Rearset/drum brake leverage

[AgentX]

Hi everyone--

Working on a project with a Royal Enfield Bullet, 1977. Left brake, right shift.

I am making it a street-trackerish sort of build, so I want to move the chairlike seating position to something more truly mid-set.

Attached below is a photo of the brake side, as stock.

Would like to use a set of universal-style rearsets (pegs with concentric levers) at what is the current pivot point of the rear brake lever. Will need to fab a bracket for mounting the shift side in the same spot, but that should be no problem.

What I'm concerned about is how much leverage I'm going to get on the rear drum. Dealing with both a shorter brake lever and a shorter actuator. Can anyone assist with science and/or experience? Will this work out? (Edit: Waitaminit. A shorter actuator is going to give me MORE leverage, no?)

The Tarozzi rearsets I was looking at from Fastfromthepast.com have an especially short actuator.

Would like to order one of Durgam_K's folding sets, which are a bit taller coming off the peg, but he's out of stock at the moment, so I started looking at the Tarozzis. (need folders to clear kickstart lever on opposite side, so the Loaded Gun and most other universal sets are out.) Neither set, obviously, is going to match the big honking brake setup on there now.

Thanks and happy holidays!

 

[DrJ]

No universal rearset brake lever will have the same lever ratio as one for a stock drum brake. It's a trade off between looks and function. The Tarozzi lever will need around twice the pressure as the stock setup to apply the same braking force.

 

[AgentX]

Thanks for the reply! In my case, it's about the location of the controls more than the looks of a trick set of pegs. Universals just seemed the easiest way to go.

Any ideas on something I could do with the ergos I want that would give function closer to the original?

 

[CCRider]

AgentX

Let's assume you need a 20lb force (or whatever number you like) to stop the bike with your foot at the stock brake pedal. Mulyiply that 20lbs by the distance from the center of the pivot to the pedal pad. Take that number (the torque at the pivot) and divide it by the distance from the center of the pivot to the brake rod. That will give you the force on the brake rod. Now take that number and multiply it by the distance from the pivot to the brake rod attachment on your new rearsets. Finally divide that nuimber by the distance from the pivot to the toe pad on your new rearsets. The result is the amount of force you'll have to apply to the new rearset to equal the 20lbs on the stock brake. This assumes that the forces applied to the brake arms are at 90* to the arms which they're probably not, but it will get you close.

CC

 

[AgentX]

Awesome info, thanks a lot. Just what I needed.

Except I don't know the dimensions of the peg sets I am looking at--little info seems available online.

 

[scm]

AgentX,

don't make it too complicated - it's just about the ratio lever to actuator length.
If this ratio remains the same, let's say 4:1, the brake will act as before.

Best regards
Sven

 

[FunJimmy]

Any way you look at it, a fulcrum is just a fulcrum. If you want to use a shorter brake pedal then every fulcrum in the system has to be shortened by the same percentage.

What is missed in this discussion is the length of the drum brake actuator. It’s really just a simple mathematical formula. If you shorten the brake pedal by 50%, then the pull rod arms at both locations must also be shortened by 50% to retain the original final leverage ratio at the drum.

 

[CCRider]

Any way you look at it, a fulcrum is just a fulcrum. If you want to use a shorter brake pedal then every fulcrum in the system has to be shortened by the same percentage.

What is missed in this discussion is the length of the drum brake actuator. It’s really just a simple mathematical formula. If you shorten the brake pedal by 50%, then the pull rod arms at both locations must also be shortened by 50% to retain the original final leverage ratio at the drum.

As SCM says, if the ratios in the brake lever stay the same then the force on the brake rod will be the same and there's no need to change the length of the brake drum actuator lever.

CC

 

[FunJimmy]

As SCM says, if the ratios in the brake lever stay the same then the force on the brake rod will be the same and there's no need to change the length of the brake drum actuator lever.

CC

Note true!

All three fulcrums add up to the total mechanical advantage applied to the braking system. Excluding any one of these in the calculation will give an inaccurate conclusion. The brake actuator arm also determines the amount or rod travel required to engage the brakes and the shorter pedal will not accommodate the required travel unless the drum brake arm is taken into consideration.

 

[CCRider]

Note true!

Actually true for force and travel as long as the ratios remain constant.

Consider a hypothetical stock brake lever with a length of 10" between pivot and toe pad and 3" between pivot and brake rod vs a custom rearset lever with 5" and 1.5" dimensions respectively.

Assuming the brake drum lever has to travel 0.5", then the stock brake lever arm also travels 0.5" which at 3" results in about a 10* rotation of the brake lever which works out to about 1.7" of travel at the toe pad.

To get 0.5" of travel at the 1.5" custom brake lever arm it has to rotate about 20* which works out to about 1.7" of travel at the toe pad of this lever also.

So as long the new lever has the same ratio of arm lengths as the old lever, things will feel about the same.

CC

 

[FunJimmy]

Actually true for force and travel as long as the ratios remain constant.

The ratios are NOT constant if you leave out the brake actuator arm. Period!

Assuming the brake drum lever has to travel 0.5", then the stock brake lever arm also travels 0.5" which at 3" results in about a 10* rotation of the brake lever which works out to about 1.7" of travel at the toe pad.

To get 0.5" of travel at the 1.5" custom brake lever arm it has to rotate about 20* which works out to about 1.7" of travel at the toe pad of this lever also.

So as long the new lever has the same ratio of arm lengths as the old lever, things will feel about the same.

According to your own math, the difference between 10* rotation and 20* is double!
Does that sound constant? Do you know what a 1.7” of brake pedal travel feels like?
It feels like you’ve got no rear brakes. Especially on rear-sets where your foot is angles down.

 

[AgentX]

Non-tech guy thinking here:

If my rearsets mean my new, shorter brake actuator pulls the brake rod a shorter travel than it used to (by virtue of being a shorter lever), albeit with the same force, I'd need a shorter arm at the brake drum to accomodate that shorter distance of travel, no?

MD

 

[CCRider]

The ratios are NOT constant if you leave out the brake actuator arm. Period!

According to your own math, the difference between 10* rotation and 20* is double!
Does that sound constant? Do you know what a 1.7” of brake pedal travel feels like?
It feels like you’ve got no rear brakes. Especially on rear-sets where your foot is angles down.

The rotation of the lever is not constant, but I don't care about that, I only care about how much I've got to move my big toe and that IS constant.

Why does the drake drum arm care what's going on at the brake lever? All it sees is a force being applied to it through a brake rod.

My math just shows that both foot brake levers apply the same force and the same travel to that brake rod for the same amount of force and travel of my big toe. The numbers were arbitrarily chosen for easy math, so they won't reflect the actual travel of a properly adjusted brake system.

CC

 

[CCRider]

Non-tech guy thinking here:

If my rearsets mean my new, shorter brake actuator pulls the brake rod a shorter travel than it used to (by virtue of being a shorter lever), albeit with the same force, I'd need a shorter arm at the brake drum to accomodate that shorter distance of travel, no?

MD

If you've got the same force in your brake rod and you try to adjust for a different amount of travel by shortening your brake drum lever, you'll reduce the braking force at the drum.

I wouldn't let all this academic discussion bother you. If you're designing your own rearsets, it would be worth doing some math, but if you're using an off the shelf unit, I'd just install it and use the stock brake drum lever and be done with it. Hundreds of others have done the same with no problems.

CC

 

[FunJimmy]

The rotation of the lever is not constant, but I don't care about that, I only care about how much I've got to move my big toe and that IS constant.

Why does the drake drum arm care what's going on at the brake lever? All it sees is a force being applied to it through a brake rod.

My math just shows that both foot brake levers apply the same force and the same travel to that brake rod for the same amount of force and travel of my big toe. The numbers were arbitrarily chosen for easy math, so they won't reflect the actual travel of a properly adjusted brake system.

CC

You refuse to understand the physics and choose to look at only 66% of the equation and that is your prerogative, but to continue to argue that the applied forces and effect of the two scenarios are the same is wrong.

The only thing that I agree with you on, is the fact that others have successfully used universal rear-sets and adapted to the change, but it wouldn’t be difficult to shorten the brake actuator arm and use a clevis rod end to correct the ratios.

The question that AgentX asked was:

What I'm concerned about is how much leverage I'm going to get on the rear drum. Dealing with both a shorter brake lever and a shorter actuator. Can anyone assist with science and/or experience? Will this work out? (Edit: Waitaminit. A shorter actuator is going to give me MORE leverage, no?)

To that question, the answer is YES, but at the expense of lever travel.

 

[CCRider]

Perhaps I am missing something. If the brake drum arm requires a certain amount of force to be applied over a travel of so many fractions of an inch, please explain to me why it matters whats happening at the other end of the brake rod.

The brake rod is a simple tension member. All it does is transfer force along its length from one pinned connection to the other.

As long as the same tension is being applied to the brake rod over the same travel, it doesn't matter if its coming from the stock brake lever, a custom lever (with the same ratio of pedal arm to actuator arm) or from a string tied to an elephants tail for that matter.

Maybe I'm wrong. If so, could you explain to me why?

CC

 

[Durgam_K]

I dont think the force needed is as important as the degrees of rotation to distance traveled, ratio. By that I mean, Think of distance from center. A 4 inch rod pivots 10 degrees. the end of the rod moves a farther distance than the half way point of the rod. Effectively twice the distance. So, to get the same distance traveled at the half way point, you may have to move 20 degrees. You reverse the theory at the brake arm. Now we need to move the arm 10 degrees to stop the bike. How far does the end of the arm have to travel???

Measure the pedal from linkage to pivot, and same for the brake arm. That is the ratio to try and maintain. Toe pad to foot placement has little to do with brake actuaction, just feel.

Look at how far the actual brake arm moves at the drum, and how long the arm itself is.
Lets say that 1 inch of travel at the end of the arm, pivots 10 degrees. that 10 degrees will stop the bike. 12 degrees will lock it up. (just saying) Less than 10 degrees will do less obviously.

I think that makes sense. LOL

 

[teazer]

Or let's try that another way round.

Stock is say a 16" foot pedal with the rod say 2" up from the pivot point. Oversimplifying things a little, but if teh pedal travels down say 1" to get the bike to stop, the rod will move forward 1/8" and that will rotate the brake arm by a certain angle..

If the new lever is say 8 inch long and the rod arm is still say 2" from the pivot, the foot pedal goes down the same 1" and the lever arm rotates twice as far as before - 2/8". and the brake will come on much earlier but will require twice the foot pressure.

So in reality, the foot pedal will only go down about 1/2" to get the same rotation at the brake with a stock rear brake arm.

Because that gives us 2x the effort to apply the brake, we would probably shorten the lever arm for the rod as much as possible to gain back some of that lost leverage, but not so much that the pedal moves down too far.

It's a juggling trick like all engineering issues. We want less effort and less movement - not possible, so we compromise and try to get reasonable effort and reasonable rotational movement.

 

[scm]

Perhaps I am missing something.

Definitely not!

Best regards
Sven

 

[FunJimmy]

Or let's try that another way round.

Stock is say a 16" foot pedal with the rod say 2" up from the pivot point. Oversimplifying things a little, but if teh pedal travels down say 1" to get the bike to stop, the rod will move forward 1/8" and that will rotate the brake arm by a certain angle..

If the new lever is say 8 inch long and the rod arm is still say 2" from the pivot, the foot pedal goes down the same 1" and the lever arm rotates twice as far as before - 2/8". and the brake will come on much earlier but will require twice the foot pressure.

So in reality, the foot pedal will only go down about 1/2" to get the same rotation at the brake with a stock rear brake arm.

Because that gives us 2x the effort to apply the brake, we would probably shorten the lever arm for the rod as much as possible to gain back some of that lost leverage, but not so much that the pedal moves down too far.

It's a juggling trick like all engineering issues. We want less effort and less movement - not possible, so we compromise and try to get reasonable effort and reasonable rotational movement.

Why do you shorten the brake arm by the same percentage as the other arms?
To retain the same mechanical advantage as the stock configuration.
No compromise.