Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

**foglefogle****Member**- Registered: 2013-11-01
- Posts: 5

Hi.

I have a problem designing a mechanism intended to deliver constant output torque. One end of a tension spring is located at any of seven locations around a circular arc. The other end of the spring is attached to the start of a curved cheek. As the spring is relocated to more distant points, it wraps around the cheek, thereby lengthening the spring. The spring tension thus increases, but the requirement is that the output torque about a defined point below the cheek should be a constant. I'm trying to determine the shape of the curved cheek to achieve this.

The following link explains in detail and, most importantly, includes a drawing. Explaining (and understanding) the mechanism without the drawing is near impossible and I can't see how to post a drawing directly here.

docs.google.com/viewer?a=v&pid=sites&sr … ZWZlMzZlOQ

Added by administrator.

Can anyone define the shape of the curved cheek, please? I'll eventually need to show how the calculations are done, if you could include those, please.

Many thanks for any help.

Offline

**foglefogle****Member**- Registered: 2013-11-01
- Posts: 5

What's the prize for posing a question nobody can solve?

Offline

**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi;

There is no prize that I know of else I would be rich.

Perhaps if you showed your work and your thoughts so far. I might be able to help if you rephrased the problem mathematically. What equations do you want solved? What methods do you want to see that solves them? What DE's have you formed and need solved?

Also, Matlab and MathcCad are designed for structural questions when you need to play with a design you know little about. Have you tried them?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

Offline

**foglefogle****Member**- Registered: 2013-11-01
- Posts: 5

Many thanks for the reply, **bobbym**.

It seems to me you're asking me to phrase the question in 'pure mathematics' terms (rather than the physical terms of my first post which requires the solver to create the applied mathematics). If I had already expressed the problem in mathematics terms, I'd have identified the required approach and solved the problem myself. The *required approach* is what I'm asking for. This is a request for *applied* mathematical skills and some very basic engineering skills, not pure mathematics skills to merely solve existing equations.

The following suggests the obvious preliminaries to the applied mathematics approach (albeit at risk of funnelling any investigators along too narrow a mental path):

The diagram is still at: http://docs.google.com/viewer?a=v&pid=sites&srcid=ZGVmYXVsdGRvbWFpbnxmb2dsZXdlYnNpdGV8Z3g6N2I3Mjk2OGM1ZWZlMzZlOQ

Let the free spring length (no tension being developed) be x

When the spring is at point 1, its length is, say, x1. Therefore, its extension is (x1-x) and its tension is k(x1-x), where k is a constant reflecting the 'stiffness' of the spring.

When the spring is at point 2, its length is, say, x2. Therefore, its extension is (x2-x) and its tension is k(x2-x)

When the spring is at point n, its length is, say, xn. Therefore, its extension is (xn-x) and its tension is k(xn-x)

# x1 = distance from point 1 to point A. The cheek is designed such that the spring is tangential to the cheek curve at A, but it doesn't wrap around any of the cheek when end B is at point 1

x2 = distance from point 2 to the corresponding point of tangency with the curve of the 'cheek', plus the length of the cheek curve between the point of tangency with the curve of the 'cheek' and point A. The spring is now wrapping around part of the cheek, which causes it to extend more than if there was no cheek.

xn = distance from point n to the corresponding point of tangency with the curve of the 'cheek', plus the length of the cheek curve between the point of tangency with the curve of the 'cheek' and point A. Unless n=1 (see #), the spring will always be wrapping around part of the cheek, which causes it to extend more than if there was no cheek.

When the spring is at point 1, the torque developed about point P is k(x1-x)t1, where t1 is the perpendicular distance from the line of action of the spring to point P.

When the spring is at point 2, the torque developed about point P is k(x2-x)t2, where t2 is the perpendicular distance from the line of action of the spring to point P.

When the spring is at point n, the torque developed about point P is k(xn-x)tn, where tn is the perpendicular distance from the line of action of the spring to point P.

For the given example, the requirement is that k(xn-x)tn is a constant for n = 1 to 7 (7 points of spring attachment).

The difficulty is that the length of the cheek curve between point A and the point of tangency of the spring with the curve of the 'cheek' is a function of the shape of the curve and the perpendicular distance from the line of action of the spring to point P is also a function of the shape of the curve. This is something of a 'chicken and egg' problem, which is what makes it so challenging.

At further risk of funnelling any investigators along too narrow a mental path, I **have** solved this graphically using 'CAD iteration' (for want of a better description), starting with a circular cheek curve, centred at P

*This produces torque about P which is **not** constant for points 1 to 7.

I then reduced the torque arms for points 2 to 7 by the difference between the torque at each of those points and the torque at point 1. This completely ignored the effect of the altered cheek curve upon spring extension (as the spring wraps around that curve).

**This produces torque about P which is **not** constant for points 1 to 7, but the deviation from constant is not as severe as the first attempt, above*.

I then reduced the torque arms for points 2 to 7 by the difference between the torque at each of those points and the torque at point 1. This again ignored the effect of the altered cheek curve upon spring extension (as the spring wraps around that curve).

This produces torque about P which is **not** constant for points 1 to 7, but the deviation from constant is not as severe as the second attempt, above**.

The process was repeated over and over, until the deviation from constant torque was so small, it could be ignored (constant torque to ten decimal places was achieved, if I recall).

The 'CAD iteration' method, above, is appallingly crude and very, very, *very* time consuming. I have many of these cheek profiles to design, for different numbers and locations of spring end points (n), spring rates (k), spring free lengths (x) and spring 'start' lengths (x1).

I am looking for an alternative, elegant and speedy mathematical method. All I need is a definition of the shape of the cheek curve and the method used to derive it.

Use any method you wish to define the shape of the curve, e.g. polar coordinates from P, vertical and horizontal coordinates from P etc etc.

Hope this helps.

*Last edited by foglefogle (2013-11-13 20:04:04)*

Offline

**anonimnystefy****Real Member**- From: Harlan's World
- Registered: 2011-05-23
- Posts: 16,037

Hi

Do we pick the position of the point A once we find the surface?

Currently, I am looking at the way of determining which tangent from a point will yeald minimum distance. It seems very difficult.

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

The knowledge of some things as a function of age is a delta function.

Offline

**foglefogle****Member**- Registered: 2013-11-01
- Posts: 5

Hi **anonimnystefy**

Yes, you can choose your own point A, provided you define where it is. The only condition is that the end of the spring attached to A is tangential to the cheek curve at point A, when the other end of the spring (labelled B in the drawing in the link) is at point 1.

By the way, the same goes for the radius of the imaginary arc centered at P, passing through points 1 to 7. Choose your own arc radius.

Although not essential, it might help if you stick fairly close to the layout and proportions of the drawing shown in the link. That way, we'll all be on a similar wavelength further down the road.

Offline

**anonimnystefy****Real Member**- From: Harlan's World
- Registered: 2011-05-23
- Posts: 16,037

Oh, so the spring is tangential to the surface! As I first got it, the spring was allowed to curve around the cheek and then went on straight.

Thanks for the additional clarification!

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

The knowledge of some things as a function of age is a delta function.

Offline

**foglefogle****Member**- Registered: 2013-11-01
- Posts: 5

**anonimnystefy**

At risk of overkill, I'll describe the spring/cheek arrangement and a few other things in fresh terms, just to be absolutely sure we're on the same course:

**(1) -** **When end B of the spring is at point 1**, the entire spring is straight and tangential to the cheek curve at A.

If you think about it from the mechanical point of view, there's no point having any wrap-around of the cheek when the spring is in this position, since such a portion of wrap-around would never unwind - it would be permanently wrapped around an extension of the cheek curve, beyond A to the right. Therefore, such a cheek extension and any spring wrap-around of that extension may as well not exist at all (as the drawing confirms by not illustrating any extension of the cheek curve).

In terms of solving the problem (IMHO), the benefit of such an arrangement is that the torque developed when end B of the spring is at point 1 is easily expressed. That then gives you the required, target torque to aim for when the spring is in the other six positions.

**(2) - ****When end B of the spring is at points 2, 3, 4, 5, 6, or 7**, the other, fine wire end of the spring is straight for a portion, then curves around the cheek for a portion until it reaches its attachment point at A. The straight portion is tangential to the cheek curve. The length of the straight portion and the extent of the wrap-around depends upon the position of end B and the shape of the curve. The curve shape for constant torque about P is what we need to determine. Regardless of the position of B or the shape of the cheek, the length of the straight portion plus the wrap-around length must be a constant (it may be made of fine wire, but we can safely assume it doesn't stretch at all).

**(3) - **The torque arm, t, is measured from P along a perpendicular to the straight line of action of the spring. It is * not* measured from the point of tangency of the fine wire portion of the spring with the cheek.

**P.S. - **The statement in the link *'The tension in the spring is proportional to its total length, multiplied by a constant, k.'* should more correctly read *'The tension in the spring is proportional to its extension, multiplied by a constant, k.'* This doesn't affect any proposed analyses, but we may as well be strictly consistent with the earlier text.

.

*Last edited by foglefogle (2013-11-14 19:37:04)*

Offline

Pages: **1**