Cantilever/Centerpull Brake Visual Calculator
 

I have written a small javascript program that serves as a visual aid to understand the power generated by cantilever brakes. You can see the effect different cantilever geometries have on mechanical advantage and you can adjust all setup variables independently.

You can see here:
http://www.circleacycles.com/cantilevers/

with background analysis here:

http://www.circleacycles.com/cantile...i-geometry.pdf

Thoughts and suggestions are appreciated.

Regards,

Benno

Awesome! Thanks for this.

This is a very cool program.

But I think I see a problem. I may be missing something, but because the cable acts only in tension, MA is proportional to sin beta. I don't know where the "inversely proportion to sin gamma" comes from -- it seems like it would require some kind of rigid linkage for this to be true.

Sin beta is the force perpendicular to the effective arm length, which is why you maximize MA by setting up the yoke length so that the straddle cable is perpendicular to the arm.

Am I misunderstanding something?

Does the attached sketch help?
Attachment 132484

Very cool program, I'm bookmarking it. It does only show the MA at the single point of contact though. I think it would be better if for a given yoke height it would consider the MA through the brake's range of motion so you could get a sense of how much pad clearance you get and what happens as the brake pads wear down.

A cable can do all kinds of bends and curves and the tension stays the same (less friction losses). Take a look here if you don't believe me. The angle gamma does not matter. The cable tension in your first diagram is incorrect; it will be the same as shown in your second diagram. Actually the only difference between the two diagrams is that gamma is fixed. If the first diagram were correct, the same calculation would apply in the second.

What does matter is the angle at which the cable exterts force on the arm of the cantilever.

Looking at the first part of my sketch, when gamma = 0, there is no tension in the vertical cable. This is not true in the second part of the sketch. You are correct that in the second part of my sketch it doesn't matter what the angle is and the tension in the vertical cable is always 1 lb.

Perhaps it would have been better if I had drawn the fixed rings as fixed pulleys. I think this would have been more clear.

Gamma = 0 is a physical impossibility if there is any tension in the vertical cable. With the smallest angle gamma physically possible, the tension in the vertical cable is still 1 lb. at incipient motion of the weights.

I think that we agree about how to analyze the second part of my sketch, and that the force on the yoke is 1 lb.

Regarding the 1st diagram: Would you agree with this vector analysis?:

Attachment 132490

No I would not agree. You are making it much too complicated. If you have a 0.5 lb weight hanging on a cable and it goes around a curve, there is 0.5 lb tension in the cable on on both sides of the curve. It is that simple.

And actually, if gamma is not constrained as in the second diagram, it will always be 45 degreesif the weights are equal. Always 90 degrees with just one weight, right? Add another equal weight and it will be 45 degrees. The only way it will be any other angle is if you constrain it, as in your second figure.

Not sure I follow your second paragraph, but I ask you to consider this:

Imagine there is no vertical center cable. Gamma will be zero, right?

Now attach a yoke to the center and slowly pull up on the yoke, like applying a brake. I can pull up with any force I desire, do you agree?

If I pull up with a small force, gamma will be small; if I pull with a larger force, gamma will be larger, no?

Gamma is 90 degrees (vertical) if you only have one weight -- the unconstrained cable will align with the weight.

And if there is no vertical center cable, the weights will fall until gamma is 90 and the weight is held by the lower cables If you then added a vertical cable and started pulling, gamma would increase until it got to 45 degrees and not get any greater becasuse the horizontal components of the cable tension is only in balance at 45. Okay, not 45, but a constant angle depending on how far apart the weights are.

Our disagreement about your example seems to be whether the lower cable tensions are added as vector sums or arithmetic sums at the vertical cable. You say vector in first figure, arithmetic in second, though I don't see the difference. I say arithmetic in both. Even if you were correct, the angle of the cable at the cantilever arm directly effects the mechanical advantage of the brake, which you do not seem to be taking into account in your model.

Let's just disagree, then.

Yes, I would like to be able to do something like this. If you read the second half of the analysis here, I talk about MA as a function of angle, for the very reasons you mention. Haven't had time to think about what the best way to do this as a calculator would be, though.

There's no "agree to disagree" in physics. This is verifiable. I am not totally confident about the vector sum vs arithmetic addition thing at the cable. Make a physical model and prove me wrong. The physical model would be pretty easy.

I am certain, however, about the sin beta thing. A force applied parallel to a lever arm does not do any leveraging. If you do not take this into account your program produces incorrect results.

For a gut check, consider a set of cantilevers with horizontal arms (I have some). As you make the yoke shorter, the cable pull is more and more parallel to the arms, and MA decreases. If the yoke cable could become horizontal, MA would be zero and you would not even squeeze the arms together.

And if the forces in the lower cables do indeed add as vectors, you would want a yoke as long as possible and more nearly parallel to the vertical cable, so that those forces don't cancel, right? But your model says the opposite.

I don't need to make a model or prove you wrong. I've tried to be as clear as I can. There's obviously something that we're seeing differently and I am unable to bridge the gap.

I take into account sin beta. It is part of the main equation, which reads (on page two):

MA = (1/sin(gamma)) * sin (beta) * PA/DO

Does this disagreement you guys are having have any real world implications? Do you agree on what happens when one raises or lowers the straddle cable yoke? Do you agree on the difference in wide profile brakes like tektro cr720 or paul neo retro vs a low profile brake like the paul touring or cane creek or whatever?

p.s. fwiw, there is a brake made by a touring bike company that pulls from the bottom of the brake arm (below the pads) so that it has an angle over 90 degrees.

D'oh! I don't know how I missed the sin beta part. It is very clear in your write up. I apologize for my obtuseness.

I am still not sure about the 1/sin gamma, though.

And yes, pacificaslim, it does have some real-world significance. If you don't have the 1/ sin gamma term, MA is maximum where the straddle cable is perpendicular to the cantilever arm, as conventional wisdom says. However, this model says that MA is always maximum at minimum yoke length for arm angles < 90 degrees; MA is constant at any yoke length for arm angle = 90, and is maximum at maximum yoke lengths for arm angle > 90. If you followed this model, you would set up your brakes differently than conventions and manufacturers recommendations.

Yes it does. One of the main points that I make in the analysis is that yoke height/straddle cable angle is the most important factor in determining mechanical advantage.

Sluggo makes the point correctly that changing the yoke height changes the angle at which the cable hits the arms, and thus changes the effective arm length of the brake and the MA.

Additionally, and I think this is where he and I diverge, I assert that there is an independent effect on the MA of the brake caused by the yoke angle (gamma). One of the consequences of this is that I assert that

MA is always greatest (for better or worse) with the lowest possible yoke,

regardless of whether lowering the yoke causes the cable to hit the brake arm more obliquely. The only exception to this is for brakes with canti angles > 90 degrees.

Others have spoken about this, I merely attempted to put all the effects in a single equation. See here:

http://www.sheldonbrown.com/cantilev...tml#mechanical

and scroll down a little ways to where he discusses yoke height.

I don't think that SB really proves this here, he merely asserts it, but I was trying to show the mechanics behind this in the sketches above.

Reworked sketch with detail
 

I redrew the sketch showing the two differing conditions and provide a close up detail of the yoke, as I think my previous sketch was open to interpretation as to exactly what physical objects were being represented.

Page 2 shows the detail as well as a vector analysis for each side.

I ask: what is the force pulling up on the cable on the left side?

Attachment 132680Attachment 132679

It should be obvious that at least this is true; when arm angle = 90, the ends of the arms do not move in or out as the arms move, so that pulling the cable 1 mm moves the arm up 1 mm no matter the yoke angle.

(it may be easier to determine MA in terms of "how far does the output move when I incermentally move the input" rather than "how much force comes out when I put a certain force in." Both methods have to give the same answer but I find the first to be easier to think about when I'm drawing diagrams.)

Benno, you rock! Totally. I'm a computer dude, and I'm more impressed with your programming skill than your mechanical engineering skill. I can't assess the latter as well.

One more comment, for now, to show that I did not just abandon the discussion.

I have reached the conclusion that the original analysis is pretty much correct. I apologize again for being slow on the uptake.

The model still operates in a way that is contrary to my expectations, and I am not entirely convinced that there is not something wrong. My next few days are very busy so I will not have much to mull, but I need to go through some more details in order to convince myself that the model is correct. This will probably just show that my intuition is incorrect. I will check back in in a few days.

And again, it is a cool program.

Not obvious to me. The end of the cantilever arm only moves 1 mm with 1 mm of cable motion if the cable is pulling at a perpendicular (thus the sin beta term). If the arms are horizontal and the straddle cable is also horizontal, in the rigid idealized system you can't have 1 mm of cable motion because the force is parallel to the cantilever arm.

Thanks Sluggo. Your thoughts are exactly what I was looking for in seeking out criticism.

Yes. When the cantilever arm is at 90 deg and the yoke height is zero, MA will be zero. In fact, MA will be zero whenever beta is zero.

The equation accounts for this: MA = 1/sin(gamma) * sin(beta) * PA/DO

I think the best way, for me, to think about MA not changing with yoke height for a cantilever arm of 90 degrees (except when beta = 0), is that in this condition gamma = beta. Thus, the equation becomes MA = PA/DO. In other words, the anchor angle and the straddle cable angle cancel each other out, leaving just the ratio between the anchor arm and the pad (or resistance) arm.