Here's the weight distribution graphic I promised:Any questions?
Here's the weight distribution graphic I promised:
Bike Weight Distribution Downhill
Any questions?
So I went and grabbed my DH bike to see how the CG location stacks up. This is an easy experiment to do if anyone feels inclined to put some real numbers on themselves… measure the weight of you and your bike. In my case this is 190.4 lbs. Then find a nice level spot where you can balance with your bar end ever so slightly on a wall for support. Put the scale under the front wheel and then weight the bike however you see fit. Record that weight. (Total weight-front wheel weight)/total weight is the percentage of weight carried by the rear wheel. You can then find the X coordinate of your CG. With my weight shifted reasonably far back but not stretching it I got 41.4 lbs on the front wheel (78.26%). The bike in question has a 1299 mm wheelbase. (1-0.7826)*wheelbase is the horizontal distance from the rear axle to your CG. In my case here that’s 282 mm. Y height of CG is harder to estimate, but I’d put mine at maybe 645 mm.
Now with the weight distribution problem it’s a little more complicated than where the CG is relative to the front axle. How much braking force can be generated is a massive factor. If we assume as much braking force as friction allows, the angle of inclination actually factors out entirely and the end result is a function of coefficient of friction. In the case of my DH bike, the equation is 0.78-0.50μ. With a coefficient of friction of 0.5, that puts 53% of the weight on the rear wheel. For a tire on dry concrete, μ is 1 so on a super grippy surface you’ve still got 28% on the rear wheel. And if there’s no grip… you’re at the 78% rear wheel bias regardless of angle of inclination.
So I went and grabbed my DH bike to see how the CG location stacks up. This is an easy experiment to do if anyone feels...
So I went and grabbed my DH bike to see how the CG location stacks up. This is an easy experiment to do if anyone feels inclined to put some real numbers on themselves… measure the weight of you and your bike. In my case this is 190.4 lbs. Then find a nice level spot where you can balance with your bar end ever so slightly on a wall for support. Put the scale under the front wheel and then weight the bike however you see fit. Record that weight. (Total weight-front wheel weight)/total weight is the percentage of weight carried by the rear wheel. You can then find the X coordinate of your CG. With my weight shifted reasonably far back but not stretching it I got 41.4 lbs on the front wheel (78.26%). The bike in question has a 1299 mm wheelbase. (1-0.7826)*wheelbase is the horizontal distance from the rear axle to your CG. In my case here that’s 282 mm. Y height of CG is harder to estimate, but I’d put mine at maybe 645 mm.
Now with the weight distribution problem it’s a little more complicated than where the CG is relative to the front axle. How much braking force can be generated is a massive factor. If we assume as much braking force as friction allows, the angle of inclination actually factors out entirely and the end result is a function of coefficient of friction. In the case of my DH bike, the equation is 0.78-0.50μ. With a coefficient of friction of 0.5, that puts 53% of the weight on the rear wheel. For a tire on dry concrete, μ is 1 so on a super grippy surface you’ve still got 28% on the rear wheel. And if there’s no grip… you’re at the 78% rear wheel bias regardless of angle of inclination.
Have a look at Matt Beer's article from PB. At 2:40 in the video you can see his weight distribution changing in real time.
That said, I don't need another Shockcraft branded drawing to tell me if I can use my rear brake on a steep trail. 30 some years of actual riding is a sufficient data point to know the answer is yes. Often and effectively.
So I went and grabbed my DH bike to see how the CG location stacks up. This is an easy experiment to do if anyone feels...
So I went and grabbed my DH bike to see how the CG location stacks up. This is an easy experiment to do if anyone feels inclined to put some real numbers on themselves… measure the weight of you and your bike. In my case this is 190.4 lbs. Then find a nice level spot where you can balance with your bar end ever so slightly on a wall for support. Put the scale under the front wheel and then weight the bike however you see fit. Record that weight. (Total weight-front wheel weight)/total weight is the percentage of weight carried by the rear wheel. You can then find the X coordinate of your CG. With my weight shifted reasonably far back but not stretching it I got 41.4 lbs on the front wheel (78.26%). The bike in question has a 1299 mm wheelbase. (1-0.7826)*wheelbase is the horizontal distance from the rear axle to your CG. In my case here that’s 282 mm. Y height of CG is harder to estimate, but I’d put mine at maybe 645 mm.
Now with the weight distribution problem it’s a little more complicated than where the CG is relative to the front axle. How much braking force can be generated is a massive factor. If we assume as much braking force as friction allows, the angle of inclination actually factors out entirely and the end result is a function of coefficient of friction. In the case of my DH bike, the equation is 0.78-0.50μ. With a coefficient of friction of 0.5, that puts 53% of the weight on the rear wheel. For a tire on dry concrete, μ is 1 so on a super grippy surface you’ve still got 28% on the rear wheel. And if there’s no grip… you’re at the 78% rear wheel bias regardless of angle of inclination.
That graphic is my old Bergamont Encore (475 reach) and measured weight distributions. DH bikes used to be more rearward, but Enduro bikes have caught up enough that some companies are using the same frames for both!
I think your tyre friction coefficients might be a bit optimistic. Coefficient of 1 and higher required a very sticky and hot tyre on a clean surface. I think 0.8 static is about where you'd be for a sticky MTB tyre on rock/concrete/asphalt. Gravel/dirt is going to vary massively and worth testing. Anything wet is going to be very low.
This is good data from tractors:
Concrete/asphalt/rock static 0.75, dynamic a bit over 0.8. Hard dirt static 0.5, dynamic (really digging in) almost 0.8 Loose static 0.4, really digging in a bit over 0.5
Problem with the back brake is it loses weight as you brake. On flat ground seated you've generally got 2/3 weight on the back wheel. But the harder you brake the more weight shifts forwards and off your back wheel. Which Braking on hard ground (0.8 traction) with the back brake you're going to peak out about 2.6 m/s^2 (0.25 G force) decelleration using the back wheel (depending on geometry). Braking on hard ground with the front wheel you're going to peak out about 8m/s^2 (0.8 G Force) (rear wheel zero weight).
Front wheel on hard ground and firm soil is ~3x more effective.
Add in a down slope and that 3x number gets bigger. To the point where the back wheel can't even hold the bike.
That graphic is my old Bergamont Encore (475 reach) and measured weight distributions. DH bikes used to be more rearward, but Enduro bikes have caught up...
That graphic is my old Bergamont Encore (475 reach) and measured weight distributions. DH bikes used to be more rearward, but Enduro bikes have caught up enough that some companies are using the same frames for both!
I think your tyre friction coefficients might be a bit optimistic. Coefficient of 1 and higher required a very sticky and hot tyre on a clean surface. I think 0.8 static is about where you'd be for a sticky MTB tyre on rock/concrete/asphalt. Gravel/dirt is going to vary massively and worth testing. Anything wet is going to be very low.
This is good data from tractors:
Concrete/asphalt/rock static 0.75, dynamic a bit over 0.8. Hard dirt static 0.5, dynamic (really digging in) almost 0.8 Loose static 0.4, really digging in a bit over 0.5
Problem with the back brake is it loses weight as you brake. On flat ground seated you've generally got 2/3 weight on the back wheel. But the harder you brake the more weight shifts forwards and off your back wheel. Which Braking on hard ground (0.8 traction) with the back brake you're going to peak out about 2.6 m/s^2 (0.25 G force) decelleration using the back wheel (depending on geometry). Braking on hard ground with the front wheel you're going to peak out about 8m/s^2 (0.8 G Force) (rear wheel zero weight).
Front wheel on hard ground and firm soil is ~3x more effective.
Add in a down slope and that 3x number gets bigger. To the point where the back wheel can't even hold the bike.
We all get that braking shifts weight forward. What I just walked through is how much weight you can end up with on the front and rear wheels in a certain scenario that involves braking as hard as grip will allow. Downslope does not factor into weight distribution. If you walk through the math you end up with a cos(theta) term in both the numerator and denominator that cancel out. 0.5 is a pretty reasonable number for loose soil such as a rake and ride and that yields about a 50/50 weight distribution on any slope with weight shifted rearward. You can easily end up with it biased even more rearward riding in the wet.
That plot does bring up an interesting thing about braking force on loose surfaces that happens to lend itself towards rear wheel braking. Wheel slip from the front wheel is generally not very tolerable especially if turning is involved but is required for max braking force. On the contrary the rear wheel can slip a ton. If you want hard braking that's less susceptible to the front wheel slipping and washing out maybe it's a good idea to bias weight rearward.
We all get that braking shifts weight forward. What I just walked through is how much weight you can end up with on the front and...
We all get that braking shifts weight forward. What I just walked through is how much weight you can end up with on the front and rear wheels in a certain scenario that involves braking as hard as grip will allow. Downslope does not factor into weight distribution. If you walk through the math you end up with a cos(theta) term in both the numerator and denominator that cancel out. 0.5 is a pretty reasonable number for loose soil such as a rake and ride and that yields about a 50/50 weight distribution on any slope with weight shifted rearward. You can easily end up with it biased even more rearward riding in the wet.
That plot does bring up an interesting thing about braking force on loose surfaces that happens to lend itself towards rear wheel braking. Wheel slip from the front wheel is generally not very tolerable especially if turning is involved but is required for max braking force. On the contrary the rear wheel can slip a ton. If you want hard braking that's less susceptible to the front wheel slipping and washing out maybe it's a good idea to bias weight rearward.
Downslope drives your entire weight distribution. This is due to your weight being above the ground line so slope changes the proportions. The only time your values would cancel out is if COG was at ground level. Which gets close for a split second after an OTB!
Moving your weight back isn't optional on steeps. If you don't do it you're eating dirt.
Here is the resulting weight shifts and max traction on a 20° (36%) downslope:
Bike Braking Dynamics
The front braking is limited by stability (rear wheel lift), the rear is limited by the interplay of forward weight shift and traction.
We all get that braking shifts weight forward. What I just walked through is how much weight you can end up with on the front and...
We all get that braking shifts weight forward. What I just walked through is how much weight you can end up with on the front and rear wheels in a certain scenario that involves braking as hard as grip will allow. Downslope does not factor into weight distribution. If you walk through the math you end up with a cos(theta) term in both the numerator and denominator that cancel out. 0.5 is a pretty reasonable number for loose soil such as a rake and ride and that yields about a 50/50 weight distribution on any slope with weight shifted rearward. You can easily end up with it biased even more rearward riding in the wet.
That plot does bring up an interesting thing about braking force on loose surfaces that happens to lend itself towards rear wheel braking. Wheel slip from the front wheel is generally not very tolerable especially if turning is involved but is required for max braking force. On the contrary the rear wheel can slip a ton. If you want hard braking that's less susceptible to the front wheel slipping and washing out maybe it's a good idea to bias weight rearward.
Yup good points! Rubber has the highest grip when there is a small amount of slip. Also your front tyre can't brake AND turn in equal amounts so even if you were braking with the front only, you better hope there aren't any turns on that trail!
And because the friction coefficient (which is different to tractive efficiency...) gets lower as you increase vertical load, if you were to have 100% weight on the front wheel (which you don't) you would still have much less braking available anyway.
Also does anyone know the actual gradient of a "steep" trail? Most people here probably don't need convincing but if you stick your phone on the ground with the surface level app going you would see its no where near as high an angle as some would think
They say Dead Dog is 32 degrees. Definitely not the steepest trail but the key is it’s constant with no catch berms for very long distances. This makes it a good test for this conversation as it’s all about what gives you maximum braking power.
Downslope drives your entire weight distribution. This is due to your weight being above the ground line so slope changes the proportions.The only time your values...
Downslope drives your entire weight distribution. This is due to your weight being above the ground line so slope changes the proportions. The only time your values would cancel out is if COG was at ground level. Which gets close for a split second after an OTB!
Moving your weight back isn't optional on steeps. If you don't do it you're eating dirt.
Here is the resulting weight shifts and max traction on a 20° (36%) downslope:
Bike Braking Dynamics
The front braking is limited by stability (rear wheel lift), the rear is limited by the interplay of forward weight shift and traction.
Slope angle factors out entirely. You need to step through it by summing the moments about the front contact patch. These sketches work fine for reference, but you need to actually do the math. It's pretty easy to see how m, g, and Cos(theta) factor out, but I had it simplify for the sake of it.
Yup good points! Rubber has the highest grip when there is a small amount of slip. Also your front tyre can't brake AND turn in equal...
Yup good points! Rubber has the highest grip when there is a small amount of slip. Also your front tyre can't brake AND turn in equal amounts so even if you were braking with the front only, you better hope there aren't any turns on that trail!
And because the friction coefficient (which is different to tractive efficiency...) gets lower as you increase vertical load, if you were to have 100% weight on the front wheel (which you don't) you would still have much less braking available anyway.
Also does anyone know the actual gradient of a "steep" trail? Most people here probably don't need convincing but if you stick your phone on the ground with the surface level app going you would see its no where near as high an angle as some would think
Yeah it's actually kind of comical and makes you feel like you can't ride anything steep some times. You may very well find you can't slow down on even maintain speed on a 25 degree decline.
Slope angle factors out entirely. You need to step through it by summing the moments about the front contact patch. These sketches work fine for reference...
Slope angle factors out entirely. You need to step through it by summing the moments about the front contact patch. These sketches work fine for reference, but you need to actually do the math. It's pretty easy to see how m, g, and Cos(theta) factor out, but I had it simplify for the sake of it.
There are several ways to analyse this and they all give the same answers. But the intermediate results are very different. I think that's what's going on. You're still talking static braking and not dynamic with the weight shift.
The bike in that graphic has a 2/3 rear weight distribution static, but rear only braking on the flat hits equialibrium at 0.21g decelleration which moves 21kg onto the front and kills your rear traction.
Everything sums about any point. Doesn't matter if it's front tyre, rear tyre, COG etc. But you can rotate the diagrams (slope angle) or rotate the vectors (use accelerations to represent slope).
The answers are all the same. Back brake can hold a bike on 10 degrees and at ~45 degrees, depending on geometry, your back wheel is floating.
Many people consider 20% (11 deg) a very steep slope. It is if you're climbing.
So I went and grabbed my DH bike to see how the CG location stacks up. This is an easy experiment to do if anyone feels...
So I went and grabbed my DH bike to see how the CG location stacks up. This is an easy experiment to do if anyone feels inclined to put some real numbers on themselves… measure the weight of you and your bike. In my case this is 190.4 lbs. Then find a nice level spot where you can balance with your bar end ever so slightly on a wall for support. Put the scale under the front wheel and then weight the bike however you see fit. Record that weight. (Total weight-front wheel weight)/total weight is the percentage of weight carried by the rear wheel. You can then find the X coordinate of your CG. With my weight shifted reasonably far back but not stretching it I got 41.4 lbs on the front wheel (78.26%). The bike in question has a 1299 mm wheelbase. (1-0.7826)*wheelbase is the horizontal distance from the rear axle to your CG. In my case here that’s 282 mm. Y height of CG is harder to estimate, but I’d put mine at maybe 645 mm.
Now with the weight distribution problem it’s a little more complicated than where the CG is relative to the front axle. How much braking force can be generated is a massive factor. If we assume as much braking force as friction allows, the angle of inclination actually factors out entirely and the end result is a function of coefficient of friction. In the case of my DH bike, the equation is 0.78-0.50μ. With a coefficient of friction of 0.5, that puts 53% of the weight on the rear wheel. For a tire on dry concrete, μ is 1 so on a super grippy surface you’ve still got 28% on the rear wheel. And if there’s no grip… you’re at the 78% rear wheel bias regardless of angle of inclination.
Let's see if we can find some common ground on flat braking.
Your bike, 1299 wheelbase, 86kg total weight, in rearward position you've got 22/64kg fr/rear split. Using a 1m COG height (round numbers, can adjust later) I get max rear braking at 3m/s^2 (0.3G). This results in a forward weight shift of 20.3kg and dynamic weights of 42F and 44R kg.
Front brake you can pull 9.5 m/s^2 (just over 1G) IFF you've got the grip. Front is a a shade over 3x as effective as the rear on flat ground.
Let's see if we can find some common ground on flat braking.Your bike, 1299 wheelbase, 86kg total weight, in rearward position you've got 22/64kg fr/rear split.Using...
Let's see if we can find some common ground on flat braking.
Your bike, 1299 wheelbase, 86kg total weight, in rearward position you've got 22/64kg fr/rear split. Using a 1m COG height (round numbers, can adjust later) I get max rear braking at 3m/s^2 (0.3G). This results in a forward weight shift of 20.3kg and dynamic weights of 42F and 44R kg.
Front brake you can pull 9.5 m/s^2 (just over 1G) IFF you've got the grip. Front is a a shade over 3x as effective as the rear on flat ground.
So I think what’s throwing you for a loop is there is a fundamental flaw in your FBD. What you have drawn is only accurate if the front wheel is pinned to the ground at the contact patch (ie wheel stuck up against a log or something). If your wheel is stuck up against a log or anything else that prevents it from being able to roll downhill then the weight distribution 100% shifts forward as it gets steeper. But this is a fundamentally different scenario than braking force. See the FBD below if you’re actually interested in sorting it out. There is one equation for coefficient of friction where they work out to be the same (when μ=tan(θ)), but that’s it. The rear wheel weight distribution equation for max braking force can be boiled down to (initial weight fraction)-(ycg/wheelbase)*μ. There are no hidden layers or complexity to it beyond that rear wheel and front wheel friction may be different in practice. You can figure out what value of μ would result in zero rear wheel weight by solving (initial weight fraction*wheelbase)/ycg. You’ll find that number to be on the very high end that you’d only see riding things like Squamish slabs.
So I think what’s throwing you for a loop is there is a fundamental flaw in your FBD. What you have drawn is only accurate if...
So I think what’s throwing you for a loop is there is a fundamental flaw in your FBD. What you have drawn is only accurate if the front wheel is pinned to the ground at the contact patch (ie wheel stuck up against a log or something). If your wheel is stuck up against a log or anything else that prevents it from being able to roll downhill then the weight distribution 100% shifts forward as it gets steeper. But this is a fundamentally different scenario than braking force. See the FBD below if you’re actually interested in sorting it out. There is one equation for coefficient of friction where they work out to be the same (when μ=tan(θ)), but that’s it. The rear wheel weight distribution equation for max braking force can be boiled down to (initial weight fraction)-(ycg/wheelbase)*μ. There are no hidden layers or complexity to it beyond that rear wheel and front wheel friction may be different in practice. You can figure out what value of μ would result in zero rear wheel weight by solving (initial weight fraction*wheelbase)/ycg. You’ll find that number to be on the very high end that you’d only see riding things like Squamish slabs.
Those diagrams are illustrations to show people, they aren't intended to be FBD, they've got too many arrows for a start and those arrows aren't to scale either. If the front wheel was pinned the forces would go through the axle instead of contact patch. But it isn't. On your FBD you've got vectors joining tip to tip under each contact patch. They should be tip-tail which then orients the resulting sum in the directions I have shown.
I'm running all my calculations on a separate spread-sheet. Throw some numbers into your calcuations and see what you get for forward weight shift on that angle static and under braking. Remember if you're braking to decelerate that modifies your gravity vector. D'Alembert's Principle.
Those diagrams are illustrations to show people, they aren't intended to be FBD, they've got too many arrows for a start and those arrows aren't to...
Those diagrams are illustrations to show people, they aren't intended to be FBD, they've got too many arrows for a start and those arrows aren't to scale either. If the front wheel was pinned the forces would go through the axle instead of contact patch. But it isn't. On your FBD you've got vectors joining tip to tip under each contact patch. They should be tip-tail which then orients the resulting sum in the directions I have shown.
I'm running all my calculations on a separate spread-sheet. Throw some numbers into your calcuations and see what you get for forward weight shift on that angle static and under braking. Remember if you're braking to decelerate that modifies your gravity vector. D'Alembert's Principle.
It doesn’t matter if they are tip to tip, tip to tail, or tail to tail so long as they are in the correct orientation because vectors are summed based on their components. Same goes for gravity. It’s broken down into components in that diagram such that everything is parallel or perpendicular and can be summed easily. Go look up the FBD of a car braking. This is basic stuff. Static is essentially the same as setting the coefficient of friction to tan(theta) so you’re changing a relationship to that is fundamental to the problem by looking at that. D'Alembert's Principle is why we can ignore the mgSin(theta) portion of the gravity vector which is what I’ve done in the calculations above. What I’ve shown you is correct. You can plug whatever weight distributions, friction coefficients, and CG heights you want in it. There is not a missing bit that causes weight to shift forward more when going down something steeper.
It doesn’t matter if they are tip to tip, tip to tail, or tail to tail so long as they are in the correct orientation because...
It doesn’t matter if they are tip to tip, tip to tail, or tail to tail so long as they are in the correct orientation because vectors are summed based on their components. Same goes for gravity. It’s broken down into components in that diagram such that everything is parallel or perpendicular and can be summed easily. Go look up the FBD of a car braking. This is basic stuff. Static is essentially the same as setting the coefficient of friction to tan(theta) so you’re changing a relationship to that is fundamental to the problem by looking at that. D'Alembert's Principle is why we can ignore the mgSin(theta) portion of the gravity vector which is what I’ve done in the calculations above. What I’ve shown you is correct. You can plug whatever weight distributions, friction coefficients, and CG heights you want in it. There is not a missing bit that causes weight to shift forward more when going down something steeper.
If you don't add vectors tip to tail you get the wrong direction. Yes this is basic stuff.
Can you throw some numbers into your diagram and tell us what you get for braking force and weight shift?
If you don't add vectors tip to tail you get the wrong direction. Yes this is basic stuff. Can you throw some numbers into your diagram...
If you don't add vectors tip to tail you get the wrong direction. Yes this is basic stuff.
Can you throw some numbers into your diagram and tell us what you get for braking force and weight shift?
How they are positioned on a piece of paper only impacts the end result if you create the summed vector by tracing a line from the first tail to the last tip. But I’m not tracing lines so it doesn’t matter.
There are only four numbers to plug in. It doesn’t necessitate a spread sheet. I’ve thrown countless numbers into it that I’ve listed here before. I can come up with weight distributions and friction values that will give pretty much any weight shift imaginable. There’s no point. This was all to show that with a rearward weight bias it’s perfectly reasonable to expect 50% of braking force to come from the rear wheel still. If you have shifted your weight rearward, which usually moves it downward too, and it’s not a super grippy surface that’s perfectly reasonable to expect.
If you don't add vectors tip to tail you get the wrong direction. Yes this is basic stuff. Can you throw some numbers into your diagram...
If you don't add vectors tip to tail you get the wrong direction. Yes this is basic stuff.
Can you throw some numbers into your diagram and tell us what you get for braking force and weight shift?
That said, if you use an initial distribution of 2/3 on the rear wheel, a CG height of 1000, a wheelbase of 1299, and a coefficient of friction of 0.5, you get 28% of weight on the rear wheel under max braking regardless of angle.
So I went and grabbed my DH bike to see how the CG location stacks up. This is an easy experiment to do if anyone feels inclined to put some real numbers on themselves… measure the weight of you and your bike. In my case this is 190.4 lbs. Then find a nice level spot where you can balance with your bar end ever so slightly on a wall for support. Put the scale under the front wheel and then weight the bike however you see fit. Record that weight. (Total weight-front wheel weight)/total weight is the percentage of weight carried by the rear wheel. You can then find the X coordinate of your CG. With my weight shifted reasonably far back but not stretching it I got 41.4 lbs on the front wheel (78.26%). The bike in question has a 1299 mm wheelbase. (1-0.7826)*wheelbase is the horizontal distance from the rear axle to your CG. In my case here that’s 282 mm. Y height of CG is harder to estimate, but I’d put mine at maybe 645 mm.
Now with the weight distribution problem it’s a little more complicated than where the CG is relative to the front axle. How much braking force can be generated is a massive factor. If we assume as much braking force as friction allows, the angle of inclination actually factors out entirely and the end result is a function of coefficient of friction. In the case of my DH bike, the equation is 0.78-0.50μ. With a coefficient of friction of 0.5, that puts 53% of the weight on the rear wheel. For a tire on dry concrete, μ is 1 so on a super grippy surface you’ve still got 28% on the rear wheel. And if there’s no grip… you’re at the 78% rear wheel bias regardless of angle of inclination.
Have a look at Matt Beer's article from PB. At 2:40 in the video you can see his weight distribution changing in real time.
That said, I don't need another Shockcraft branded drawing to tell me if I can use my rear brake on a steep trail. 30 some years of actual riding is a sufficient data point to know the answer is yes. Often and effectively.
That graphic is my old Bergamont Encore (475 reach) and measured weight distributions. DH bikes used to be more rearward, but Enduro bikes have caught up enough that some companies are using the same frames for both!
I think your tyre friction coefficients might be a bit optimistic. Coefficient of 1 and higher required a very sticky and hot tyre on a clean surface. I think 0.8 static is about where you'd be for a sticky MTB tyre on rock/concrete/asphalt. Gravel/dirt is going to vary massively and worth testing. Anything wet is going to be very low.
This is good data from tractors:
Concrete/asphalt/rock static 0.75, dynamic a bit over 0.8.
Hard dirt static 0.5, dynamic (really digging in) almost 0.8
Loose static 0.4, really digging in a bit over 0.5
Problem with the back brake is it loses weight as you brake.
On flat ground seated you've generally got 2/3 weight on the back wheel. But the harder you brake the more weight shifts forwards and off your back wheel. Which
Braking on hard ground (0.8 traction) with the back brake you're going to peak out about 2.6 m/s^2 (0.25 G force) decelleration using the back wheel (depending on geometry).
Braking on hard ground with the front wheel you're going to peak out about 8m/s^2 (0.8 G Force) (rear wheel zero weight).
Front wheel on hard ground and firm soil is ~3x more effective.
Add in a down slope and that 3x number gets bigger. To the point where the back wheel can't even hold the bike.
We all get that braking shifts weight forward. What I just walked through is how much weight you can end up with on the front and rear wheels in a certain scenario that involves braking as hard as grip will allow. Downslope does not factor into weight distribution. If you walk through the math you end up with a cos(theta) term in both the numerator and denominator that cancel out. 0.5 is a pretty reasonable number for loose soil such as a rake and ride and that yields about a 50/50 weight distribution on any slope with weight shifted rearward. You can easily end up with it biased even more rearward riding in the wet.
That plot does bring up an interesting thing about braking force on loose surfaces that happens to lend itself towards rear wheel braking. Wheel slip from the front wheel is generally not very tolerable especially if turning is involved but is required for max braking force. On the contrary the rear wheel can slip a ton. If you want hard braking that's less susceptible to the front wheel slipping and washing out maybe it's a good idea to bias weight rearward.
Downslope drives your entire weight distribution. This is due to your weight being above the ground line so slope changes the proportions.
The only time your values would cancel out is if COG was at ground level. Which gets close for a split second after an OTB!
Moving your weight back isn't optional on steeps. If you don't do it you're eating dirt.
Here is the resulting weight shifts and max traction on a 20° (36%) downslope:
The front braking is limited by stability (rear wheel lift), the rear is limited by the interplay of forward weight shift and traction.
Yup good points! Rubber has the highest grip when there is a small amount of slip. Also your front tyre can't brake AND turn in equal amounts so even if you were braking with the front only, you better hope there aren't any turns on that trail!
And because the friction coefficient (which is different to tractive efficiency...) gets lower as you increase vertical load, if you were to have 100% weight on the front wheel (which you don't) you would still have much less braking available anyway.
Also does anyone know the actual gradient of a "steep" trail? Most people here probably don't need convincing but if you stick your phone on the ground with the surface level app going you would see its no where near as high an angle as some would think
They say Dead Dog is 32 degrees. Definitely not the steepest trail but the key is it’s constant with no catch berms for very long distances. This makes it a good test for this conversation as it’s all about what gives you maximum braking power.
Slope angle factors out entirely. You need to step through it by summing the moments about the front contact patch. These sketches work fine for reference, but you need to actually do the math. It's pretty easy to see how m, g, and Cos(theta) factor out, but I had it simplify for the sake of it.
Yeah it's actually kind of comical and makes you feel like you can't ride anything steep some times. You may very well find you can't slow down on even maintain speed on a 25 degree decline.
Most people feel a lot of exposure over a scree slope and consider them too step to ride, and they're only 30 degrees or so.
There are several ways to analyse this and they all give the same answers. But the intermediate results are very different. I think that's what's going on. You're still talking static braking and not dynamic with the weight shift.
The bike in that graphic has a 2/3 rear weight distribution static, but rear only braking on the flat hits equialibrium at 0.21g decelleration which moves 21kg onto the front and kills your rear traction.
Everything sums about any point. Doesn't matter if it's front tyre, rear tyre, COG etc. But you can rotate the diagrams (slope angle) or rotate the vectors (use accelerations to represent slope).
The answers are all the same. Back brake can hold a bike on 10 degrees and at ~45 degrees, depending on geometry, your back wheel is floating.
Many people consider 20% (11 deg) a very steep slope. It is if you're climbing.
Let's see if we can find some common ground on flat braking.
Your bike, 1299 wheelbase, 86kg total weight, in rearward position you've got 22/64kg fr/rear split.
Using a 1m COG height (round numbers, can adjust later) I get max rear braking at 3m/s^2 (0.3G). This results in a forward weight shift of 20.3kg and dynamic weights of 42F and 44R kg.
Front brake you can pull 9.5 m/s^2 (just over 1G) IFF you've got the grip. Front is a a shade over 3x as effective as the rear on flat ground.
How about this, In n Out Burger with only your rear or only your front brake? Out run not included.
So I think what’s throwing you for a loop is there is a fundamental flaw in your FBD. What you have drawn is only accurate if the front wheel is pinned to the ground at the contact patch (ie wheel stuck up against a log or something). If your wheel is stuck up against a log or anything else that prevents it from being able to roll downhill then the weight distribution 100% shifts forward as it gets steeper. But this is a fundamentally different scenario than braking force. See the FBD below if you’re actually interested in sorting it out. There is one equation for coefficient of friction where they work out to be the same (when μ=tan(θ)), but that’s it. The rear wheel weight distribution equation for max braking force can be boiled down to (initial weight fraction)-(ycg/wheelbase)*μ. There are no hidden layers or complexity to it beyond that rear wheel and front wheel friction may be different in practice. You can figure out what value of μ would result in zero rear wheel weight by solving (initial weight fraction*wheelbase)/ycg. You’ll find that number to be on the very high end that you’d only see riding things like Squamish slabs.
Those diagrams are illustrations to show people, they aren't intended to be FBD, they've got too many arrows for a start and those arrows aren't to scale either. If the front wheel was pinned the forces would go through the axle instead of contact patch. But it isn't.
On your FBD you've got vectors joining tip to tip under each contact patch. They should be tip-tail which then orients the resulting sum in the directions I have shown.
I'm running all my calculations on a separate spread-sheet. Throw some numbers into your calcuations and see what you get for forward weight shift on that angle static and under braking. Remember if you're braking to decelerate that modifies your gravity vector. D'Alembert's Principle.
This one? https://www.trailforks.com/trails/innout-burger/
Average slope 23% (13 degrees) , I found max slope on the graph to be 54% (28 degrees) . But GPS tracks are going to miss all the fun bits.
https://www.trailforks.com/trails/innout-burger/
It doesn’t matter if they are tip to tip, tip to tail, or tail to tail so long as they are in the correct orientation because vectors are summed based on their components. Same goes for gravity. It’s broken down into components in that diagram such that everything is parallel or perpendicular and can be summed easily. Go look up the FBD of a car braking. This is basic stuff. Static is essentially the same as setting the coefficient of friction to tan(theta) so you’re changing a relationship to that is fundamental to the problem by looking at that. D'Alembert's Principle is why we can ignore the mgSin(theta) portion of the gravity vector which is what I’ve done in the calculations above. What I’ve shown you is correct. You can plug whatever weight distributions, friction coefficients, and CG heights you want in it. There is not a missing bit that causes weight to shift forward more when going down something steeper.
If you don't add vectors tip to tail you get the wrong direction. Yes this is basic stuff.
Can you throw some numbers into your diagram and tell us what you get for braking force and weight shift?
How they are positioned on a piece of paper only impacts the end result if you create the summed vector by tracing a line from the first tail to the last tip. But I’m not tracing lines so it doesn’t matter.
There are only four numbers to plug in. It doesn’t necessitate a spread sheet. I’ve thrown countless numbers into it that I’ve listed here before. I can come up with weight distributions and friction values that will give pretty much any weight shift imaginable. There’s no point. This was all to show that with a rearward weight bias it’s perfectly reasonable to expect 50% of braking force to come from the rear wheel still. If you have shifted your weight rearward, which usually moves it downward too, and it’s not a super grippy surface that’s perfectly reasonable to expect.
That said, if you use an initial distribution of 2/3 on the rear wheel, a CG height of 1000, a wheelbase of 1299, and a coefficient of friction of 0.5, you get 28% of weight on the rear wheel under max braking regardless of angle.
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