DDWFTTW - Tests.

[ynot]

Try an extreme case like a round coffee tin with the plastic cap on both ends or two large wheels with a thin axle between them.




The angle comes from the rate that the string unwinds so it doesn't matter where the string goes after that as long as it doesn't upset the ballance. In the case of just the spool there is a simple mechanical construct that John pointed out.

You are absolutely correct! 45 degrees is a delusional load of crap created by not testing at large enough ratio differences. My apologies.

ETA - And an incorrect intuition.

 

[John Freestone]

You are absolutely correct! 45 degrees is a delusional load of crap created by not testing at large enough ratio differences. My apologies.

ETA - And an incorrect intuition.

No problem, ynot. It's great you get stuck in and do some experimentation.

Dan, you may have missed my last post. I found that I had put sin^-1 instead of cos^-1, although I translated that for clarity and got the one I meant "arc-cosine" (acs). But you have the arcsin. I'd like to sort out which it is, and I want to learn your method anyway, so would you check it and give some more steps/diagram or whatever? I might've got some of my trig wrong.

Here's my working:

P.S. Is it still considered good practise these days to use that cos^-1 notation? I think I remember being taught that because the acs, asn, sin are so easy to mix up.

 

[Michael C]

Dan, you may have missed my last post. I found that I had put sin^-1 instead of cos^-1, although I translated that for clarity and got the one I meant "arc-cosine" (acs). But you have the arcsin. I'd like to sort out which it is, and I want to learn your method anyway, so would you check it and give some more steps/diagram or whatever? I might've got some of my trig wrong.

I read through those posts a bit too fast yesterday. I think you are right: according to my calculations, the correct formula is

"angle a = arccos(tr/sr)".

I don't have time to do a complete explanation of my method now; maybe I'll manage that tomorrow.

 

[Dan O.]

Dan, you may have missed my last post. I found that I had put sin^-1 instead of cos^-1, although I translated that for clarity and got the one I meant "arc-cosine" (acs). But you have the arcsin. I'd like to sort out which it is, and I want to learn your method anyway, so would you check it and give some more steps/diagram or whatever? I might've got some of my trig wrong.

You have it right. I checked my results by assuming a horizontal pull where the coefficient is 1.0 = sin(0)




(adding extra wheels to show that it is the angle of the pulled string and not the angle where the string meets the spool)

This can be solved by either looking at forces (force pulling cart forward = horizontal component of pull on string).

F_forward = F_pull • rt / rs
F_horizontal = F_pull • cos(a)


It can also be solved by looking at the change in length of the string given a change in position of the cart and solving for dl/dp = 0. (which will have to wait for my next post)

 

[John Freestone]

Wow. I don't quite grok that, but I can see how it should make sense! It's interesting to see the pull angle separated from the winding, like ynot's. It's the force forward = F_pull * rt/rs that I don't see, and then how you put them together to get the 'stall'. I can see that if Fforward = Fhorizontal, then the other bits give the final equation. That suggests that this condition is what I need to see is the solution, but I don't. It looks like Fforward should equal zero, because that's what defines the stall, unless I'm misunderstanding Fforward. Could you explain it a bit more sometime?

Glad to get the asn,acs question sorted. Looks like we both had a little nap there!

I look forward to seeing your method, too, Michael. I think I'd better go and do the empirical bit and prove it to myself. I'm a cotton-reel-come-hither virgin.

 

[Dan O.]

It's the force forward = F_pull * rt/rs that I don't see, and then how you put them together to get the 'stall'. I can see that if Fforward = Fhorizontal, then the other bits give the final equation. That suggests that this condition is what I need to see is the solution, but I don't. It looks like Fforward should equal zero, because that's what defines the stall, unless I'm misunderstanding Fforward. Could you explain it a bit more sometime?

Perhaps I skipped a bunch of steps. I was transfering the pull of the string to a torque on the main wheel (torque = force x distance) so we get T = F_pull • rt (the cross product is just a multiply because the force and distance are at right angles) and then convert this torque to the tangential force of the spool on the road F_forward = T/rs. But also there are a bunch of other force vectors involved that I just ignored because they all cancel out.

At each of the smaller wheels there is a pull force in the direction of the string (both ways). The torque on those wheels is balanced because the string on each side is pulling with the same force at the same radius but in opposite directions. Each segment of string between two wheels is also pulling on the cart body through the axle of the wheel but since the force is the same at both ends and in opposite directions, these forces also cancel out. The only forces on the cart that we are left with are the horizontal and vertical components of the string where it meets the first wheel and the torque on the main spool which translates into a forward pull through traction where the spool contacts the surface.

We are also making a simplifying assumption that the cart is heavy enough that the pull on the string isn't going to lift any part of it off the surface so all the vertical force vectors can simply be ignored.


ETA: Forgot to add that it is when F_forward (the force pulling the cart away) equals F_horizontal (the force pulling the cart back), the forces acting on the cart are balanced so the cart won't move. When one of these forces is greater than the other, the cart moves in that direction.

 

[John Freestone]

Thank you so much Dan, that's brilliant.

Perhaps I skipped a bunch of steps. I was transfering the pull of the string to a torque on the main wheel (torque = force x distance) so we get T = F_pull • rt (the cross product is just a multiply because the force and distance are at right angles) and then convert this torque to the tangential force of the spool on the road F_forward = T/rs. But also there are a bunch of other force vectors involved that I just ignored because they all cancel out.

Mostly stuff I don't know about, but I could guess the whole step made sense as one. The force trying to roll the wheel is the ratio of those radii between where we're pulling on it and its circumference. Since it's not a deformable object and we're ignoring inertia, etc., it wouldn't matter at what angle we applied the force to a wheel, it tends to roll one way or the other, so it's just the ratio of the radii. I've never worked with torque equations, but clearly they aren't too difficult in this case!

At each of the smaller wheels there is a pull force in the direction of the string (both ways). The torque on those wheels is balanced because the string on each side is pulling with the same force at the same radius but in opposite directions. Each segment of string between two wheels is also pulling on the cart body through the axle of the wheel but since the force is the same at both ends and in opposite directions, these forces also cancel out. The only forces on the cart that we are left with are the horizontal and vertical components of the string where it meets the first wheel and the torque on the main spool which translates into a forward pull through traction where the spool contacts the surface.

Yes I kind of intuited that as one, but it's useful to see how you can analyse it.

We are also making a simplifying assumption that the cart is heavy enough that the pull on the string isn't going to lift any part of it off the surface so all the vertical force vectors can simply be ignored.

Yes.

ETA: Forgot to add that it is when F_forward (the force pulling the cart away) equals F_horizontal (the force pulling the cart back), the forces acting on the cart are balanced so the cart won't move. When one of these forces is greater than the other, the cart moves in that direction.

I saw that this must be the missing step.

In fact, I see now that my confusion arose because it wasn't completely clear to me that the string can only cause F_forward to be to the left, no matter what angle its tangent is at within the machine. Your method of isolating the angle from the rotation makes it clear that there is no other effect at the wheel other than the ratio of the radii. To get F_forward to force the wheel to the right, we have to imagine that thread radius getting smaller and smaller until it passes the centre and pulls the other way. All we're left with for the opposing force is F_horizontal a horizontal vector component of the external pull force, and again it is simplified by separating the two effects (rolling and pulling), because a string can't push, so we know that these two must oppose each other.

That's been very instructive for me, Dan. Cheers.

Ynot - I've just realised that your Testing thread has got very theoretical. Hope you don't mind. You started it!

 

[ynot]

Ynot - I've just realised that your Testing thread has got very theoretical. Hope you don't mind. You started it!

It has been a great learning exercise so thanks to Dan O and yourself for your efforts. My layman’s method is to extend the line of the string past the spool to a point on the circumference of the wheel. If this point is beyond the vertical (from the string side) it will roll up the string (backwards). If at the vertical it will stall (won’t roll) and if on the string side from the vertical it will roll off the string (forwards). Don’t think that explains what I mean very well so here’s a pic . . .

As I said earlier, this doesn’t fully apply to a paddlewheel in water because the paddlewheel is partially submerged and doesn’t run just on the surface of the water.

 

[ynot]

I put my TT and cart together again to show a visiting friend and made a video with the TT running at a slower speed for those that want to compare calculations at different speeds. I have also replaced the “large” tether arm with a thin bar. Are there any other tests anyone wants done while it’s together again?

http://www.youtube.com/watch?v=5wijrV-1pHw

 

[Dan O.]

ynot
Location: New Zealand

I should have looked at that before. Of course your cart's going to appear to be working when it spins that way. You're on the other frigg'n side of the world!!



So, when are you going to mount the solid disk on an axle so we can see more of the ball rolling on a spinning disk?

 

[Michael C]

I put my TT and cart together again to show a visiting friend and made a video with the TT running at a slower speed for those that want to compare calculations at different speeds. I have also replaced the “large” tether arm with a thin bar. Are there any other tests anyone wants done while it’s together again?

http://www.youtube.com/watch?v=5wijrV-1pHw

The test already shows the cart starting from a stationary position and accelerating past wind speed. What more do the doubters want? Well, it's obvious that those who think the tether arm somehow makes the cart go forward (Humber!) won't be impressed, but perhaps you could do something for the "kinetic energy" camp. How about running exactly the same test at the same speed, but leave it going longer? Leave it going for several more minutes, so people can see that a terminal velocity is attained and held: in your video I can't work out if the cart is still accelerating at the end, or if it has already reached its terminal velocity.

Maybe for a longer test you could do a voice-over, or at least replace the nerve-grating motor sound with some suitable music?

 

[tsig]

The test already shows the cart starting from a stationary position and accelerating past wind speed. What more do the doubters want? Well, it's obvious that those who think the tether arm somehow makes the cart go forward (Humber!) won't be impressed, but perhaps you could do something for the "kinetic energy" camp. How about running exactly the same test at the same speed, but leave it going longer? Leave it going for several more minutes, so people can see that a terminal velocity is attained and held: in your video I can't work out if the cart is still accelerating at the end, or if it has already reached its terminal velocity.

Maybe for a longer test you could do a voice-over, or at least replace the nerve-grating motor sound with some suitable music?

Take it outside and run it in the wind. Why is that so hard for you to do?

 

[spork]

Take it outside and run it in the wind. Why is that so hard for you to do?

Build one and you'll see. Why is that so hard for you to do?

By the way, I'll be the first in line to tell you the thousands of things that are wrong with your demonstration when you do.

 

[ynot]

ynot
Location: New Zealand

I should have looked at that before. Of course your cart's going to appear to be working when it spins that way. You're on the other frigg'n side of the world!!



So, when are you going to mount the solid disk on an axle so we can see more of the ball rolling on a spinning disk?

I did the “marble” test just to show how low friction, rolling kinetic enery could cause a thing to almost hover on a moving surface for some time (ball bearing was even better). At that stage I was concerned that the cart might be conserving kinetic energy and gradually slowing down. As my cart has run for over 10 minutes on the TT without losing any speed this concern has been answered (for me). (it doesn’t). The TT/cart designs I’m using were only meant to be test models but they work well enough (for me) that I don’t have to build better models.

 

[mender]

The longer I have to wait for the right conditions to test my cart outside, the more respect I have for Jack Goodman. I missed my one good opportunity by about 15 minutes and about 30 degrees of wind direction.

 

[ynot]

The test already shows the cart starting from a stationary position and accelerating past wind speed. What more do the doubters want? Well, it's obvious that those who think the tether arm somehow makes the cart go forward (Humber!) won't be impressed, but perhaps you could do something for the "kinetic energy" camp. How about running exactly the same test at the same speed, but leave it going longer? Leave it going for several more minutes, so people can see that a terminal velocity is attained and held: in your video I can't work out if the cart is still accelerating at the end, or if it has already reached its terminal velocity.

It takes ages to upload videos to Youtube so I keep them as short as possible. I let them run to beyond where I believe the cart has reached terminal speed. I’ve tested that terminal speed is constant for over 10 minutes. Those that say I could be lying could also say that a video showing this could be faked.

Maybe for a longer test you could do a voice-over, or at least replace the nerve-grating motor sound with some suitable music?

Or you could simply turn your sound off.

 

[ynot]

Here’s a video of a crude, smaller, tiered TT to change the relative speed. It was put together very quickly so “never mind the quality, feel the width”. It shows you don’t need to build a big TT to observe the effect.

http://www.youtube.com/watch?v=DWcjmQ8Vz88

(don't forget to turn your sound off ;-)

 

[ynot]

A smaller cart wheel (5.5cm dia). Sorry about the background glare and the now unbalanced TT (stupid tiered demo).

http://www.youtube.com/watch?v=VOlCTCrJjx8

 

[tsig]

Build one and you'll see. Why is that so hard for you to do?

By the way, I'll be the first in line to tell you the thousands of things that are wrong with your demonstration when you do.

You are the one making the claim not me. Why do you shift the burden of proof. Could it be that your cart doesn't work in the wind?

 

[mender]

tsig, it just ain't that easy to have the right wind for testing. Going downstairs and turning on the treadmill to get a consistent air/ground speed difference is.

I put in my wind order at least a month again. So far the best is a wind that started the cart moving slowly along the ground from a standstill. Kinda hard to get decent test conditions up here with all the ice and snow too.