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Guido
07-21-2006, 01:54 AM
Given: Suspend exactly 1,000 ft. of 0.250 in. diameter, perfectly straight, stranded cable down a wellbore. At the top, pinch the cable between the faces of two flat faced wheels, each 7.6394 in. diameter. With no slippage of the cable with respect to the wheels, count exactly 500 revolutions of each wheel while pulling the 1,000 ft. of cable vertically, from the wellbore.

Question 1: Same scenario as given, except, instead of pulling the 1,000 ft. vertically from the wellbore, have the cable contact one wheel over a 90 degree arc of it's circumference and be pulled at a right angle to the centerline of the wellbore. Now, what measuring wheel diameter would be required for it to rotate exactly 500 times for the same 1,000 ft. of 0.250 in. cable pulled?

Question 2: Any super duper formulas for calculating measuring wheel diameters with regards to cable diameters and the arc degrees of measuring wheel contact? Our designs call for contact angles of from 80 to 90 degrees.

G

Evan
07-21-2006, 02:14 AM
?????

It makes no difference if it is wrapped 1 degree or 180 degrees. Why would it?

If you are measuring the length keep in mind that 1000 feet of 1/4" cable weighs 105 lbs. As it is pulled the suspended weight grows less and so does the cable stretch.

lynnl
07-21-2006, 11:05 AM
Cable stretch? ...hmm! What's the temperature at bottom of well? Top of well? Is the temp gradient constant?

How long has cable been hanging in well?

Boy this could get messy!

pcarpenter
07-21-2006, 11:14 AM
Yeah...and what Evan mentioned only takes into consideration *elastic* stretch. New cable will stretch quite a bit on a permananet one-shot basis. My ATV winch instructions prescribed stretching it before it was ever used, so that can be a solution to that problem.

So can use of some synthetic winching products that are out there now. They use Kevlar or something and are *very* lightweight and don't splinter etc. They are subject to abraision, though and do require a nice smoothly polished hawse type fairlead. They are also pricey, but are really cool stuff. If ever I need replace the cable on my winch, it will be with this stuff.

Here is one example:

http://www.rockstomper.com/catalog/recovery/ropes.htm

Mcgyver
07-21-2006, 12:07 PM
?????

It makes no difference if it is wrapped 1 degree or 180 degrees. Why would it?

If you are measuring the length keep in mind that 1000 feet of 1/4" cable weighs 105 lbs. As it is pulled the suspended weight grows less and so does the cable stretch.

I think what he's getting at is when its pulled straight up, the circumference of the wheel * # revs = exactly how much cable is moved, whereas when it passes over the circumference, the cable must change shape slightly, outside stretched, inside compressed- ie is the distanced moved based on the inside radius (the circumference of the wheel) or the circumference of a line somewhere in the middle of the cable - not sure what the answer is?

Guido is this a practical question, or one of those "it be a billionth longer because of more surface area exposed to sunlight" puzzles? :D

Evan
07-21-2006, 12:13 PM
The answer is that it makes no difference at all. The cable makes contact with the wheel at a tangent and leaves at a tangent which is a straight line. What happens in between doesn't matter. If the cable compresses on one side closest to the wheel it also decompresses the exact same amount before it breaks contact so the net effect is zero.

Think of it this way: The cable is pulled at exactly one foot per second. Regardless of how it contacts the wheel, wrapped or not, the circumference of the wheel must rotate at one foot per second. Time x speed = distance.

nheng
07-21-2006, 12:35 PM
Just went thru this recently and it does matter for high accuracy measurements. The only solution is to calibrate against a reference length that has been mechanically measured by hand. The reference length then needs to be run thru the measuring system under working conditions and a cal factor developed.

Evan
07-21-2006, 12:53 PM
A difference can only occur if there is slip. No slip was specified.

astroracer
07-21-2006, 01:19 PM
A difference can only occur if there is slip. No slip was specified.
Like Evan said, it will make no difference. His wheel diameter has an exactly 24" circumference. One revolution of the wheel (with no slip) will meter out exactly 2' of cable, whether there is a single point contact or he has it wrapped around the wheel twice, one rev equals two feet...
Mark

J Tiers
07-21-2006, 01:34 PM
The internals of the cable are moving against each other if it is "uncoiling" from a pulley vs being pulled straight.

The stresses involved extend past the last point of contact, and I would expect a measureable difference in the vicinity of the pulley if the cable is loaded.

For the ridiculously fussy files:
Not only that, but there will be some energy absorbed in rotating the pulley, friction of wire on pulley, internal wire friction of bending, and so forth. The only place that energy comes from is added pull on the cable, which means that coming off the re-directing pulley, it will be stretched slightly more than it is before that. So measuring at the pulley has an unknown error amount, depending on how much of the stretch affects the pulley rotation.

As to whether that will affect the measurement? I don't know.

There are lots of error sources besides that. The basic stretch of the loaded cable, which is length dependent, is a big error either way.

Probably the accuracy is best described as "close", but not ever exact. There is no way to account for everything.

I think I would go for the vertical cable, and "nipping" pulleys. You can get a decent calculated correction for cable stretch as a function of load and amount of cable out. You can calibrate that system more easily than any more complex system, and have more confidence that you are really correcting for known properties of the system..

nheng
07-21-2006, 01:35 PM
Due to compression caused by the wrap, the length payed out is less than the actual length (somewhere near the "core" ). What the magic number is for steel cable I don't know.

Guido shows his wheel diameter down to 0.0001" and uses the words "exactly 500".

John R
07-21-2006, 01:54 PM
when the cable touches the wheel at its tangent point it will not be bent, streatched, compressed or in any other way deformed and it will be moving 2 ft per rev. ..when it exits at the top tangent (or any tangent) it has to be moving at the same speed....if it were slower there would have to be wire pilling up somewhere and if it were faster the wire would have to be stretching

Evan
07-21-2006, 02:07 PM
If you pay out or reel in the cable at one foot per second it make no difference whatever what happens to the cable around the wheel, it still approaches and leaves the wheel at one foot per second. With no slip the wheel must have a radial velocity of one foot per second at the circumference.

If slip enters the picture then everything changes but that is a different can of worms and probably cannot be calculated. There may be some rules of thumb that have been empirically determined.

J Tiers
07-21-2006, 02:48 PM
when the cable touches the wheel at its tangent point it will not be bent, streatched, compressed or in any other way deformed and it will be moving 2 ft per rev. ..when it exits at the top tangent (or any tangent) it has to be moving at the same speed....if it were slower there would have to be wire pilling up somewhere and if it were faster the wire would have to be stretching

Not quite...

If there is any difference of tension across the pulley, there is a difference in stretch AND THERE IS SLIP. same thing that causes your flat belt to "talk to you" under a heavy cutting load.

In some special cases it may be true that a pulley NOT merely contacting a straight wire gets an accurate measurement....... but in most any case there is *some* difference, even if only due to the extra tension required to supply the lost energy of bending in the cable. You cannot escape that unless you don't bend the cable.

Given the choice of "nipping" wheels or one that has the cable bent around it, one CLEARLY has fewer variables, so why fight it?

BillH
07-21-2006, 04:25 PM
Just like a new bicycle, few hours using it, all the cables stretch, re-adjust, good for a year.

Guido
07-21-2006, 05:10 PM
For 12 or more years now, we've played with these scenarios, always ending with what we learned from an old time 'wireliner' fabricator/machinist. He figured the centerline length of the cable remains unchanged and most nearly represents the true length of the cable. Cable being pulled vertically and measured, with zero slippage, between two flat wheels of 2 ft. circumference will generate 500 revolutions of each wheel per 1,000 ft. length of smaller diameter cables. I believe everyone agrees the centerline length here, is equal to the length of the cable as measured on the north, south or whatever side of the vertical, straight cable.

When the cable is played over a flat surfaced measuring wheel, (in contact for 90 degrees of circumference), the centerline of the cable will remain one half the cable diameter away from the wheel surface. In effect the circumference now needing measured, (pitch?), is represented by the radius of the measuring wheel plus half the diameter of the cable. This means in our case, the radius of the wheel will be reduced by 0.125 in. resulting in the centerline of the cable being at 7.6394 in. from the wheel center, generating a theoretical centerline circumference of 24 inches/rev. for measuring purposes.

The cables we are measuring consist of two layers of "plow steel' wires wound opposite to each other. The core consists of an insulated, coax, stranded conductor, insulated and wrapped in one layer of thin tinfoil. The effects of temperature changes and cable stretch are very minimal under 1,000 ft. well logging depths and 60 pounds of tool weight. The cable has to 'squirm' as it rolls over the measuring wheel, ie. the outside 'stretches' and the inside 'compresses'. Over years of useage, the individual strands of a cable will show surface wear where they are in contact and movement with, each other. The OD of a cable will show wear with useage and can affect the accuracy of a measuring wheel.

The old wireliner machinist always wants to provide finished wheel diameters oversize by 0.03 to 0.05 in., put in service and accuracies observed. Sometimes we'll have him take off a 'layer or two of dust' so that at 1,000 ft. true depths, the depth totalizer will indicate between 999.0 and 1,001.0 ft.

G

Evan
07-21-2006, 05:47 PM
In effect the circumference now needing measured, (pitch?), is represented by the radius of the measuring wheel plus half the diameter of the cable. This means in our case, the radius of the wheel will be reduced by 0.125 in. resulting in the centerline of the cable being at 7.6394 in. from the wheel center, generating a theoretical centerline circumference of 24 inches/rev. for measuring purposes.

That is a fallacy. All of the cable moves together. As I said, with no slip a cable moving at one foot per second must turn the wheel at one foot per second. It cannot be otherwise.

For it to be otherwise the cable must be moving faster than the wheel since if the center of the cable is the only "true" measure of the speed of cable then the smaller diameter wheel and the outer surface of the cable in contact with the wheel would have to be moving slower. That obviously cannot happen without some significant slip. If there were no slip then the difference in relative motion between the center of the cable and the wheel surface would reach a maximum at 45 degrees of a 90 degree wrap. This would amount to a slowing that would offset the outer portion vs the center of a .250" cable wrapped 90 degrees on a 2' wheel by .050" in three inches or about half an inch per yard. Not going to happen. The cable must slip.

LES A W HARRIS
07-21-2006, 05:49 PM
This means in our case, the radius of the wheel will be reduced by 0.125 in. resulting in the centerline of the cable being at 7.6394 in. from the wheel center, generating a theoretical centerline circumference of 24 inches/rev. for measuring purposes.
G

Shouldn?t that be 3.8197? Radius?
C= Pi*D? And so forth.
Anyway, to minimize slip etc would wheels with a 0.125? radius groove in the center, with the bottom of the grove at the calculated dia be better?

J Tiers
07-21-2006, 10:38 PM
That is a fallacy. All of the cable moves together. As I said, with no slip a cable moving at one foot per second must turn the wheel at one foot per second. It cannot be otherwise.

For it to be otherwise the cable must be moving faster than the wheel since if the center of the cable is the only "true" measure of the speed of cable then the smaller diameter wheel and the outer surface of the cable in contact with the wheel would have to be moving slower.

They have different diameters, AND they are moving at different speeds. How could it be otherwise?

You are up against a mechanical conundrum, and you picked the wrong end.

The cable need not slip, AND the cable is not moving at the same speed throughout, AND the same amount of cable comes off the pulley per second across the cross-section!!!!!!!

IT IS ALL TRUE, even though it CANNOT BE true..... well, yes it can.

The deal is, however, that the cable becomes LONGER on the outside, and SHORTER on the inside.

It stretches on outside, and bunches on the inside, and, these bunchings and stretchings are exactly right to make the AVERAGE point move at the average speed. It comes out of the "equation" when it unwraps.....

Actually, what that says is simply that it is wrapped around a pulley....... no more than that....

The ANGULAR VELOCITY is constant.... and, taken with the radius to any point, "happens" to give a correct speed.

However, unfortunately, the cable thickness comes into the average radius. He is dead right about that. So the actual surface velocity at the wheel surface (flat wheel) is LESS than teh cable by the difference in radius.

However, YOU are correct in that the cable and pulley are traveling at the same speed where they are in contact.

But,due to the "bunching" that is effectively LESS than the speed at the center (average point) of the cable.....

In reality, there is also stretch....the source of the energy lost in friction etc, is tension. That forces the "off-coming" tension to be higher than the "wrapping-on" tension. So there will be some stretch and hence some possible error.

So what is so duifficult about the simple contact wheel on a stright vertical part of the cable?

It gives a good reading, and requires the least correction and the simplest correction.

Leigh
07-21-2006, 11:04 PM
And of course there's compression of the cable around the measuring wheel, which flattens the inner cable surface and the outer cable surface, thus moving the center of the cable closer to the wheel, introducing another error. This effect is really obvious if you try to measure tubing using this technique, but it will occur with stranded cables and even with solid wire, although the force required is much higher.

And as JT said, it would be easier and more accurate to measure using a device placed along a straight length of the cable. Then you don't worry about the radius of the direction-change wheel at all.

Scatterplot
07-21-2006, 11:10 PM
the centerline of the cable will remain one half the cable diameter away from the wheel surface.

Quoted for truth. I made a quick picture that might convert the unbelievers :)
http://img.photobucket.com/albums/v240/bertmcmahan/Pulley.jpg

Here we have a drawing simplified to not include temperature changes, weight of the cable, etc.

It shows what a "real world" pulley would look like. If the wire was of effectively zero diameter- like a thread or piece of small string/twine- this would be negligible. But a .250 wire will, like above mentioned, stretch on one side of the pulley and bunch up on the other side. In the pic, R1 is the radius of the pulley, and R2 is the radius from the centerline (the right place to measure) to the center of the pulley.

It often helps me to imagine things happening way afar off at extremes far beyond this question. We have to set limits sometimes- but we can obviously know that there will be no changes in behavior for any and all wire sizes we choose, so we can know that whatever function describes this setup, it will be continuous all the way, and never change into another eqation. Think of it in simple terms.

Say we have a five foot diameter pulley. It's just hovering in the air, remove all inconsequential facts from the imagination for now. If we were to tape a piece of thread onto the vertical edge of the pulley and rotate it around 180 degrees, then marked the edge, we would have just shy of 8 feet of string. That's easy to imagine- it's in the world of the theoretical, where cables have neither weight nor dimension other than length. Now lets' go to the other extreme- somewhere on the other side of the .250 mark in the actual problem. Say now we have a cord that is another 5 feet in diameter. Forgetting about how in the world it could flex this far, repeat the experiment. Tape it to the side of the pulley and rotate 180 degrees. Now what is the length of cord that is wrapped around the pulley? Clearly it is more than 8 feet! In effect, the Centerline of the pulley can be considered a theoretical thread again- but wrapped instead around a 7.5 foot pulley, allowing for it's own shape. This here is the question presented to us. How long is this new cable? Theoretically it is close to 12 feet long. Over a half a revolution, this is quite large- then again the diameter difference is quite large. Reduce the diameter difference from zero diameter to .250 diameter and we have a small change- then multiply it by 500. I can't answer your question, Guido, but I can attest to the fact that it is indeed a valid one to ask.

Evan
07-22-2006, 12:27 AM
Sorry guys. In the spirit of carrying things to extremes let us consider the limit case.

We have a cable of diameter n. It is infinitely compressible and stretchable when bent. Our wheel has an infinitely small diameter. The cable is "wrapped" 90 degrees over this "wheel", essentially a perfect 90 degree corner. If the cable is pulled at one foot per second how much cable passes the corner at the corner in one second?

This reminds me of Zeno's paradox and in fact it is a variation of it. With such an infinitely compressible cable we then find ourselves with a cable that has been compressed to zero length at the corner if we consider the center of the cable to be the portion that has the true velocity. Therefore it can have no velocity if it has no length. However, it still passes the corner. How can that be?

Any other size of wheel and cable is still subject to this same apparent paradox. Yet it isn't really a paradox.

edited to add the qualifier "when bent"

Scatterplot
07-22-2006, 01:48 AM
Sorry guys. In the spirit of carrying things to extremes let us consider the limit case.

We have a cable of diameter n. It is infinitely compressible and stretchable when bent. Our wheel has an infinitely small diameter. The cable is "wrapped" 90 degrees over this "wheel", essentially a perfect 90 degree corner. If the cable is pulled at one foot per second how much cable passes the corner at the corner in one second?

This reminds me of Zeno's paradox and in fact it is a variation of it. With such an infinitely compressible cable we then find ourselves with a cable that has been compressed to zero length at the corner if we consider the center of the cable to be the portion that has the true velocity. Therefore it can have no velocity if it has no length. However, it still passes the corner. How can that be?

Any other size of wheel and cable is still subject to this same apparent paradox. Yet it isn't really a paradox.

edited to add the qualifier "when bent"


Honestly not trying to be a jerk or anything... but I don't see your point. Could you explain a little more? Edit- kinda seeing it...

I would assume that since pi*d= the circumference (your "d" not the universal one", you rotate 90 degrees... that the answer to your question is simply pi*d*.25=distance (of unbent rope) moved past the corner. I don't understand how your statement supports your position. In anything rotating the very CENTER of the rotating part is not moving any- yet the part as a whole is spinning. Considering infinitesimal movements of the cord along the individual point of the infinitesimal-diameter pulley, I don't see how this changes anything at all. Could you perhaps clarify?

Edit again-
OK I'm gonna make another go here- we say that the velocity at that pointis zero. The equation relating angular velocity (say "w") and velocity (v) is v=rw. As r goes to zero (note: r is NOT an ever-decreasing pulley radius, it is the distance from a point where we need to know the velocity and the point of no motion) we see that velocity goes to zero. Then again the velocity hits zero for zero time as well, doesn't it? It's the same in a wheel of a car- the part of the wheel that touches the ground hits a velocity of zero when it's touching the ground and only when it's touching the ground- the car still moves, and we see that the velocity of the individual particle on the edge of the wheel is indeed zero when it touches the ground- the velocity curve would just touch the time axis, then move away. It can hit zero ONCE, but since it stays there for zero time... well we see the problems of using ACTUAL limits. My point above was to not use the actual limits imposed, but approaching the limits. We could run into similar problems saying you had a cord of infinite diameter.

Evan
07-22-2006, 02:19 AM
Limit cases are useful in examining problems. In this problem it is specified that there is no slip. If we allow the line to compress on the inner side in contact with the wheel then it must slip as I have said before. The slip occurs as the line compresses when it begins to bend around the wheel, stops slipping while the bend is constant and the slip reverses as it unbends when leaving the wheel. If the line compresses at it bends then the portion of the line in contact with the wheel must move relative to the surface of the wheel. That is slip.

If no slip against the wheel is allowed then the case I described holds. The line will not be allowed to compress, only stretch. The line of contact with the wheel will then be the part of the line that that has the average linear velocity of the line since if no slip is allowed then no compression may take place, only stretching.

If no slip is allowed then the line must maintain a one to one relationship with the surface of the wheel where it touches the line. The surface velocity of the wheel will be the linear velocity of the line.

Millman
07-22-2006, 02:38 AM
Scatterplot's drawing says it all. The centerline changes as the wire rope is drawn over the pulley, depending on the dia. of the rope and pulley. What is the exact profile, radius in the pulley groove? Noone knows. To do the formulae correctly, what is the exact composition of the wire rope? Noone knows. The formulae are fun...but do not apply, except for, if all is known. Evan, for your mill counterweight, I would use a BF pulley, as your rope will last longer. Just had to throw that in!

dicks42000
07-22-2006, 02:40 AM
Marv;
Here's a cool little practical problem....Evan, J. Tiers, & Scatterplot all seem to be heading towards a diferential equation to describe the wrapping & un-wrapping of our well logging cable.
Can you concoct a formula & program to solve this? I might even have to unearth the dusty calculus & differential equations texts....
Back to beer & building a stand for my notcher tonight....have fun.
Rick

Scatterplot
07-22-2006, 02:51 AM
Limit cases are useful in examining problems. In this problem it is specified that there is no slip. If we allow the line to compress on the inner side in contact with the wheel then it must slip as I have said before. The slip occurs as the line compresses when it begins to bend around the wheel, stops slipping while the bend is constant and the slip reverses as it unbends when leaving the wheel. If the line compresses at it bends then the portion of the line in contact with the wheel must move relative to the surface of the wheel. That is slip.

If no slip against the wheel is allowed then the case I described holds. The line will not be allowed to compress, only stretch. The line of contact with the wheel will then be the part of the line that that has the average linear velocity of the line since if no slip is allowed then no compression may take place, only stretching.

If no slip is allowed then the line must maintain a one to one relationship with the surface of the wheel where it touches the line. The surface velocity of the wheel will be the linear velocity of the line.

Ah! I get what you're saying now. I think the point of discussion here is whether or not that slip will mean anything in terms of overall motion of the cord. I would say no- probably not. I might have to move back from the theoretical for a minute though- perhaps that instead of a slip that occurs, it is simply a "bunching," i.e. the entire "bottom" side of the cord does Not make contact with the wheel, but instead makes sort of (in profile) a sine-wave looking shape? Like imagine a scissor-lift type of device, laid on it's side- a bunch of X's- XXXXXXXXXX Each point of contact between X's is hinged, and the cross of the X is hinged. The whole shape could change it's length on each side overall without having a total contact along the bottom surface.

Evan
07-22-2006, 02:51 AM
I should amend my previous statement slightly. In the no slip case compression is permitted in the radial direction of the line but not in the linear direction. In other words, the line may flatten but the inner radius may not compress in the linear direction. To the degree that the line is flattened linear stretching and compression are eliminated.

Millman
07-22-2006, 02:57 AM
SLIP seems to be the word of the day. When you design something, a certain amount of slip, stretching, modulus, is included. That's why all the formulae in the world aren't worth the paper they're printed on. They are only basic "Starting Parameters".

Evan
07-22-2006, 04:12 AM
perhaps that instead of a slip that occurs, it is simply a "bunching,"

The line is under tension at all times. If it were to somehow bunch as it wrapped onto the wheel then it would have to stay that way until unwrapping. At that point the constant tension would pull out the bunches. Bunching is no different than compression. Since the friction of the line against the wheel asymptotically approaches zero as the line spools off at a tangent the line will slip against the wheel before it loses contact.

The only way that zero slip can be achieved is by keying the line to the wheel, in essence a form of gear or like a cog belt. If you study sheaves that have been used for wire line for a long period you will see that the line wears the sheave in a characteristic pattern that matches the strands of the line. For that to occur it must be slipping.

Millman
07-22-2006, 04:18 AM
[[zero slip can be achieved is by keying the line to the wheel, in essence a form of gear or like a cog belt.]] Even that has SLIP and sliding pressures on a miniscule scale. We're not going to achieve perfection here, so that proves my point that the formulae are only a starting point. Any round or flat object is never being pulled at the exact dimensions according to it's structure and surfaces. There always has to be an anchoring point of reference, either on the inside or outside, or a combination of all. That anchoring point of contact is the determining factor of the strength or weakness. of any and ALL CONNECTIONS. Like I said, they're just a starting point.

Evan
07-22-2006, 04:36 AM
That is also my point. The original statement of the problem stipulated no slip. I was pointing out what must happen if no slip were the reality. However, slip must occur. Because of the friction of the line to the wheel it is too simplistic to take the axial center of the line as the effective diameter of the wheel. That would work if the line were bent to the same radius while free of contact but when in contact the line will be prevented from compressing to a degree by friction with the wheel as well as being flattened by radial compression caused by tension on the outer radius. Both of these actions will move the effective radius closer to the surface of the wheel than the normal center of the line.

How much it moves will depend on many factors including the size and strength of the line, friction against the sheave, load on the line, velocity of the line, temperature of the line, the type of line, the material of the line and a few others that I can't think of just now.

Simply using the center of the line as an approximation will get you in the ball park but if accuracy of 1 part in a thousand is required I don't expect that the approximation will be good enough.

Forrest Addy
07-22-2006, 06:18 AM
The problem is indeterminant if you're looking for better than two significant accuarcy. The people who make elevators use proximity devices at each floor because of stretch, slip, thermal expansion, and other variables. There's no holding the car within 1/4" as fills or empties therefore the it's dynamically positioned.

If the measuring wheel takes any part of a wrap you have to figure in the wire's neutral axis into the wheel radius. If the wire simply runs past it then the old pi x D formula works.

I used to work as a sailing yacht outfit. The owner would get out his cheap transit and measure the chain plate and fitting eyes from the water line, measure the mast and spreaders then order shrounds and rigging sets from a specialist supply house. There was always +/- 3/4" variation thanks to custom cinstruction but Gary's shrowd sets always fit right on the money: all the turnbuckles finished about 1/3 taken up. Stop and work out the problem remembering there is mast rake and assymetry to crank into the figures.

Anyway, you won't be able to run out 1000 feet of wire and get precisely 1000 feet on the counter no matter how accurately the measuring wheel is made unless you calibrate the system under different loads and make up a chart.