[lecepe]

I already searched and didn't find anything useful.
I know that it isolates braking forces completely from suspension if it's designed well. I've heard that a thousand times, but how is it designed well? It has to be a paralelogram i think, but how does the caliper need to move? Is it supposed to move paralel to the mainframe trhoughout the travel like the rear axle moves on the yeti dh9?

Thanks
Leonardo

 

[Echo]

It couples the caliper to the front triangle.

The reason it's necessary with single pivot designs is, when the brakes are applied, it basically locks out the suspension because the caliper pulls the swingarm.

If it's coupled to the front triangle then that pulling force is neutralized, allowing the suspension to work.

I'm not sure how the other characteristics come into play though, such as angle of the floating arm in relation to axle path.

 

[rbx]

take for example a single pivot bike...when you brake the contac patch between the rear tire and the ground create a force along the ground multiplied by the distance=leverage to the main pivot that will cause compression(makes the whole suspension rotate)...now with a floating caliper that pivot become "virtual" somewhere in front of the bike and at infinite space so the the ground doesnt have any leverage to act upon the suspension..

 

[MikeOK]

See this link:

Bogus brake lockout

It makes sense to me, what do you guys think?

 

[Echo]

Originally posted by MikeOK
See this link:

Bogus brake lockout

It makes sense to me, what do you guys think?

That's a cute way to look at it from a lab, but anyone who's ridden a bike knows otherwise.

I'm quite certain that there are several hundred people in the industry who could tear that report apart from a technical and physics standpoint, if it were worth their time.

 

[Westy]

Originally posted by MikeOK
See this link:

Bogus brake lockout

It makes sense to me, what do you guys think?

I read this last week and found it very intriguing. I was not sure if I believed so I sat down and did a bunch of calculations.

I have not had to this kind of work for years so I could be completely wrong but I think my conclusions are fairly accurate. I found that in general braking on a single pivot bike would not lock out the suspension but cause some extra compression. But the weight transfer from braking could counter act this. I think that you could come up with an ideal pivot location that is completely neutral under braking where the weight shift and braking forces cancel each other out. But that would totally go out the window if you used the front brake, not to mention that that pivot location would cause crappy pedaling.

A floating brake will help to reduce these forces and allow a pivot to be designed more around pedaling than braking. I think that suspension design is more of an art than a science or the perfect design could be easily calculated.

 

[MMcG]

This is interesting stuff....can you guys provide some examples of squishies that employ the floating brake technology?

I'd be very interested in what these frames/bikes are.

Thanks!

 

[MikeOK]

Originally posted by MMcG
....can you guys provide some examples of squishies...

LOL! now you're talking my language!

Seriously though, I've always found suspension on bikes interesting. When I was racing motocross I used to go to the track on practice days and tune my suspension all day. I agree with Westy, "suspension design is more of an art than a science". I usually tinker until everything just "feels" right.

From experience I have found that you can analyze all you want, then go out and try to put what you thought you understood into practice and it just doesn't work. In other words, I think I always understand this simple concept of suspension until I try to use it. Then I usually nearly endo on my first long jump heheh...

 

[MikeOK]

Originally posted by Westy
A floating brake will help to reduce these forces and allow a pivot to be designed more around pedaling than braking.

I went back and read the article more thoroughly. I have given it more thought, too, and although I am not a physicist I still can not see how a floating brake can have any effect on the action of the suspension on a single pivot point design. It seems to me that the shifting of the rider's weight would be the main effect, or?

 

[Westy]

Originally posted by MikeOK
I went back and read the article more thoroughly. I have given it more thought, too, and although I am not a physicist I still can not see how a floating brake can have any effect on the action of the suspension on a single pivot point design. It seems to me that the shifting of the rider's weight would be the main effect, or?

My understanding of the floating brake goes like this. When rear brakes are applied on a single pivot the brakes impart a torque onto the swingarm and pulls back on the axle. The floating brake negates the effect of the torque on the swingarm, but it will actually increase the force on the rear axel. On a bike where the axel is at the same height as the pivot all forces are cancelled out. But when the pivot is higher than the rear axel this rearward force causes some compression of the swingarm. It is much easier if you compare a pivot at the same height of the axel with a very high pivot design. At this point though the forces involved will not matter much when you consider the effects of weight transfer, especially when the front brake is applied.

It is very interesting when you look at the effects of the front brake on the fork. Since the front suspension acts linearly the front brake is already pretty much isolated, depending on head angles. But you can use a four bar mechanism to use front brake feedback to reduce brake dive. Some Moto GP bikes use this and it was kind of the idea behind the old E-Fork design. I don't think it has been used on a DH bike before, probably because of the amount of travel involved would require a large brake rotor.

My brain hurts, I'm getting a beer.

 

[Steve from JH]

RBX has it about right and Westy has it partly right.

You have to look at what's happening down at the rear contact patch. The trajectory of that point is rotational around the pivot. Draw a line from the pivot to the contact patch and then take a perpendicular. That perpendicular is the instantaneous trajectory.

On a monopivot the line down to the contact patch is quite steep and so the trajectory has a lot of rearward as well as an upward direction to it. That means that the braking force, acting horizontally back along the ground, pushes the wheel up and back. It causes compression.

The compression counters the weight shift and so it reduces the amount that the frame pitches forward from weight shift to the front. But the compression also messes up the way your suspension handles bumps. The wheel will tend to skid and chatter.

Adding a floating link creates a four bar linkage that only functions when the brake is on. The pivot of that linkage is at the intersection of the extended swingarm and link lines. If the link is parallel, it's off at infinity.

Now when you calculate the trajectory of the contact patch it will be less rearward and more vertical, maybe even completely vertical. So you reduce the compression or eliminate it completely.

Now your suspension will react freely to bumps, just like you were coasting, but there's nothing much to counter the extension from weight shift. So the rear shock will extend and the frame will pitch forward more and that affects the steering. Which is better is up to you.

Another benefit of floating brakes links is supposed to be a kind of anti-lock effect. The caliper is pushed backwards around the rotor during compression, which produces more braking power. The opposite happens during a rebound. So you get more stopping power on the front side of a bump and less on the back. That's good because you have more traction during the compression and less during the rebound.

I should say that Ken Sasaki, who wrote the bit about bogus claims, doesn't believe any of this. I disagree with him on this and on a lot of other stuff.

 

[rbx]

steve---i understand how to calculate braking effect about the main pivot or I.C with a level ground..but help me clear some things if you would be so kind

1.when the tire hits a rock the face of that rock becomes your contact patch and your force is the "normal" in relation to the face of the rock..now you calculate the normal multiplied by the 90degree distance to the mainpivot in order to obtain your torque?

2.also you said a vertical path has almost no effect on suspension
is it because the force has no more pivot point(parralel links)so its has no real leverage?

3,just curious which parts of K.S works you did not agree with?

 

[Steve from JH]

Originally posted by rbx
steve---i understand how to calculate braking effect about the main pivot or I.C with a level ground..but help me clear some things if you would be so kind

1.when the tire hits a rock the face of that rock becomes your contact patch and your force is the "normal" in relation to the face of the rock..now you calculate the normal multiplied by the 90degree distance to the mainpivot in order to obtain your torque?

2.also you said a vertical path has almost no effect on suspension
is it because the force has no more pivot point(parralel links)so its has no real leverage?

3,just curious which parts of K.S works you did not agree with?

1. Yes, if by the "normal" you mean the perpendicular distance from the new ground line, tangent to the bump contact point, to the pivot point.

2. There are two ways to look at it. If the trajectory of the contact patch is vertical and the ground line is horizontal and the force acts along the ground line, well, no force has an effect at 90 degrees to its direction of action. Or you could say that a level parallel linkage creates the same effect as an infinite level swingarm with a pivot at the "end". There would be no brake torque if the pivot point could lie on the ground line, and parallel lines can be said to meet at infinity. So that infinite pivot lies on the ground line.

3. Besides the disagreement I mentioned about the claim for an anti-locking effect with floating links, our disagreements seem to center on how the ground force in relation to the contact patch is treated. KS (and DW and Ohio) want to translate the ground force up to the axle. I want to keep the force acting at the contact patch. I think of the contact patch as basically rotating around the pivot as though there were a rigid connection between the two. They do not.

 

[rbx]

thanks for clearing that up

its been buging me for a long time...

which books do you suggest that deserve a good read?

 

[ohio]

Originally posted by Steve from JH

Another benefit of floating brakes links is supposed to be a kind of anti-lock effect. The caliper is pushed backwards....

Steve, I agree with everything you wrote except for this part. This only occurs if you DO NOT have a perfect parallelogram in your 4-bar system, so that your IC is actually BEHIND the rear axle. This is the way the old GT LTS/STS performed at certain points in it's travel, and Brake Therapy has attempted to use this method at times. In my opinion it is less than ideal, as it amplifies the extension effects of force transfer ("weight" transfer... same thing).

Also, our disagreement in terms of the role and behavior of the contact patch is not all that great. Under locked braking, we are in complete agreement. Under dynamic braking, my opinions diverge from yours, but even then our results are not vastly different.

RBX and Steve, one last thing to keep in mind in regards to the discussion about brake induced torque about the main pivot is that while you are both correct, the condition where the pivot is inline with the tangent of the contact patch (zero braking induced torque) only occurs about 13"-14" into the travel, even with the lowest pivot points. Obviously this is different when not on level ground, but have fun calculating actual contact patch tangent, which is an function of bump geometry, moment of inertia, speed, tire geometry, and tire pressure.... and a dynamic function of wheel rotation rate. I, um, just guesstimate.