"Brake jack" - an explanation.

[WolfCreekPsycho]

I would have thought there would also be an addtional force not in your picture.... i have added the Yellow and Blue arrows.

Using the rear wheel to stop the bike (halt momentum) means that the rider and rest of the bike will want to keep moving forward. However that cannot take place as the bike and rider are all attached to the rear wheel and swing arm. What can happen is that the force of rider and front of bike will try to actuate on the pivot point between the rider and the item doing the braking, causeing force through the swingarm pivot. You have another pivot point in this equation and that is the front wheel on the ground. When you apply the rear brake it attempts to turn the frame holding the rear wheel in the same direction as the tyre is rotating ie towards the ground (shown by blue arrow). As the front wheel on the ground prevents the swingarm and frame from rotating around it has to put that force through the swingarm linkage causing it to flex.


Just my 2 cents in a great thead

 

[S.]

Hey S.

For calculating braking i use the force applied on the rear wheel and friction of tire patch contact.
So basically i get a resulatante force which is usualy above 45 degre and up toward the cyclist.
Im intersted in this mongoose by analysing brake forces i come to the conclusion that this bike should be pretty neutral under braking.


What do you think

green arrow is force appiled on the rear wheel, blue arrow is the frictional force and red is the resultante of these two force:

Not really sure where you're getting that vertical force from, unless you're trying to calculate the CHANGE in the normal reaction force at the tyre. If that's the case, you're trying to perform a quantitative analysis using qualitative techniques (ie get a specific number as an answer, using only conceptual understanding. This really can't be done, it's not accurate at all. If you want to be able to calculate just how neutral a bike is (and keeping in mind that "neutral" is not always optimal) then you need to take into account a whole lot of factors, such as position of the centre of mass (using cartesian coordinates, ie not just the height), the shock rates front and rear, the wheelbase, the influence of varying geometry under weight shift, the gradient of the slope the bike is on, etc. As I mentioned about the neutral/optimal thing, you need to also work out whether you want to analyse the bike relative to a net input (ie including rider weight shift forwards and the fact that rear suspension inherently extends here), or whether you want to assume that whatever changes occur SOLELY from weight shift (ie not considering other compressive/extensive forces acting on the suspension under brakes) as the benchmark, and concentrate on the forces/moments directly applied to the bike by the braking system. Either method is valid, but you have to be aware of what the difference in outcomes means.

What you also need to consider is that on that Mongoose, although the AXLE moves relative to the front triangle like a singlepivot, the BB (and thus the rider/CoM) also move relative to the front triangle.

I would have thought there would also be an addtional force not in your picture.... i have added the Yellow and Blue arrows.

Using the rear wheel to stop the bike (halt momentum) means that the rider and rest of the bike will want to keep moving forward. However that cannot take place as the bike and rider are all attached to the rear wheel and swing arm. What can happen is that the force of rider and front of bike will try to actuate on the pivot point between the rider and the item doing the braking, causeing force through the swingarm pivot. You have another pivot point in this equation and that is the front wheel on the ground. When you apply the rear brake it attempts to turn the frame holding the rear wheel in the same direction as the tyre is rotating ie towards the ground (shown by blue arrow). As the front wheel on the ground prevents the swingarm and frame from rotating around it has to put that force through the swingarm linkage causing it to flex.


Just my 2 cents in a great thead

Whilst there is a moment applied directly to the rear wheel (by means of the brake caliper/axle couple), if you're going to calculate that on its own (rather than implicitly, from the tyre's contact patch) then you have to resolve the horizontal force on the tyre contact patch as though it was acting through the axle. Otherwise you effectively have that moment on the swingarm twice - once as force x wheel radius, and a second time as a couple moment about the axle. NFI where you've pulled that yellow arrow from though - there is no force, resultant or otherwise, acting like that. I get the feeling you may be forgetting that this is essentially a dynamic analysis, and that reaction forces can be caused by acceleration/deceleration of masses as well as static loads.

 

[sygote]

Instant centre theory:
Notably, the bike on which Fabien Barel won the 2004 world DH championships on had a brake linkage designed specifically to increase the level of (pro-)squat far beyond what normal bikes generate. Riding the production version of this bike, you can feel a huge tendency for the rear end to dive when the rear brake is applied. Given that no owners of those bikes seem to have any problem with the extreme brake setup, one might logically assume that it’s not actually that bad, and that other bikes with considerably less brake induced squat can hardly be any worse off, and thus are perfectly fine to ride – although not necessarily as comfortable as they could be..[/b]

this is not true there are to points where the floating rear brake can be attached and the top one which is what i ride with does dot squat but when attched to the lower point the bike squates excessively. sorry jus thought i would point that out

 

[S.]

this is not true there are to points where the floating rear brake can be attached and the top one which is what i ride with does dot squat but when attched to the lower point the bike squates excessively. sorry jus thought i would point that out

Uh, so then it WOULD be true. He runs the floater in the lower point... it's because of him that that point is there in the first place.

 

[alpinestar12]

have you ridden it in the squat- inducing hole much sygote? or anyone else with a D.O.P.E system on their Kona? I think it would be good, seeing that your forks would be diving and it would keep the bike level and the geometry stable. Maybe it would'nt be as good in really bumpy braking sections, as the suspension would pack down. anyway, tell us how the brake linked kona'a ride.

 

[rbx]

Not really sure where you're getting that vertical force from, unless you're trying to calculate the CHANGE in the normal reaction force at the tyre. If that's the case, you're trying to perform a quantitative analysis using qualitative techniques (ie get a specific number as an answer, using only conceptual understanding. This really can't be done, it's not accurate at all. If you want to be able to calculate just how neutral a bike is (and keeping in mind that "neutral" is not always optimal) then you need to take into account a whole lot of factors, such as position of the centre of mass (using cartesian coordinates, ie not just the height), the shock rates front and rear, the wheelbase, the influence of varying geometry under weight shift, the gradient of the slope the bike is on, etc. As I mentioned about the neutral/optimal thing, you need to also work out whether you want to analyse the bike relative to a net input (ie including rider weight shift forwards and the fact that rear suspension inherently extends here), or whether you want to assume that whatever changes occur SOLELY from weight shift (ie not considering other compressive/extensive forces acting on the suspension under brakes) as the benchmark, and concentrate on the forces/moments directly applied to the bike by the braking system. Either method is valid, but you have to be aware of what the difference in outcomes means.

What you also need to consider is that on that Mongoose, although the AXLE moves relative to the front triangle like a singlepivot, the BB (and thus the rider/CoM) also move relative to the front triangle.



Whilst there is a moment applied directly to the rear wheel (by means of the brake caliper/axle couple), if you're going to calculate that on its own (rather than implicitly, from the tyre's contact patch) then you have to resolve the horizontal force on the tyre contact patch as though it was acting through the axle. Otherwise you effectively have that moment on the swingarm twice - once as force x wheel radius, and a second time as a couple moment about the axle. NFI where you've pulled that yellow arrow from though - there is no force, resultant or otherwise, acting like that. I get the feeling you may be forgetting that this is essentially a dynamic analysis, and that reaction forces can be caused by acceleration/deceleration of masses as well as static loads.

So the mongoose (with its floating BB)should have more or less brake squat then a normal single-pivot?

 

[S.]

So the mongoose (with its floating BB)should have more or less brake squat then a normal single-pivot?

If by "normal single pivot" you mean one with the pivot in the same location as the Mongoose (which is VERY high), I'd be inclined to say the Mongoose would have less squat, but such a bike would have a much more rearwards axle path relative to the centre of mass blah blah... they're a relatively complicated system to analyse though, and I've never bothered to measure one up and see.

 

[Northern Hemi]

Barels fork of choice may have something to do with his need for a rear that lowers into its travel as he brakes.
The 888, when compared to a Boxxer W/C, dives quite a bit.
Unless the new 888 has sorted this issue out, I would say that the whole high-squat design may be in place to make up for the inefficiencies of the fork.

 

[rbx]

If by "normal single pivot" you mean one with the pivot in the same location as the Mongoose (which is VERY high), I'd be inclined to say the Mongoose would have less squat, but such a bike would have a much more rearwards axle path relative to the centre of mass blah blah... they're a relatively complicated system to analyse though, and I've never bothered to measure one up and see.

Sorry i meant a low position single pivot ala kona.

 

[angrygecko]

I thoroughly concur that reading the post has made me so smart that I am now dumber than I was 5 minutes ago..

Aaand my head hurts!

Can't we just go ride now??

 

[Northern Hemi]

I'd like to just go ride now, but it's -40 out...

 

[S.]

Sorry i meant a low position single pivot ala kona.

You'd have to do a full numerical analysis to find out - the situations are significantly different in this case, so you'd have to work out a value for the percentage pro-squat in each case and compare em side by side, rather than being able to look at the pivot position and say "yeah that has more" like you can easily enough do with a pair of conventional singlepivot bikes.

 

[toodles]

Barels fork of choice may have something to do with his need for a rear that lowers into its travel as he brakes.
The 888, when compared to a Boxxer W/C, dives quite a bit.
Unless the new 888 has sorted this issue out, I would say that the whole high-squat design may be in place to make up for the inefficiencies of the fork.

Barel runs the BOSS 888s which use a different damper. Apparently they're pretty heavily compression damped.

 

[fatass]

You're using BIG words now! (very fking big words at that)

 

[bromontryder]

- What determines the amount of squat (or jack, but that will be dealt with later) is determined by two things: the height of the pivot (or instant centre, explanation of instant centre to follow shortly), and the length of the swingarm. The height of the pivot/IC is what determines the vertical distance (which is the moment arm as mentioned before) between the pivot and the axle, and so a higher pivot = longer moment arm = more squat generated by horizontal axle forces. The length of the swingarm changes the effects of the couple moment (which is effectively a constant for whatever calculation you’re doing) by altering the moments generated by the vertical force on the axle and the force of the shock on the swingarm, about the main pivot. For example, if, as above, the axle was 0.45m from the main pivot and the vertical force on the axle was 500N, the moment generated by that (which has to be equaled by the shock’s force multiplied by its normal distance [0.15m] from the pivot) would be 225N.m.

If this is determined by two things, then why did you simply use .45m x 500nm, just swingarm length or "IC Length" with 4-bars, and not IC Height anywhere?


Exactly what is the difference between IC length and IC Height. Also, IC Height is height relative to what exactly? And where are we suposed to use IC Height in the calculation?

 

[WolfCreekPsycho]

If this is determined by two things, then why did you simply use .45m x 500nm, just swingarm length or "IC Length" with 4-bars, and not IC Height anywhere?


Exactly what is the difference between IC length and IC Height. Also, IC Height is height relative to what exactly? And where are we suposed to use IC Height in the calculation?

..whoa... how long was the handle on your shovel ?!!

 

[bromontryder]

..whoa... how long was the handle on your shovel ?!!

What does that mean?

 

[Bodin]

What does that mean?

We need a thread called "Grave Dig - an explanation."

Google "forum gravedig" and in a short time, you'll understand.

 

[S.]

If this is determined by two things, then why did you simply use .45m x 500nm, just swingarm length or "IC Length" with 4-bars, and not IC Height anywhere?


Exactly what is the difference between IC length and IC Height. Also, IC Height is height relative to what exactly? And where are we suposed to use IC Height in the calculation?

Already answered your PM, but that was just to show how the couple moment works in terms of the existing compression (sag) of the swingarm, not the net effect on the swingarm, cos for starters to get a net effect you'd need a numerical CoM position. There is no IC height given anywhere in that article, and "IC length" isn't a term I've ever heard (maybe you meant length from IC to axle?). Like I said, it's a qualitative explanation with a few arbitrary calculations thrown in to make a point.

 

[bromontryder]

We need a thread called "Grave Dig - an explanation."

Google "forum gravedig" and in a short time, you'll understand.

Maybe I should have started a new thread so you could all tell me to use the search function

“fully active in all circumstances” claim (with the understanding that "fully active" means "shock absorption completely unaffected by braking").

So to be "fully active..." you need three things?
-Two of the links which are perfectly parallel with one another all throughout the travel.
-a CC which is fixed with respect to at least on of the pivots.
-an IC distance which is infinitely far away (in front or in back of the axle).

???