque cannot describe the Torsional Rigidity of the torque transmitting element, aka The Shaft. It therefore does not provide information about the shaft.

The entire diatribe about how simple it is to determine Torsional Rigidity is a Red Herring with absolutely no basis in fact.

philbur
04-21-2010, 05:40 AM
Evan, do you accept that it is a fact that Torsional Rigidity is defined as the applied torque divided by the resulting angular displacement.

Yes or No.

If the answer is yes then clearly your previous post is complete nonsense. The nonsense is self evident so I shall not bother to itemise each point, unless you would like me to.

However just one example.


We have one participant that would have you believe that Torsional Rigidity is a simple calculation unrelated to the physical properties of the member transmitting torque. That is entirely incorrect.
One very, very, very, very, very last time:

Torional Rigidity = torque divided by the angular displacement.

This is not a “calculation” it is a definition. Where are the other properties in this definition. If the other properties are not in the definition then you do not need to know them to determine the value. Simple, no Snow no Bull Sxxx.

So now I guess your answer to my question will have to be “no”, and so we go on. New balls please.:D

Phil:)

Black_Moons
04-21-2010, 07:29 AM
Is that the same plank of wood you made the lathe bed out of ?

.

lol, Made my day!

But of course, Can't keep your lathe bed straight if your wood twists too easily. And gotta know to the micron how much it twists!
At least, when you work to the level of precision evan does.

Evan
04-21-2010, 08:56 AM
Evan, do you accept that it is a fact that Torsional Rigidity is defined as the applied torque divided by the resulting angular displacement.

Yes or No.

If the answer is yes then clearly your previous post is complete nonsense. The nonsense is self evident so I shall not bother to itemise each point, unless you would like me to.


You pose what is called a false dilemma. It isn't up to me to define Torsional rigidity nor is it up to you. I have no obligation to choose anything. I choose to follow the definition of what constitutes Torsional Rigidity as it is presented by example in the paper above. If you don't agree then take it up with the researchers that are using the term incorrectly, in your eyes. I doubt they would agree that their paper is "nonsense".

Personally, since a wide variety of papers and related research uses the term in the same manner as detailed above that is the definition that I will use too. Good luck convincing the rest of the world otherwise.

J Tiers
04-21-2010, 09:34 AM
Actually, Evan. YOUR bringing in of MATHEMATICAL ANALYSIS METHODS is the red herring. And that makes you the red Queen here.

The original example you brought up was a TORQUE METER, which required "calibration". And at the same time, YOU said that bringing in analysis methods (ships shaft example) was a red herring.

obviously sine it is no longer convenient to retain that position, you have changed your ever-changing argument.

Now, the paper extract you have rammed into this discussion is an analysis, not a measurement. Their difficulty of determining rigidity is basically an analysis limit, not strictly a measurement limit.

if you want to determine the torsional rigidity of a "system", which may or may not include a shaft...... the simplest and most direct method is, AS YOU YOURSELF AGREED, to hold one end, twist the other, and see what the torque to twist relationship is.

This has NO LIMITS of cross-section.....

You could as well test a shaft, a building, a system of rubber pads and levers, or whatever.

Bottom line here is that your ship sank pages ago, and you are circling the last of the wreckage, up to your neck with no lifejacket, denying that your feet are in the water.

You seem to be getting very used to being in that position.

John Stevenson
04-21-2010, 10:25 AM
You pose what is called a false dilemma. It isn't up to me to define Torsional rigidity nor is it up to you. I have no obligation to choose anything. I choose to follow the definition of what constitutes Torsional Rigidity as it is presented by example in the paper above.

Laymans translation.

Engage reverse Scotty...........

.

Evan
04-21-2010, 10:29 AM
The original example you brought up was a TORQUE METER, which required "calibration"

I never defined it as "Torsional Rigidity". You are out of arguments Jerry.


YOUR bringing in of MATHEMATICAL ANALYSIS METHODS is the red herring.

What an odd train of thought. Especially when the analysis has everything to do with what I described and uses precisely the same method. I never ruled it out. I asked "Now, what does it tell us about what is in the box? No assumptions are permitted. You must justify your answers based on the available evidence."


It's really simple. There is no known solution to calculating Torsional Rigidity for most possible cross sections. It must discovered empirically with full knowledge of the shaft and it's materials and cross section.

Evan
04-21-2010, 10:40 AM
Laymans translation.

Engage reverse Scotty...........

I have noticed that the only time you start posting these amusing witticisms is when you haven't a clue as to the subject matter at hand.

Do you realize that you have posted comments similar to this over 100 times now? How amazingly insightful and informative. Or, perhaps it is a sign of incipient OCD* since these comments are only directed at me.



*Obsessive Compulsive Disorder, treatment is available.

philbur
04-21-2010, 12:35 PM
Evan, why are you not prepared to give a yes or no answer? It’s a simple, unambiguous choice - is it in you opinion true or false. Of course you have no obligation, but your silence speaks volumes. I can only guess that your difficult is that having to answer a direct question prevents you from hiding you lack of knowledge behind page after page of irrelevant technical descriptions that you clearly don't understand.

Phil:)



You pose what is called a false dilemma. It isn't up to me to define Torsional rigidity nor is it up to you. I have no obligation to choose anything. I choose to follow the definition of what constitutes Torsional Rigidity as it is presented by example in the paper above. If you don't agree then take it up with the researchers that are using the term incorrectly, in your eyes. I doubt they would agree that their paper is "nonsense".

Personally, since a wide variety of papers and related research uses the term in the same manner as detailed above that is the definition that I will use too. Good luck convincing the rest of the world otherwise.

John Stevenson
04-21-2010, 12:45 PM
Yes
Yes
No
Yes

.

Evan
04-21-2010, 12:56 PM
Look up the definition of the fallacy "False Dilemma".


BTW, is this really you?

http://ixian.ca/pics7/phil.jpg

philbur
04-21-2010, 01:28 PM
Yes, that was me in my prime.:D

So what are the other options besides yes or no. It’s a simple technical question, not a metaphysical conundrum.

Phil:)


Look up the definition of the fallacy "False Dilemma".


BTW, is this really you?

http://ixian.ca/pics7/phil.jpg

Evan
04-21-2010, 02:20 PM
False Dilemma

Description
Either A or B is true. If A is true, B is therefore false. C is not an option.

The other person is offered a choice where rejecting one item acts as a selection of the other.

Example
Either you are with me or against me.

We have to spend less on hospitals, otherwise we won't be able to afford education improvements.

Discussion
This is based on the assumption that the choices offered are the only choices. By focusing on the choice, the decision to be made, the other person is distracted from the fact that there may be other alternatives.

This is usually presented as two choices, although more may sometimes be used.

Classification
Assumptive, Distraction, Falsehood

Also known as
Bifurcation, False Dichotomy

---------------------------------------------
I will not fall for such cheap tactics.

John Stevenson
04-21-2010, 02:34 PM
Where does "Grasp on reality" come into this ?

philbur
04-21-2010, 02:52 PM
So I ask again, what are the other choices ..... possible in your case "I don't know"?

Phil:)


False Dilemma

Description
Either A or B is true. If A is true, B is therefore false. C is not an option.

The other person is offered a choice where rejecting one item acts as a selection of the other.

Example
Either you are with me or against me.

We have to spend less on hospitals, otherwise we won't be able to afford education improvements.

Discussion
This is based on the assumption that the choices offered are the only choices. By focusing on the choice, the decision to be made, the other person is distracted from the fact that there may be other alternatives.

This is usually presented as two choices, although more may sometimes be used.

Classification
Assumptive, Distraction, Falsehood

Also known as
Bifurcation, False Dichotomy

---------------------------------------------
I will not fall for such cheap tactics.

Evan
04-21-2010, 07:26 PM
By asking this sort of question framed in this manner you have identified yourself as a debate troll. It is a standard tactic used by those that don't care at all about the subject matter, only thier own self gratification in appearing to "win" the debate.

Even though I have had a variety of debates over the years with other members on this forum, none have ever used this blatantly obvious tactic which immediately tags you for what you are.

You aren't as smart as you think.

J Tiers
04-21-2010, 11:06 PM
I never defined it as "Torsional Rigidity". You are out of arguments Jerry.



What an odd train of thought. Especially when the analysis has everything to do with what I described and uses precisely the same method. I never ruled it out. I asked "Now, what does it tell us about what is in the box? No assumptions are permitted. You must justify your answers based on the available evidence."


It's really simple. There is no known solution to calculating Torsional Rigidity for most possible cross sections. It must discovered empirically with full knowledge of the shaft and it's materials and cross section.


I will give you an A+ for changing the subject, reversing your stance, and basically trying to sell "dehydrated water".....

I suppose you see nothing wrong with it, but in the most astounding way you have attempted to couple two things that cannot be coupled in any way whatever......

It is jaw-dropping..... and almost as jaw-dropping that you see nothing out of the ordinary.....

1) Evan proposes a "torque box" which has no description of the innards AND THEREFORE NO WAY TO CALCULATE ANYTHING, LET ALONE THE TORSIONAL RIGIDITY.

2) Then he says this: "It's really simple. There is no known solution to calculating Torsional Rigidity for most possible cross sections. It must discovered empirically with full knowledge of the shaft and it's materials and cross section.'.....

Now, the original subject wasn't this at all, it was the issue of a "phase torque meter", which was alleged, by "him we have made mention of", to supply an accurate reading with no knowledge of the shaft whatever, but which naturally needs calibration.....

A gentle reminder that calibration supplies a scale factor in the form of the "torsional rigidity" of the shaft was met with denials and obfuscation......

Then the 'torque box" appeared, which we weren't allowed to see the innards of.....

Somehow this undefined, and therefore non-calculable, device has generated a "triumphant" conclusion (just what has been concluded is in doubt) based on the fact that "There is no known solution to calculating Torsional Rigidity for most possible cross sections. ".

Of course, the obvious course of simply measuring was denied and denied and evaded..... But suddenly it is back....... now it is EVAN's idea, and it has blossomed into "THE ANSWER" (to what it isn't clear)................ ""It's really simple. There is no known solution to calculating Torsional Rigidity for most possible cross sections. It must discovered empirically with full knowledge of the shaft and it's materials and cross section."

of course we really don't need teh cross-section if we have a lever and a weight, but so be it....... "empirically discovering" it would be the same thing as "measuring".......

Maybe this makes sense to the red Queen....... It makes no sense to anyone else......

As for "torsional rigidity", it makes no difference what EVAN called it, it IS "torsional rigidity", so discussing it under that name is perfectly correct.

news Flash....... Lewis Carrol has been dead a long time, and the "lookingglass world" was fictional, very probably fueled by opium..... So don't try to resurrect it here...... but it's too late to complain about that now..... This whole thread has a surreal "looking glass" quality, and I expect to see the jabberwok any moment...... We seem to already have something that will "gyre and gimble in the Wabe"..... maybe it's a 'tove", it has a "v" in it anyhow....

Yeeeeesh........ I go away for a while and make parts, only to come back and find what this has become......

Evan
04-22-2010, 12:34 AM
1) Evan proposes a "torque box" which has no description of the innards AND THEREFORE NO WAY TO CALCULATE ANYTHING, LET ALONE THE TORSIONAL RIGIDITY.


Precisely. That is the entire point.


2) Then he says this: "It's really simple. There is no known solution to calculating Torsional Rigidity for most possible cross sections. It must discovered empirically with full knowledge of the shaft and it's materials and cross section.'.....


And of course we cannot do that so that avenue is closed.



Of course, the obvious course of simply measuring was denied and denied and evaded.....


No it wasn't. Simply meauring from outside tells you nothing specific.



But suddenly it is back....... now it is EVAN's idea, and it has blossomed into "THE ANSWER" (to what it isn't clear)................ ""It's really simple. There is no known solution to calculating Torsional Rigidity for most possible cross sections. It must discovered empirically with full knowledge of the shaft and it's materials and cross section."



As is often the case your understanding of the discussion is inverted.
Measuring isn't back, it's impossible. That is my point.



of course we really don't need teh cross-section if we have a lever and a weight,

Yes you do if you want to quantify the characteristics of the torque member. You see, the point is that you cannot do so without knowing the cross section and since you don't know that you cannot describe it no matter what you do externally. That of course follows because many instances of the problem cannot be solved even if you do have full knowledge.

Your post is completely meaningless since you failed entirely to grasp my point and have actually taken it to mean the exact opposite. That of course would make no sense. Do you stand on your head often?

J Tiers
04-22-2010, 01:10 AM
And of course we cannot do that so that avenue is closed.


Measuring isn't back, it's impossible. That is my point.


Yes you do if you want to quantify the characteristics of the torque member. You see, the point is that you cannot do so without knowing the cross section and since you don't know that you cannot describe it no matter what you do externally. That of course follows because many instances of the problem cannot be solved even if you do have full knowledge.



First, re-read Philbur's posts

Second step: Justify the EXTREMELY odd statements above.

while doing that, please explain again why you need to know anything other than "so much torque produces this much twist" , at however many values seem required.....

Unless your device is time-varying in a random fashion, which we will presume is excluded unless you choose to claim it now, those statements amount to denying a considerable part of reality..... S it will be interesting to see your justification.

You may be touching a piece of the deck, but that doesn't mean the ship didn't sink a while ago....... which explains why you are treading water now.

While you try to come up with a coherent reason why physics, logic, and engineering reality does not apply to your part of the universe, I got stuff to do.............

Evan
04-22-2010, 01:47 AM
while doing that, please explain again why you need to know anything other than "so much torque produces this much twist" , at however many values seem required.....


If you don't get it by now you never will.

dp
04-22-2010, 02:20 AM
Here are some more incontrovertible illustrations that Torsional Rigidity cannot be quantified by a simple measurment of angular deflection.


Ho, whoa - that's a given. How can this be disputed?

Bob G
04-22-2010, 03:35 AM
How many other viewers of this thread are only following it to see how far someone will go to defend a statement like [throw something together like that in a few hours. Cheap and easy.]
This is from an otherwise intelligent person with many years of experience. His first post [#9] to a thread that has now reached [#148.]

I am [80] years old and just once I wish I could hear the words [I AM WRONG] from an obvious BS’er.

Have fun,
Bob G

Evan
04-22-2010, 03:53 AM
Torsional Rigidity is an intrinsic property of a material. The meaasurement of the deflection vs applied load doesn't measure the intrinsic Torsional Rigidity of the material since it is the geometry of the material that has the greatest affect on the measured result.

The geometry of a construct of any material is not a property of that material. Note that the Torsional Rigidity of a particular material is a direct function of the Shear modulus of that material per unit area. Geometry is not specified.

A test that generates a graph of deflection vs load produces a graph that represents the Torsional Rigidity of an UNKNOWN cross section of an UNKNOWN geometry of an UNKNOWN material.

As such, it does not inform one about the nature of the torque transmitting element nor is it possible to deduce that information from the particular curve generated.

Evan
04-22-2010, 03:56 AM
I am [80] years old and just once I wish I could hear the words [I AM WRONG] from an obvious BS’er.


In a recent and completely unrelated thread I not only admitted my mistake but apologized as well. If I thought I were wrong I would say so.

John Stevenson
04-22-2010, 04:20 AM
Evan might not be right but he's never wrong.

High reverse Scotty..................

.

Evan
04-22-2010, 04:23 AM
101 and counting...

J Tiers
04-22-2010, 09:29 AM
Torsional Rigidity is an intrinsic property of a material. The meaasurement of the deflection vs applied load doesn't measure the intrinsic Torsional Rigidity of the material since it is the geometry of the material that has the greatest affect on the measured result.

The geometry of a construct of any material is not a property of that material. Note that the Torsional Rigidity of a particular material is a direct function of the Shear modulus of that material per unit area. Geometry is not specified.

A test that generates a graph of deflection vs load produces a graph that represents the Torsional Rigidity of an UNKNOWN cross section of an UNKNOWN geometry of an UNKNOWN material.

As such, it does not inform one about the nature of the torque transmitting element nor is it possible to deduce that information from the particular curve generated.

A very nice (but wrong) answer to a problem which is NOT involved here.....

We do NOT care what the properties of the MATERIAL are........ in the "abstract"... That isn't relevant. (and torsional rigidity is, of course, not a "property" of a material, but of a material AND a geometry)

The torsional rigidity of so many square cm of steel, or bone, or balsa wood, will be radically different depending on how that steel is arranged..... hence the dependence on geometry.

However, you have now made a direct statement, frozen and unalterable above. It is now up to you to defend that, because it is incorrect in the view of most.

Please explain in detail, showing your work, exactly why the direct measurement of the angular deflection vs torque is INCORRECT..... and why the REAL "torsional rigidity" will have some other value totally unrelated to that measurement.

This would seem to be rather difficult , since the definition of "torsional rigidity" is the angular deflection vs torque....... But we are always willing to see a good argument as to why "down is really up". "left is right", "blue is really red", etc.

hitnmiss
04-22-2010, 10:16 AM
A test that generates a graph of deflection vs load produces a graph that represents the Torsional Rigidity of an UNKNOWN cross section of an UNKNOWN geometry of an UNKNOWN material.

As such, it does not inform one about the nature of the torque transmitting element nor is it possible to deduce that information from the particular curve generated.

Don't tell that to my torque wrench!

Carefully chosen words allow you to continue to debate your extremely convoluted points...

"Nature"?? I suppose that could be debated as doesn't tell me the exact geometry of the torque transmitting member. Agreed but who the hell cares? If it's 1" dia of lead or .1" dia of steel does the mechanic frightening the bolt with the torque wrench care? It does tell me that at this displacement of the torque arm I get this much torque on the bolt, which is the whole point.

Barrington
04-22-2010, 10:44 AM
A discussion of 'torsional rigidity' is always potentially hazardous, because there are two slightly different definitions.

The use of the term 'torsional rigidity' for torque divided by the angle of twist seems to be almost universally used in the real world, but in the academic world 'torsional rigidity' is defined as the torque divided by the angle of twist per unit length - i.e. it is influenced by cross section and material only.

To add to the confusion, some authors then differentiate between 'torsional rigidity' and 'torsional stiffness', and use the latter to refer to torque/twist. So: (torsional stiffness) = (torsional rigidity)/length. - These authors will then define 'torsional compliance' as the inverse of 'torsional stiffness' rather than the inverse of 'torsional rigidity' etc. etc.

All this is fine and dandy until the unsuspecting Googlist merges results from both camps...

Don't even start on modulus of...

Clear - as mud ??:D

Cheers

.

J Tiers
04-22-2010, 09:04 PM
(Edited for mis-spelling typos)
Well, Barrington seems to bring out the point that Evan is using the academic definition, because it suits his argument AT THIS TIME...... while he apparently used the enginerring definition earlier, BECAUSE IT SERVED HIS POINT AT THAT TIME.

Here, for his and your delectation and enjoyment, is a compilation of contradictory posts by Evan....

Note that early on, up to post 29, he said you could do it all by measurement, and NOW, somewhere after post #29, he reversed his position, he has taken up the "Albigensian heresy" and says that using measurement is "impossible" (his word)

It does not seem very necessary to take anything he says about the matter very seriously. An excellent example of why "the shoemaker should stick to his last"......

For the purposes of full disclosure, I mention that I have in some cases added bold and italic to egregious points... but not every bold or italic item is bold etc by my action.

Post #11
Meauring the twist doesn't require knowing anything about the shaft, only the

load.
Post #20
Nope. All you need to know is how much twist is produced by a particular

amount of torque on the output. That can be a purely relative value that is determined by an

experienced operator who can tell when the system is close to it's limit. All the other possible

methods face exactly the same issue if an absolute value of torque is required to be known and

that is somehow measuring the actual torque on the output.
post #23
Nope. All it is is knowing the relative alignment of each end of the shaft to the other. That does not reveal the properties of the shaft nor does it matter. That isn't what is being measured. It is a value that is directly coupled to the torque load and can be calibrated in absolute terms without even knowing what size or shape the shaft is.

Post #25
In case you missed it here it is again.

All you need to know is how much twist is produced by a particular amount of torque on the output. That can be a purely relative value that is determined by an experienced operator who can tell when the system is close to it's limit. All the other possible methods face exactly the same issue if an absolute value of torque is required to be known and that is somehow measuring the actual torque on the output.


It still doesn't require knowing anything about the properties of the shaft.



By the way, we don't even need to know that much. All we need to know is that there is a timing difference between when the marks pass the sensors and that also doesn't need to be quantified. The most we need to know is that X timing difference corresponds to N torque.

Post #27
You can measure the torque on the output end of the shaft with a dyno. That is all you need to know to calibrate the system. If you don't understand that I can't make it any simpler. That doesn't give you information about the properties of the shaft since it doesn't even tell you the amount of deflection that takes place. You don't need to know that.

The only information that you obtain is a phase difference between input and output. That is enough.

Post #29
I said no such thing. Go back and read the postings Jerry. The dyno is for calibration which is precisely what I said in two different posts.

For the third time:

All the other possible methods face exactly the same issue if an absolute value of torque is required to be known and that is somehow measuring the actual torque on the output.



Yes it was. It's called a propellor. That is exactly how some dynomometers work.

copied from Post #145


Quote:

1) Evan proposes a "torque box" which has no description of the innards AND THEREFORE NO WAY TO CALCULATE ANYTHING, LET ALONE THE TORSIONAL RIGIDITY.

Precisely. That is the entire point.

Quote:
2) Then he says this: "It's really simple. There is no known solution to calculating Torsional Rigidity for most possible cross sections. It must discovered empirically with full knowledge of the shaft and it's materials and cross section.'.....

And of course we cannot do that so that avenue is closed.

Quote:
Of course, the obvious course of simply measuring was denied and denied and evaded.....

No it wasn't. Simply meauring from outside tells you nothing specific.

Quote:
But suddenly it is back....... now it is EVAN's idea, and it has blossomed into "THE ANSWER" (to what it isn't clear)................ ""It's really simple. There is no known solution to calculating Torsional Rigidity for most possible cross sections. It must discovered empirically with full knowledge of the shaft and it's materials and cross section."


As is often the case your understanding of the discussion is inverted.
Measuring isn't back, it's impossible. That is my point.


Quote:
of course we really don't need teh cross-section if we have a lever and a weight,

Yes you do if you want to quantify the characteristics of the torque member. You see, the point is that you cannot do so without knowing the cross section and since you don't know that you cannot describe it no matter what you do externally. That of course follows because many instances of the problem cannot be solved even if you do have full knowledge.

Your post is completely meaningless since you failed entirely to grasp my point and have actually taken it to mean the exact opposite. That of course would make no sense. Do you stand on your head often?