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Fasttrack
05-17-2008, 12:03 AM
Lets say I've got a 36" solid rod with an axis of rotation through the center of the rod and perpindicular to the length. I want to know, in terms of tension, what kind of force it expierences.

I thought I worked up an equation for it, but the numbers I'm getting seem to be a little bit askew. I googled it but didn't come up with the formula I wanted.

Basically I need to spin a rod very fast. Like I said, the axis of rotation is perpindicular to the length of the rod and in the center. I need to get as high a tangential velocity as possible without blowing something up to give you all a little background for this machine. I'm in the intitial design process trying to decide if it is even feasible to get the velocity I need in a mechanical system.

Alright well I've said enough. Let me know if you've got a handy dandy equation ...

mechanicalmagic
05-17-2008, 12:11 AM
Lets say I've got a 36" solid rod with an axis of rotation through the center of the rod and perpindicular to the length.

So, is this a rod shaped propeller?

Is there a hole in the material, or other shape that would degrade the strength.?

oldtiffie
05-17-2008, 12:18 AM
First thing to do is to define hoop stress.

Will this help?

http://en.wikipedia.org/wiki/Hoop_stress

Fasttrack
05-17-2008, 12:21 AM
Nope - Say its an 18" long rod with a diameter of 1/2" and welded to a hub. I figured on 36 because I will have another rod on the otherside for balance but they will be attached via a hub.

Not exactly a propellor, more of a centrifuge. I need to accelerate an electronic package I've designed to very very high speeds.


OT - "will this help" Not very much. I used the term "hoop stress" in order to help give a mental picture of what is going on and, mainly, attract the engine guys here who know about designing flywheels and etc :o

Evan
05-17-2008, 12:26 AM
Look for flywheel formulas. Same thing. Many high energy density flywheels use in ride through power supplies are drum shaped.

Fasttrack
05-17-2008, 12:28 AM
Well I guess thats another question, am I better off with a solid flywheel or just a rod? Obviously a rod is much cheaper to produce and easier to work with. Like I said, this project will really be pushing the envelope for me. Evan -what do you think about spining an 18" radius at 2000 (or higher!) rpm? Can it be done?

oldtiffie
05-17-2008, 12:31 AM
The forces are huge as are the mechanics and physics!!

http://en.wikipedia.org/wiki/Centrifuge

http://en.wikipedia.org/wiki/Special:Search?search=centrifuge&fulltext=Search

Also check Machinery's Handbook.

Google?
http://www.google.com.au/search?hl=en&q=centrifuge&btnG=Google+Search&meta=

I've reached my limit here.

Evan
05-17-2008, 12:35 AM
Here are some hints. The optimum shape for a disc flywheel is the one below. It results in constant stress on the material because of the thickness distribution. This will apply equally to an arm as it does to a disc.

http://vts.bc.ca/pics3/flywheel.jpg

Unfortunately the above shape is a pain to make. Fortunately a close approximation is both easy and nearly as good.

http://vts.bc.ca/pics3/flywheel2.jpg

Tinkerer
05-17-2008, 12:37 AM
I guess I'm not as good at reading between the line. So... what are the other dimension of this 36" rod. what size is the hole thru it?

Ok you answered that will I was typing no hole but a hub... So your making a centrifuged.

oldtiffie's Your fast with the wickedpeda...:p

Evan
05-17-2008, 12:41 AM
Just calculate the centrifugal "force" and compare it to the tensile strength of the material used for the arm or disc. If the shape approximates the upper one the tension will be linearly distributed with radius. The number of gees times the density will tell you when the limit is reached. For a safety margin then back it off by 50%.

BTW, is this homework?

Tinkerer
05-17-2008, 12:45 AM
I'd think about using a flat bar instead of a round... and balanced off end for whatever your load (package) is so it did not shake it self apart. Not really knowing how big the package is but what about modifying a lawn mower blade to hold it... you could cut the grass while you give it a whirl. :D

J. R. Williams
05-17-2008, 12:47 AM
The maximum speed of the flywheel is dependent on the material available and the quantity of money avilable. The earlier post shows the classic high speed solid flywheel design. The Swiss used them in an energy storage system to drive passenger busses many years ago. Look at the speeds produced in the modern jet engine and they use high alloy forged wheels. Define the velocity required for your project and then some valid numbers can be calculated, along with the energy required to accelerate the mass.
JRW

Fasttrack
05-17-2008, 12:59 AM
Wow thanks for all the replies guys. Alright, here it goes.

1) Evan - thats what I was curious about, how the force of tension is distributed through the radius of the part. Cost becomes an issue when we start talking about flywheels in large diameters since it requires a rather large blank of material. Plus, it requires a big lathe! :) Nope, its not homework. I'm begining an "advanced" physics lab next semester where we essentially pick a project and then present our results to the entire faculty. Kind of intimidating! Anyway, my plan is pretty ambitious at this point but should be interesting. Most students end up doing something sort of lame, like measuring the effect of acceleration on newtonian and non-newtonian fluids. :rolleyes: It's like the college equivalent of the "volcanoe experiment".

2) Does anyone know how tension is distributed in a round, square, rectangular, etc bar? something like 1/2 r^2 ? I can't remember now...

3) It definitely needs to be balanced! I'd like to spin this just as fast as possible.

4) I don't know for sure what speed I need, or rather I have some wiggle room. The faster the better. The minimum speed is 36,000 inches per minute. This an 18" radius wheel spinning at 2000 rpm and the package on the outside.

Some more details:
The "package" is fairly light, figure an absolute max of 150 grams. I will need to run at least three wires to the package. A reference, a voltage, and a signal. The signal wire needs to be shielded as it will be transmitting a high frequency to a series of digital flip-flops. The faster the velocity, the lower frequency signal required (which means cheaper components and less noise) and the more accurate the data. This is a tricky procedure at best, since I have to deal with accuracies of the electronics and the effect that enormous accelerations will have on the components.

aostling
05-17-2008, 01:55 AM
Roark's Formulas for Stress and Strain has a formula for the maximum tensile stress in a rod rotating about a pivot at one end

s (psi) = numerator/denominator


numerator = W*L*(w**2)


denominator = (2)*(386.4)*A



W = weight of rod, lb.
L = length of rod, in.
w = rotational velocity (radians/sec) = pi*(RPM)/30
A = rod cross-section area, in**2

So here L is the radial length from the pivot, half the length of your rod which is pivoted at its center instead of its end.

oldtiffie
05-17-2008, 02:10 AM
OK.

A bit of academic regimen/rigour is required here instead of perhaps pulling numbers and opinions out of the air.

I'd guess that the OP has a good grip on math and physics.

He should be able to derive his specific application from the general case/principles involved by substituting his own values as regards inputs and limitations.

As "fly-wheels" has come to the fore, try these:

http://en.wikipedia.org/wiki/Flywheel

http://en.wikipedia.org/wiki/Special:Search?search=fly-wheel&fulltext=Search

As in all flywheels, there will be potentially considerable dynamic fores not the least of which may be those of:

- gyro-scopic action/s:
http://en.wikipedia.org/wiki/Gyroscope

and:
http://en.wikipedia.org/wiki/Special:Search?search=dynamic+balance&fulltext=Search

Swarf&Sparks
05-17-2008, 02:24 AM
Presuming this is a repetitive/cyclic stress, I'd be designing for a max strain of < 0.5%. High tensile steels have a tensile strength of up to 60,000 PSI, IIRC.

Evan
05-17-2008, 02:32 AM
The most important consideration in this instance will be air drag if a rod or other non aerodynamic shape is used. For a flywheel that size if it isn't aerodynamic the losses due to churning will be the limiting factor. To even approach the tensile limit of ordinary materials the overall shape will have to be an unbroken smooth shape of some sort. Have a look at high speed lab centrifuges for design hints.

http://vts.bc.ca/pics3/centrifuge.jpg

oldtiffie
05-17-2008, 02:43 AM
My "gut" feeling is much the same as Evan's.

It will require very high orders of dynamic balance as well as design for minimising "stress-raisers".

I was going to suggest using an "on-Campus??" or "in-house" centrifuge with equal loads equally applied and dynamically balanced.

Perhaps some of the aero-dynamics would be over-come if it were "spun-up" in a vacuum.

As hinted in a previous post, it would have been preferable if the project/subject was identified as a personal/educational/"for work" project.

rantbot
05-17-2008, 03:06 AM
My, what a lot of singing and dancing.

The acceleration of a point rotating at a speed W radians/sec at a radius R meters from some axis is a=R(W**2), a in meters/sec squared, naturally.

Tony
05-17-2008, 03:11 AM
"hoop stress" -- if I understand you correctly -- doesn't
really apply here so looking up hoop stress formulas will
lead you down the wrong path.

if you're only interested in rod failure (which really is going
to be the least of your problems!) it doesn't really matter
to the rod that it is spinning. all that spinning does is
set up an acceleration and a force that will try to stretch
your rod (look up 'centrifugal' force) -- use this number
in one of the basic stress/strain formulas. be sure to add
the weight of the electronic package, connectors, wires,
etc.

again, all you are doing here is (from the rods point of view)
is stretching it.

remember that physics problem where you had a ball tied
to a string and you spun it over your head as fast as you
can and were asked how fast till the string broke? same
problem.

this is assuming that you spin it up to speed slowly.. ie
you don't need to get from 0 to 2000 rpms in a 1/10th
of a second. if that were the case you'll need to worry
about bending in the rods.

the real problem is going to be that hub (the welds) and
the bearing assembly.

that and keeping your wires from getting all twisted up :)

the rest of the problem is going to be decided by your
electronics and the purpose of your experiment -- how
"balanced" is "balanced"?

Evan's cross-section is correct from a stress point of
view -- but not very good for a flywheel (since the point
is to store energy) -- again, since you're not trying
to store energy, "flywheel equations" will also lead you
down the wrong path.

your best bet, if all you are looking for is tangential speeds,
is to use something really lightweight -- maybe even an
aluminum disk -- and have it dynamically balanced before
you mount it (bolted!) to your hub. drill holes in the plate
to balance it or add through bolts on the opposite side of
your package. the disc wouldn't need to be turned to a
perfect circle -- as long as it is properly balanced.

the enemy here is vibration & what it'll do to your machine.

wow, that was long-winded.

-Tony

Ian B
05-17-2008, 03:12 AM
Fasttrack,

18" diameter and 2.000rpm doesn't sound like a problem. We're running turbine-driven centrifugal gas compressors here offshore. The rotors are about 2 feet in diameter and they run at 16,000rpm day in, day out. These are industrial machines, not especially high-tech.

The rotors look a bit like ventilated brake discs and weigh something like 100Kg each. Never heard of one failing.

Regards,

Ian

Ian B
05-17-2008, 03:19 AM
Oh, on the vibration issue; the rotors are dynamically balanced by the vendors, but then we get hold of them :-)

In service, deposits build up in the gas passages (we're compressing wet natural gas, mostly methane). We inject various chemicals to stop this, but buildup still happens. Every so often, a bit of foulant breaks off and we see vibration. If left unattended, this tends to wear the floating pad capsule bearings to the point where we have to fit a rebuilt bearing (at $10k a pop). That sorts most of the vibration out. Every so often, we change the entire compressor stage out for a factory rebuilt one.

The rotors are made of some kind of stainless steel.

Ian

ckelloug
05-17-2008, 05:16 AM
So Fasttrack,

This is what you are doing?
http://i161.photobucket.com/albums/t202/ckelloug/fasttrack.png

I assume that you are trying to demonstrate special relativity by skewing the frequency of an oscillator!

The hint for the stress distribution is to remember that the formula for centripetal force as mentioned by Rantbot is mw^2*r. (I cheated and looked it up on the web as my employment of late usually involves rotating images rather than masses). This is only good for a particle of a given mass on a string.

So, being the good engineers we are, we integrate. So first we must find mass in terms of r.

A*rho*dr should be the mass at r. Where phi is the density and A is the cross sectional area. This should be sufficient as long as the shape is symmetric.

Thus F=integral(A*rho*r*w^2*dr) from 0 to r. which should mean F=A*rho*r^2*w^2 /2.0 for one half of the assembly.

Total Force should then be A*rho*r^2w^2; Stress however is force divided by area so rho*r^2*w^2 appears to be the stress for a rotating object that is of fixed cross section.

I had a few too many last night so I may have made an arithmetic mistake but considering what you appear to be up to, I suspect that you can derive the correct stress. I got N/m^2 as the unit of the above so it's not insane but check the equation before using it. . .

At any rate, have fun and good luck with whatever you are actually doing. P.S. What degree are you studying
for?

Regards,
Cameron

bob ward
05-17-2008, 05:21 AM
What about going to 18" diameter and 4000 rpm, or 12" diameter and 6000 rpm? At 12" diameter a machined disc becomes more feasible, or maybe use an automotive flywheel and skip some machining. Plus you know a modern flywheel is good for 6000 rpm, no worries at all.

Set up the flywheel on the end of a 1 1/2" dia countershaft, select 2 V belt pulleys to provide a countershaft speed of 6000 rpm from the electric motor you had in mind.

Switch on from a safe distance and see what happens.

DICKEYBIRD
05-17-2008, 08:22 AM
You guys are way over my head with all your high level cipherin' & guzintas (as Jethro Bodine would say) but bob ward's solution looks to be the easiest & safest solution. A good used one at the junkyard should be next to nothing. Just don't mention that it's going in a college engineering project....that'd drive the price way up there.;)

Also, isn't it coincidental how the optimum flywheel cross-section Evan posted mirrors the shape of a "NACA duct." Strange things at work here I tell ya!


http://vts.bc.ca/pics3/flywheel.jpg

Rustybolt
05-17-2008, 09:17 AM
just duct tape whatever your testing to the chuck of a large CNC lathe.
I kid. I kid.

A.K. Boomer
05-17-2008, 09:45 AM
Also, isn't it coincidental how the optimum flywheel cross-section Evan posted mirrors the shape of a "NACA duct." Strange things at work here I tell ya!




pretty observant of you DB, I just got a blast of that kinda music you hear when they show extraterrestrials in movies and crap...:) (also very similar to the tunes they play when you die and your body's going to heaven)

rantbot
05-17-2008, 09:50 AM
Flywheels have no more to do with the problem than hoop stress does. Flywheels are designed to store energy, test packages held on the ends of sticks are not. Entirely different problems with entirely different solutions.

The easiest way to generate high G loads on a test object is to drop it on the floor. The subsequent load is, of course, not continuous. For a continuous load a centrifuge of some sort is needed. In-between would be something like a rail gun.

Evan
05-17-2008, 10:18 AM
Fasttrack wrote:

4) I don't know for sure what speed I need, or rather I have some wiggle room. The faster the better. The minimum speed is 36,000 inches per minute. This an 18" radius wheel spinning at 2000 rpm and the package on the outside.
Rantbot wrote:


Flywheels have no more to do with the problem than hoop stress does. Flywheels are designed to store energy, test packages held on the ends of sticks are not. Entirely different problems with entirely different solutions.
It appears from what Fasttrack has written a flywheel is indeed what is needed. The fact that it isn't intended to store energy is irrelevant. It will just as the medical centrifuge does.

I agree that the use of an automotive flywheel is likely the best approach. However, to be safe I would have it spin tested to at least 50% greater rpms than it will operate at in the experiment. I would also provide a scatter shield and would operate it in the vertical alignment to reduce the hazard. Finding bearings able to withstand a continuous 4000 to 6000 rpm will also be a challenge as they will develop a significant amount of heat in the size needed.

hitnmiss
05-17-2008, 10:41 AM
I don't have much to add except most of the above is spot on.

I was just thinking I shifted my old 327V8 in my 55 chevy once at 7200 rpm.

I looked up the diameter of the flywheel... 14"

pi x D=44"

7200 x 44=316,672 in per min

You only need 36,000 ipm?

No worries! (unless I screwed up my figurin)

DICKEYBIRD
05-17-2008, 11:25 AM
Wouldn't gravity have an effect on any measurements if it were aligned vertically? I'm just sayin'...

Fasttrack
05-17-2008, 12:27 PM
Boy a whole bunch of great posts!

Ok... aostling, that was exactly what I was looking for originally. I'm fairly certain I've seen that before in a strength of materials book but right now, I don't have access to any books. They are all buried on the bottom of a bunch of college related paraphenalia, like clothes and food ;)

The issue here is not the acceleration but the velocity. That was why I was opting for a larger diameter wheel rotating slower. The larger the diameter, the closer it approximates linear velocity. However, a few inches will make little difference in this case so I expect a smaller, faster "flywheel" is more feasible.

Thanks for the real world examples. I should have thought about the lawnmower blade at least. That was an obvious example for someone who's worked on many a lawnmower!

Cameron - Yep, thats exactly what I intend to do. At the specified speed, I will see a difference between two precision oscillators only one order of magnitude greater than the error. Incidentally, it will take something like 5 years at that speed to notice a difference of one second :D Luckily, I don't have to look for one second differences. This is why higher speed is better, the shift will be much more pronounced at higher speeds.

I tried integrating, as you did Cameron. A(rho)r is really just the total mass, so it becomes MRw^2 for the entire assembly. I got some rather absurd numbers. I calculated a force of 341,000 N. Perhaps I made a mistake in my arithmetic or my conversions somewhere. I'll do it again today and see what I get. I was working on about 4 hours of sleep yesterday, so its entirely possible!

I'm studying physics, btw. My area of interest is in high energy physics and its been a few years since I've crunched numbers for mechanics. I tested out of mechanical and intro level E&M courses so the last time I had those was in high school. I remember how to approach problems and the basics, like centripedal force, for instance, but ... well I'm no engineer :D Without a book I don't remember all the equations for stress, strain, moments, etc. Thats why I tried integrating, that seems like the most straightforward and "physical" approach.

Alright, well I gotsta go work! Later this evening I'll crunch some more numbers.

Fasttrack
05-17-2008, 12:39 PM
Oop I forgot! Evan and everyone else who brought up the issues of safety and aerodynamics:

Maybe it is because I'm studying to be a physicist and we have a habit of reaching for a vacuum pump at the slightest provocation, but I intend to operate the physical part of the apparatus in vacuum. My plan was to use a compressed gas container of some sort, maybe an LP tank or an old aircompressor tank to make the enclosure. This would provide safety from flying shrapnel and allow me to draw a reasonable vacuum. The vacuum pump would be on the entire time so I expect some o-rings on the "drive shaft" would be sufficient to create a partial vacuum that would dramatically reduce the aerodynamic concerns.

Tony
05-17-2008, 01:58 PM
special relativity?
why didn't you say so... :)
36000ipm is only about 35mph.

can you run your experiment in a car?

how long do you plan to run for?

-Tony

ckelloug
05-17-2008, 02:12 PM
Fasttrack,

I'm an engineer but the wrong kind for your problem. I did have to take special relativity however. I'm almost sure I did something boneheaded in the derivation I posted. It seems like you have to integrate twice, once from the center to get the force required to keep each point moving in a circle and again to figure out the maximum stress at the hub. I haven't thought this through really thoroughly however.

I'm to the point where I have to look everything up too. I think Aostling and his solution are ahead of the game in that he's an aerospace engineer with the reference library on the shelf for dealing with these problems the easy way.

Actually doing the integrals and getting the answer will make your presentation to the physics faculty a lot more impressive. The trick is to make sure you get the same answer as the engineering book ;) Congrats on thinking up a really interesting demo!

If I can be of any help, please don't hesitate to post.

--Cameron

Evan
05-17-2008, 04:30 PM
I have this nagging feeling that the acceleration produced in a rotating system will produce relative motion that averages to zero. It doesn't satisfy the classic Einstein thought experiment in which linear acceleration is indistinguishable from gravitation. Determining that one is in a rotating system would be trivial. Since the acceleration vector is constantly changing I would expect there to be no net difference. It isn't the same as satellites orbiting the earth as they are in free fall around the curved space of the gravity well and are not accelerated at all.

I don't think the solution to the twins paradox will save this either as it depends on asymmetry of accelerations to produce the differential time dilation.

I may be wrong. For the sake of your experiment I hope so.

Fasttrack
05-17-2008, 10:14 PM
Actually Evan, that is the main ... umm "experimental" part of this crazy plan.
Special relativity only holds for inertial refernce frames. Of course, space is really not flat at all so thats where General Relativity comes in. Luckily I'm taking a class devoted specifically to general relativity next semester, but it is my current understanding that special relativity holds for "local" areas that approximates flat or euclidean space.

I expect that you will be right, however. Even with a relatively large diameter flywheel... it is still very very small :)



edit: Thanks Cameron! I'll most likely have many more questions.

Tony - I'd like to run it in excess of 24 hours. That will be dictated partially by the amount of heat generated in the bearings and what kind of virbations I notice (i.e. do I feel confident enough to leave it run without someone watching it...)

ckelloug
05-17-2008, 11:14 PM
Fasttrack,

From the form of the lorentz transform for time, I am assuming that your experiment will work under this rationale:

While linear dimensions are only dialated in the direction of motion, time isn't a linear dimension and therefore must be uniformly dialated by a motion in any direction. If there aren't oddball affects from the acceleration of motion in a circle, I think it will work though my memory is much to hazy to remember caveats in the theory.

I remember from my special relativity class a decade or so ago using Helliwell's book (and having Helliwell as prof) that the experiment has been shown to work with atomic clocks and airplanes. Airplanes travel in approximately circular motion around the earth and they got a dialation result by flying an atomic clock on an airplane for a long distance and comparing it with one that was not flown.

It has been ages since I thought at all about this but memorizing the Lorentz transform was mandatory for passing that class. Engineers, physicists, mathematicians, and biologists alike had to take it.

After we passed, we got t shirts that had a picture of a brain with the caption "this is your brain" followed by a picture of a vertical line captioned "This is your brain at 0.999999c: Any questions?"

Regards,

Cameron

dp
05-17-2008, 11:56 PM
This appears to be something a bicycle does all the time. I must be missing something.

oldtiffie
05-18-2008, 12:42 AM
This appears to be something a bicycle does all the time. I must be missing something.

Yup - seems you can do lottsa stuff on a bicycle!!

But, a look at Machinery's Handbook is well worth the effort as it is a gold-mine in these sorts of things. There's too much for me to scan and post today else I'd do it.

boslab
05-18-2008, 02:38 AM
telephone exchanges as was used a spinning gyro type back up [power], they were spun up and topped up by a drive motor/ mag clutch, if grid power failed the mag clutch would latch a dynamo and the spin would generate power for the telephone system, apparently several rotors exploded due to creep, all the maths worked out but,...bang, put down to the analysis of the steel, seggregation creep and non metalic inclusions.
solution was to stick them under great big concrete covers.
moral is even if it satisfys all the numbers be careful of spinning energy storage, flywheels do have a habit of disintegrating, the numbers and maths can be found in loads of places but 'the theory of machines' black is as good as any.
all the best
mark:eek:

Evan
05-18-2008, 03:48 AM
While linear dimensions are only dialated in the direction of motion, time isn't a linear dimension and therefore must be uniformly dialated by a motion in any direction. If there aren't oddball affects from the acceleration of motion in a circle, I think it will work though my memory is much to hazy to remember caveats in the theory.
The applicable caveat is likely the Einstein-Penrose-Rosenburg rotation that shows time dilation is never observable from any frame of reference. The only instance when it may be observed is when the clocks occupy the same inertial frame after acceleration.

As for the clocks around the world experiment, what it measured is not just the effect of accelerating the clock in a moving airplane.


Hafele and Kiting experiment [9] strongly support the
existence of a Universal Inertial frame. In this experiment four cesium beam
atomic clocks were flown around the world twice, once eastward and once
westward to test the theory of relativity. The time difference between the flying
clock and the clock, which stayed on the earth's surface, was compared. The
flying clock lost 59 - + 10 nanosecond during the eastward direction and gained
273 - + 7 nanoseconds during the westward trip. If we analyze a airplane flight in
around the world trip it is seen that most of the time the airplane fly in uniform
velocity .In order to fly in circular, track the airplane seldom make an angular correction of
it's flying direction, this and the time the airplane change it's height
because of elevation forces are the only time the airplane experience radial
acceleration, which is negligible compare to the flight period.
The result can only be explained by taking into account the universal inertial
frame. The earth has a rotational velocity, in the eastward direction flight the
velocity of the plane is positively added to the earth velocity and increased the
velocity of the clock on the airplane relative to the universal inertial frame and the
clock lose time relative to the clock which stayed on the surface .In the westward
trip the airplane velocity added negatively to the earth velocity this decrease the
velocity of the clocks on plane relative to the universal inertial frame and the clocks
gained time relative to the clock which stayed on the surface .

Because of the correspondence between the rotational angular velocity of the Earth and the flying speed of the aircraft the actual measurement is of the Earth's velocity through space.

This paper, although poorly translated is very on topic.

http://arxiv.org/pdf/physics/0105008

Swarf&Sparks
05-18-2008, 01:00 PM
Hmm, maybe it's late, maybe I'm old, tired and stupid:
but that makes about as much sense as Schroedinger's cat.

Evan
05-18-2008, 04:13 PM
As does much of both quantum mechanics and relativity theory. Neither are intuitive.

It's like this:

Airplane flies east or west and stays on the same side of the Earth relative to the sun since it is flying at the approximately same velocity as the Earth is rotating. The result is that the velocity of the aircraft is adding or subtracting from the movement of the Earth through space. The problem with relativity is figuring out what something is relative to.

Do a search on the "slow moving clock problem" in relativity.

Swarf&Sparks
05-18-2008, 04:17 PM
But we won't know til we open the box :cool:

Fasttrack
05-20-2008, 11:57 PM
Whew ... I've been working 17 hour days and haven't had time to keep up with the board.


I originally approached the issue of the relative motion averaging to zero by considering the motion to be in one dimension only. Then the motion is that of a mechanical oscillator relative to some stationary point. If you assume that the two reference frames are equal and indistinguishable then one frame moves forward and stops, then the next frame moves forward and stops so the relative motion averages to zero. But we've made several assumptions.

I'm pretty sure there would be a dilation, but it will be unique to rotational motion. I haven't had time to actually work out the math on paper, but I think I can see what to do. I've also emailed one of my professors who will be teaching the course on general relativity so I'll see what he has to say on the matter.


... on an unrelated note:
Evan - I finally found the bit of "pipe" (I think cored round is probably a better term) that I've been looking for to make my rotary table. I also made a deposit on a new lathe so when I get that I'll be off to the races and I'll be able to put that bearing to good use!

Bruce Griffing
05-21-2008, 02:00 AM
As a person with some experience doing experiments in physics, I have a couple of suggestions. First, define very carefully your hypothesis and the mathematics that support it. An experiment is a test of a hypothesis. Don't start thinking about stress or how to build equipment until that part is complete. Before you start the next step, bounce your hypothesis and experimental approach off professors, graduate students and any one who will listen. You will learn a lot in this part of step one. Second, don't build anything you don't have to build. For example, if you can use an existing centrifuge - don't build something that emulates one. It is a waste of your time, despite the fact that it would be fun. You don't want to spend any of your time on something that does not lead to your desired result for one big reason - unanticipated effects or outcomes. For example, in you description of the experiment you planned, have you considered the physical effects of your "centrifuge" on the electronic components? Will capacitances change? Will that impact the result? Using a crystal? Any effects there? More often than not, some unanticipated effect will more than drown out the desired signal. So you want to leave most of your time available to sort out these effects. Most of the time spent in a successful experiment is spent after the initial setup is running. This time is used to make revisions to the experiment to solve problems that arise. Often several approaches are tried and abandoned before a successful approach is developed. Give yourself time for this.
Good luck!!

Fasttrack
05-21-2008, 09:31 PM
Thanks Bruce

I think I mentioned somewhere that there is a huge effect on the electronics. In fact, it will cause a greater change in frequency than any relativistic effects, but there is a well defined empirically derived equation for the effect of acceleration on crystal oscillators. Crystal oscillators were the basis for the atomic clocks that were flown around the world and are highly accurate (if you get good ones). The packages I'm looking at are a little pricey at 30 bucks a pop but they have a high level of precision - the error should be several order of magnitudes less than the effect from relativity.

Advanced lab is a pretty free form class and the professors encourage students to pick something and basically play with it. In the field of physics that I am interested in, most of the "experiments" are based on hunches or loose theories. Unfortunantly, experimentalists sometimes don't see eye to eye with theorists and go there own way on occassion.

Plus I've got all summer to fool around on my own time ;)

Evan
05-21-2008, 11:50 PM
I've been thinking some more on the problem. First of all, time dilation can only happen in respect of a different frame of reference. There is absolutely no way to measure time dilation that is occuring in your frame of reference. This makes the question "Is the object on the centrifuge in a different frame of reference?". Surprisingly, I don't think so.

If we observe the centrifuge from outside the radius of rotation then no matter where we observe it from the average velocity in respect of the observer equals zero. If we observe it from within the radius of rotation the average velocity still equals zero and a special case exists as well. If we observe from the center of rotation then the average velocity and the instantaneous velocity are zero. All the object in the centrifuge has then is relative angular velocity and that cannot cause time dilation since it isn't a vector quantity. In fact, if we sit at the center of rotation and rotate as well then there isn't even relative angular velocity yet in none of the observations have we changed our average velocity (we have not accelerated) relative to the object in the centrifuge.

The problem with a null result is that it proves nothing other than the fact that you failed to find a non-null result. The most a null result can do is set limits on a positive result.

Bruce Griffing
05-22-2008, 01:57 AM
The twin paradox occurs in the case of circular motion as well as linear motion. IOW, time dilation occurs when the acceleration is associated with orbital motion. Here is a great reference:

http://arxiv.org/PS_cache/gr-qc/pdf/0703/0703090v2.pdf

oldtiffie
05-22-2008, 04:23 AM
Lottsa hoop-related strees in these links:

http://en.wikipedia.org/wiki/Hoop_snake

http://members.optusnet.com.au/~blueprnt/myth.html

http://www.google.com.au/search?hl=en&q=hoop+snake&btnG=Google+Search&meta=

Evan
05-22-2008, 09:46 AM
Bruce,

That is avery interesting paper. However, I detect a problem, at least in the general presentation of the examples. It constantly makes statements that require time dilation be observed in one frame from another. This is in direct contradiction to a prediction made in the 1950s by Roger Penrose and independently by James Terrell.

It elegantly proves that the Lorenzt-Fitzgerald contraction cannot be observed from any frame of reference. By corollary, neither can time dilation. In order to make any measurement of time dilation the clocks used must be in the same frame of reference.


Can You See the Lorentz-Fitzgerald Contraction? Or: Penrose-Terrell Rotation (http://www.atomki.hu/fizmind/specrel/penrose.html)

I also have an uncomfortable feeling about the idea of a N-gon tending toward a circle in this instance. It reminds me of Xeno's Paradox. As n grows larger the acceleration to make the "course changes" diminishes in inverse proportion. It can then be argued that in the limit case of n=infinity=circle the acceleration required is the sum of an infinite amount of infinitely small accelerations = zero.

Also, in the appendix the author invokes changes in acceleration even though it was stated earlier that it makes no difference.


When the traveling twin accelerates at
each vertex of the polygon they will see the Earth clock at
that vertex advance more quickly during the acceleration.
These advances accumulate so that when the traveling
twin is again next to the Earth twin both will agree the
clock carried around the polygon has recorded less time
than the one held by the Earth twin. Exactly the same
thing happens in the limit as the polygon path becomes
a circular path

At the least it is a bad example because it contradicts earlier statements and also seems to invoke yet another paradox.

Interesting stuff though.

Bruce Griffing
05-22-2008, 11:22 AM
Evan-
There is another perspective on SR that makes this clearer (hopefully). In this perspective you travel through space-time at a constant velocity wrt a given frame. As a parallel, imagine that you are travelling north at 100mph. In one hour you move 100 miles north. If you change direction to NorthEast, your progress North in one hour is now diminished by the square root of 2. Back to space time. At rest, you make progress through the time dimension a distance of ct for a time t. But if your velocity through space increases in ANY direction, that velocity increase comes at the expense of your progress through the time dimension. So your clock slows down wrt the "fixed frame".
I have not read the Penrose paper, but my guess is that he is referring to something a little more esoteric when he says observable. There are legions of examples of observation of the time dilation effect. The classic case the is decay of muons incident on the atmosphere. Based on the known decay rate, muons should not get as far as they do. They travel farther than expected because their clocks run slower wrt our "fixed" frame. The paper I referenced also give examples of experimental confirmation.

Evan
05-22-2008, 03:10 PM
Certainly there are numerous examples of time dilation. However, they all depend on the observation being made in the observers frame of reference. Time dilation cannot be observed directly in another frame of reference. In other words, you can't somehow peek at the clock in another frame to see if it is running slower. It's values must be communicated to your frame which then makes it subject to the Penrose rotation as well as simultaneity.


So your clock slows down wrt the "fixed frame".

And that is a subject of much debate. The "fixed" frame is like the holy grail. Much sought after but never confirmed as found. Some ideas serve well such as using the distant stars but even that is only an approximation.

Any "though experiment" that invokes the possibility of somehow being able to observe time dilation in respect of two different frames is placing us as observers outside of both frames and into an impossible frame with a world line that intersects the other frames regardless of their relative velocities.

This is seen in the question about "what is the relative velocity of two photons traveling in opposite directions?". The answer is that they don't have a relative velocity since their world lines don't intersect.

It's all very interesting to me and I shall have to contemplate the implications of that paper you presented as it does raise questions that are new to me (and apparently to many others as well).

Thanks for that link.

Bruce Griffing
05-22-2008, 05:14 PM
I took a look at the Penrose paper and it has to do with the visual appearance of objects moving relative to the observer. It has nothing to do with detecting time dilation. Other methods, like radioactive decay, which provide their own clock, have been used to detect time dilation. Penrose does not say that it is impossible to measure time dilation, which is the subject of Fasttrack's proposal. I see no reason why in principle, for circular motion, that you cannot observe it directly by electronic means. However, I do see a great number of problems in doing a successful experiment - it won't be easy.

Evan
05-22-2008, 08:52 PM
It's the reason for the Penrose rotation that prevents one from observing time dilation in the accelerated frame. Let's say there is a pendulum swinging on the accelerated spacecraft. Time is slowed but the entire frame is contracted so it still seems to have the same period to an observer in the accelerated frame.

We try to observe it through a window in the spacecraft as it passes us. The pendulum is contracted so it has a shorter distance to swing. If we could see the contraction we would realize that the period, although the same, is for what appears to us a shorter swing meaning the time is dilated. But, because of the Penrose rotation all we see is the pendulum displaying the same period and the same degree of swing when we first measured it in our frame.

In the example given in the paper it's clear this is an issue of simultaneity which affects all possible observations, not just the geometry of reflected light.

An example of this in the "real" universe is an object falling into a black hole. If time dilation in the object's frame were observable then it would appear to come to a halt at the event horizon and never actually cross the horizon since time is dilated to infinity at the velocity of light.

Fasttrack
05-22-2008, 11:09 PM
Both Bruce and Evan are correct, in so far as it goes. For time dilation to be observed, the object must return to the initial frame of reference. That is the solution to the twin paradox; one twin never changes reference frames and does not expierence relativistic effects. The second twin leaves the first reference frame, hurtles around the universe and then returns to the original reference frame and suddenly the effects are realized.

I'll post more as it relates to my "experiment" soon.

Fasttrack
05-22-2008, 11:31 PM
This has turned out to be quite an interesting problem! The spacetime metric is distorted by the spinning disk - in fact it is distorted due to the stress and strain within the body, not the mass specifically.

The degree to which the spatial geometry or temporal geometry is changed I have not determined. If it has been determined, its mired somewhere in general relativity with many qualifiers and advanced mathmatics. I'll see what more I can dig up on it. The rotating disk was mentioned in Einstein's 1916 paper but for his purposes, he assumed that there was no temporal distortion, but gave no explicit reasons for this. According to Max Born, a spinning disk is a contradiction (relativistically anyway) since the circumfrence contracts but the radius does not. (i.e. it is a non-euclidean circle lying in euclidean space)

Bruce Griffing
05-22-2008, 11:36 PM
Fasttrack-
"For time dilation to be observed, the object must return to the initial frame of reference."

Not so. Consider atmospheric muon decay. The muon does not slow down until it decays - thus sending a signal to the observer. In the case of the flying atomic clocks, the clock was returned to land only for convenience - with proper equipment it could have been still flying. Consider Thomas precession - another example of the object not returning to the observers reference frame.
I am not sure why you make this statement. --??

Paul Alciatore
05-23-2008, 01:56 AM
special relativity?
why didn't you say so... :)
36000ipm is only about 35mph.

can you run your experiment in a car?

how long do you plan to run for?

-Tony

That's funny, I get 375 MPH. A bit of a different ball game as regards using a car. But I guess there a few out there. And finding a stretch of road may be just as big of a challenge.

dp
05-23-2008, 02:11 AM
36,000 inches per minute is 3000 feet per minute. 1 mile per minute is 5,280 feet per minute, or 60 mph. 36,000 inches per minute is 34.090 mph.

Evan
05-23-2008, 08:56 AM
Not so. Consider atmospheric muon decay. The muon does not slow down until it decays - thus sending a signal to the observer. In the case of the flying atomic clocks, the clock was returned to land only for convenience - with proper equipment it could have been still flying. Consider Thomas precession - another example of the object not returning to the observers reference frame.

In the case of the muon we don't even know it exists until it decays. That is when it enters our frame of reference. Prior to that it's world line does not intersect ours. We then can infer what it's lifetime must be to decay at that altitude but we cannot do so by direct measurement.

We cannot send a signal from the flying clocks for the purpose of comparison. It will be subject to the same relativistic effects as the clocks themselves making comparison impossible. It is for that exact reason that every year a portable atomic clock is flown here to Williams Lake in order to calibrate the triple atomic clock stack at the Loran C station. There is no way to send a signal from Boulder Colorado to Williams lake for that purpose. The two locations are in different frames of reference. This is all a matter of simultaneity at work.

I am considering the Thomas precession but am not sure what the applicability is to other aspects of relativity, yet.

Bruce Griffing
05-25-2008, 12:46 AM
Evan-
I have a high opinion of your work on many things - but in this case we just disagree. It is simply not necessary for the an object to return to the rest frame in order to observe time dilation. Consider the airplane already discussed. I do agree that it is much more practical to land to do comparison measurements. Think about it a different way. Do an Einstein style thought experiment. Imagine a flight that goes on for a very long time. The time differential would build up to a point where it the errors associated with transmission would not drown out the effect. So I stand by my statement that it is not necessary for the object to return to the rest frame to observe time dilation. I assure you that I am not speculating on this matter.

Evan
05-25-2008, 08:53 AM
I assure you that I am not speculating on this matter.
Neither am I. It isn't a matter of the size of the error or inaccuracies in transmission of a signal.

Here is an explanation of why it is impossible.

http://galileo.phys.virginia.edu/classes/252/time_dil.html

I should point out that while you can observe a difference in the clocks you have no way of determining which clock is correct or time dilated.

[more]

As for the LORAN station clocks, they are held to a maximum error of 20 nanoseconds per year. That's the amount of time it takes for light to travel about 20 feet. In order to calibrate those clocks the reference clock must then be within that distance to be considered as being in the same frame. In fact, the exact length of the cable used to connect the reference to the triple clocks matters since it also causes a delay in transmission.

The basic fact is that time is different for different locations, even if they are only a few feet apart. Since you can't be in two places at the !exact same time! there is no way to make a comparison of times in two different locations that is any more accurate that 2X the speed of light delay between those places even if there is no relative motion. If there is relative motion all that can be observed in reality is the doppler effect of any received signal. While this is proportional to the dilation amount it isn't an observation of the actual dilation effect.

Bruce Griffing
05-25-2008, 02:04 PM
Evan-
The reference you list is of course correct - for the case of linear motion. In that case, it is not possible to detect the time dilation without the moving clock turning around and coming back toward the fixed observer. It does not have to stop however, to observe the effects of time dilation if the clock does make a return pass. In short - your example is correct for the "fly by" case, but that is not the case being discussed. The fact that it is impossible for this case to detect time dilation does not prove the general case impossible.
As to your point on the speed of light - of course the speed of light introduces a delay. To the extent that the distance is known, this can be corrected. For real situations, like clock calibration, it is of course easier to avoid the correction by putting the clocks together. Again, this does not prove anything.

Evan
05-25-2008, 04:59 PM
As to your point on the speed of light - of course the speed of light introduces a delay. To the extent that the distance is known, this can be corrected. For real situations, like clock calibration, it is of course easier to avoid the correction by putting the clocks together. Again, this does not prove anything.

In the case of clocks stationed on the earth or flying around it the delay is unknown and cannot be accurately characterized. We can only set upper and lower bounds on it. The velocity of light is only constant in a vacuum. As soon as the electromagnetic wave must travel through the atmosphere the travel time becomes unpredictable. Not only does the density of the atmosphere vary but the path the radio signal takes varies as well. The only solution is to bring the clocks together which is why the clocks at the LORAN station are calibrated to another clock in the room. The same applies to a gravity well although that is at least easier to characterize. In fact, the gravity well effect completely overwhelms the time dilation due to velocity of the clocks on GPS satellites. They run slower because of their relative velocity but since they are further out of the Earth's gravity well the net result is that they run fast.

Even if the distance is known and we attempt to compensate we cannot distinguish the nature of the curvature of space as caused by gravity. We do not know and cannot know what the gravipotential difference is between two arbitrary points in space. These are not nitpicks as the discovery of numerous "Einstein rings" caused by gravitational lensing demonstrates.

Even in the case of a fly by as you suggest, the Penrose rotation prevents us from viewing the clock in the contracted frame. All we will see is the frame rotated rather than contracted. The same will apply to any signal with the equivalent of rotation being a doppler shift.

The rules of simultaneity clearly make it impossible to observe another frame of reference and see what someone in that frame sees. We will see something but what we see is unique to our frame of reference relative to the other and an absolute determination of what is visible in the other frame cannot be made.

No two locations in space/time share the same frame of reference.

Also, for any two frames of reference that have intersecting world lines there is a third frame that will see both of the first two as being identical even while both see each other as different.

Bruce Griffing
05-25-2008, 05:22 PM
Evan-
My point is that is is not necessary for the moving clock to stop to observe time dilation. Consider the following example -

A spaceship passes earth at high speed. As it passes on the outgoing leg, it transmits a time signal. It continues to travel at high speed but after a long flight, it turns around and comes back at the same speed. The same protocol is used to transmit a time signal as the ship passes earth (judged from the moving frame). In each case, the time signal will be received. The trip is planned so that roughly 100 years of time difference build up. Now, there is an error associated with the "drive by" SR effects, but it is tiny compared to 100 years. If the protocol is carried out correctly, that error will cancel between the first and last signals. But the cancellation is not important. The point is that time dilation is observable without the moving clock coming to rest. The case of orbital motion - which is the original point of this thread - is even more special in that the distance to the moving clock is constant. This creates lots of simplifications.

You mention gravitation. Acceleration, due to gravity or otherwise is the source of the time difference. It is the acceleration that makes the twins in the twin paradox different.

Evan
05-25-2008, 06:07 PM
You mention gravitation. Acceleration, due to gravity or otherwise is the source of the time difference. It is the acceleration that makes the twins in the twin paradox different.

What makes them different is what makes it a general relativity problem. The traveling twin occupies two different non inertial frames of reference, one going away and one coming back. As soon as we add a third frame of reference the situation changes since we have two frames to compare to a third. That still doesn't permit us to

A satellite orbiting the Earth is in an inertial frame of reference as it is falling through an equipotential path in curved space. It undergoes no acceleration as it orbits. A rotating mass on a stick, which is what this thread is about, is a non inertial frame of reference and and the signal sent from the mass will be equally red and blue shifted resulting in a null difference.

There is more but I have to pound some fence posts for a neighbor.

aostling
05-25-2008, 06:26 PM
Bruce,

While Evan is pounding posts (instead of posting them) may I ask what your background is? I've visited 252 of the 254 counties of Texas, including your Bell County. Temple struck me as a quiet place, fortunately bypassed by I-35. It has a nice courthouse square and church, a big VA complex, but no college of higher learning where I might expect you to be on the faculty. Are you retired there?

Bruce Griffing
05-25-2008, 07:37 PM
Alan-
I am recently retired here. I am trained as a physicist and had a career in technology development in the electronics industry. Temple is indeed a quiet town. I enjoy the warm weather at least 9-10 months a year. The other months are quite hot - thank God for air conditioning. I have a nice shop attached to the house and enjoy working there. I did work in the Phoenix area for about 16 months and owned a house in Chandler. I sold it about two years ago.

Bruce Griffing
05-25-2008, 08:12 PM
Evan-
On your point about orbital motion - A satellite does undergo acceleration because its vector velocity changes. But it is a special kind of acceleration in which the force is always perpendicular to the direction of motion. The net result is a changing velocity vector, but a constant velocity magnitude. It is the magnitude of the velocity that is traded against progress in the time dimension in SR, so there is time dilation (however small in real cases).
As to the problems involving general relativity, any problem involving acceleration will need GR to understand completely. But the case of an clock accelerating to a velocity, traveling at that velocity and returning can, in most cases, be delt with using SR. Most of the effect on the flying atomic clocks is due to SR. That part is just the velocity part discussed above.

Evan
05-25-2008, 10:27 PM
Here is a very interesting paper that describes what happens to the clocks in the "fly around the Earth" experiment. The result is the opposite of what would be expected in a simple case of SR time dilation and illustrates why we cannot determine what is actually happening in the accelerated frame from the non-accelerated frame.



As with the comparison of a clock brought into a gravitational field from far away to that of a clock constructed within the field, the result of equation (9) is the same one we get if we construct a clock in an inertial reference frame and then impart the circular motion to it. We can determine the rate of slowing due to (9) in the same manner we used for a clock placed in motion from its rest frame. This is done in equation (4), and we arrive at the result that the accumulated time on a clock undergoing circular motion is less than that accumulated on a stationary clock and is given by t’ = tg-1
A clock constructed in the lab frame, and then placed on the edge of a rotor, will slow down due to the acquired energy. When removed from the rotor and placed again in the lab frame, the clock will speed up to its original, inertial rate. Similarly, a clock constructed on the moving rotor would run at the same lower rate as the lab frame clock placed on the rotor. When removed from the rotor, both clocks will speed up by the same amount, reflecting the more inertial, lower energy environment. In the case of the west-bound flying clock, we are taking a clock that has been constructed on a rotor (the rotating earth), and then moved to a more inertial frame. Thus the clock speeds up as compared to an earth bound clock. We allow the west-bound clock to circumnavigate the globe until it is again at the same location on the earth as the earthbound clock. The west-bound clock will have accumulated more time than the earthbound clock. Being in a more inertial state, the clock rate speeds up as it approaches metaphysical time.
Imagine that we were to place a clock on a rocket, and launch it into space at the point of earth’s perihelion in its orbit around the sun. We keep powerful rockets pushing on this clock so that it remains at this point in space as the earth pulls away in its year-long journey. From the earth’s reference frame, this clock will have been accelerated mightily, and will have attained a tremendous velocity (30 km/sec) with respect to the earth. After a year has elapsed, and the earth is approaching perihelion and the space clock once again, we read the elapsed time on the space clock and compare it to one of our earth-bound clocks. We find that, even though the space clock was accelerated to a high velocity as measured with respect to us, it has recorded more time since we’ve been gone, not less as would be expected of a clock placed in motion.
Even after we adjust the time recorded on the earthbound clock for the slowing due to the earth’s rotation and the presence of the clock in the earth’s gravitational field, we still have accounted for less time than recorded on the space clock. The reason is that a clock stationary at the point of earth’s perihelion is in a more inertial reference frame than a similar clock in orbit around the sun. Thus this clock runs closer to metaphysical time, and, therefore, runs faster than the earthbound clock. The same would be true if we could set a clock in space and wait for the entire Milky Way to complete one revolution. As the solar system drags back around after millions of years, the space clock will have recorded much more time than a clock kept on the earth for the duration. Again, we have neglected or pre-corrected for all gravitational effects. The space clock, set free of the rotational grasp and effects of the Milky Way, will have been one step closer to fully inertial, metaphysical time.


http://renshaw.teleinc.com/papers/london1/london1.stm


It also deals with some of the more esoteric concepts such as abberation which is closely related to the Penrose rotation.



BTW, the paper obviously does support the concept that rotation does produce time dilation, but not necessarily in the expected sense.

Fasttrack
05-25-2008, 10:40 PM
Incidently, Penrose has some very fascinating books published. I've just started The Road to Reality - its a nice review of my theoretical class.

I've been pretty busy lately and have not been able to do much work on my idea or, indeed, follow this thread too closely. I'm headed to NM tomorrow to pick up some machinery that I purchased, but after that excitement is over hopefully I can get up to date on whats been going on ;)

Bruce Griffing
05-25-2008, 10:58 PM
Evan-
The paper suggests exactly what you would expect - that clocks moving in circular motion slow down. A fact that has been experimentally confirmed. Much of the discussion in that paper is related to the acceleration part of the process, which as discussed before, requires GR. The corrections for GR are small, but it is instructive to look at the acceleration process to understand the sign of the correction. The key question is this - does the acceleration make the object go faster WRT the "rest" frame or slower. This is important because as a practical matter, we do not really have a rest frame here on earth. If it goes faster, the clocks slow down. If if goes slower, the clocks speed up. But for the rotor experiment originally discussed, this does not matter. The rotor experiment can be treated as if we are at rest. Done correctly, the clocks on the moving rotor will slow down.