Your COG height, is that 650mm from the ground or 650 from the BB? Your back wheel is about 750mm diameter so most of your body...
Your COG height, is that 650mm from the ground or 650 from the BB? Your back wheel is about 750mm diameter so most of your body, especially the heavy parts, are significantly higher than that.
Here's a posed pic from MTBR ages ago, not me: This would be about 1m above ground:
Rearward Riding Position
The whole point of running a single fixed position in a diagram is to find physics we agree on. From there we can complicate it as much as you want.
Using 78% static rear distribution (86kg total) on flat ground with a CG of 650mm (above ground) and COF of 0.5 here are my dynamic results. Max rear braking 2.3 m/s^2 (0.23G), front weight shift is 10kg giving you 54kg on the rear wheel and 32kg on the front. That's dynamic weight distribution of 63% rear.
Tilt the bike 20 deg and you need a force of 0.34G up the slope to hold the bike.
Agree or disagree this far?
I measured to approximately that point based on body position relative to a table. I’m not very tall. There is certainly some error in guessing CG height, but it’s +/- not inherently low.
There are some inconsistencies in your numbers. For max braking with both brakes with μ of 0.5 the weight should be 53/47 rear to front. Additionally, the g’s that can be attributed to the rear wheel are always going to be equal to the fraction of weight on the rear wheel times μ. So if you’ve got 63% of your weight on the rear wheel in this scenario it should be accounting for 0.315 of the 0.5 g’s available. 53% and 0.265 of the 0.5 is coming from the rear wheel. At a 20 degree angle it is correct that you need 0.34 g’s to hold you so just the rear brake won’t. And yes you can get all 0.5 g’s from the front wheel, but we don’t ride the front wheel down steep things unless we want to go OTB. As @TEAMROBOT said, front brake only would be Russian roulette. Either completely fine or spectacular crash. I could redo the calculations for just rear brake only and it should give a slightly higher amount of braking available from the rear, but will always be less than what front only could do in real world scenarios.
That fills in a few gaps. I can see how an engineering career isn't going to work if you don't like physics and analysis.I ride steep...
That fills in a few gaps. I can see how an engineering career isn't going to work if you don't like physics and analysis.
I ride steep and techy shit. Those steep bits you mention above where the back brake does SFA. That's what I'm talking about.
@ebikepartyshirt I think I know what's going on here. You and I are talking about the track parts well over 20 degrees that your back brake is totally useless. Try to ride those with a back only brake and you're going to hospital. The others are talking about a whole track that has some steep bits but you can still use your back brake on 90% of it. Or something.
I like both those things, I left engineering because I found my name on a drawing for a part that I'd never seen before, doing a job it wasn't fit for, in a critical part of the structure of a personnel lift. And had seen the same lack of care taken in documentation/safety across multiple companies that I worked for, could have happened at any of them... I wasn't about that.
Not having wounding conversations with munters is just a positive by-product.
I measured to approximately that point based on body position relative to a table. I’m not very tall. There is certainly some error in guessing CG...
I measured to approximately that point based on body position relative to a table. I’m not very tall. There is certainly some error in guessing CG height, but it’s +/- not inherently low.
There are some inconsistencies in your numbers. For max braking with both brakes with μ of 0.5 the weight should be 53/47 rear to front. Additionally, the g’s that can be attributed to the rear wheel are always going to be equal to the fraction of weight on the rear wheel times μ. So if you’ve got 63% of your weight on the rear wheel in this scenario it should be accounting for 0.315 of the 0.5 g’s available. 53% and 0.265 of the 0.5 is coming from the rear wheel. At a 20 degree angle it is correct that you need 0.34 g’s to hold you so just the rear brake won’t. And yes you can get all 0.5 g’s from the front wheel, but we don’t ride the front wheel down steep things unless we want to go OTB. As @TEAMROBOT said, front brake only would be Russian roulette. Either completely fine or spectacular crash. I could redo the calculations for just rear brake only and it should give a slightly higher amount of braking available from the rear, but will always be less than what front only could do in real world scenarios.
You've gone further than the example I've presented. I showed back only braking but you've gone to front and rear together. Which was never on the table.
I'm trying to keep this simple. Static weight distribution, then rear only and front only. Keeping it simple so we can check the physics matches.
But we agree on 0.34g to hold a bike on a 20 degree slope. So if the back wheel can't do 0.34g the bike will keep accelerating.
I like both those things, I left engineering because I found my name on a drawing for a part that I'd never seen before, doing a...
I like both those things, I left engineering because I found my name on a drawing for a part that I'd never seen before, doing a job it wasn't fit for, in a critical part of the structure of a personnel lift. And had seen the same lack of care taken in documentation/safety across multiple companies that I worked for, could have happened at any of them... I wasn't about that.
Not having wounding conversations with munters is just a positive by-product.
That's the sort of thing you leave a company for. But not normally a whole industry.
I like both those things, I left engineering because I found my name on a drawing for a part that I'd never seen before, doing a...
I like both those things, I left engineering because I found my name on a drawing for a part that I'd never seen before, doing a job it wasn't fit for, in a critical part of the structure of a personnel lift. And had seen the same lack of care taken in documentation/safety across multiple companies that I worked for, could have happened at any of them... I wasn't about that.
Not having wounding conversations with munters is just a positive by-product.
What about the guy who put wood chips in his pellet bbq though
You've gone further than the example I've presented. I showed back only braking but you've gone to front and rear together. Which was never on the...
You've gone further than the example I've presented. I showed back only braking but you've gone to front and rear together. Which was never on the table.
I'm trying to keep this simple. Static weight distribution, then rear only and front only. Keeping it simple so we can check the physics matches.
But we agree on 0.34g to hold a bike on a 20 degree slope. So if the back wheel can't do 0.34g the bike will keep accelerating.
It is simple with both brakes. It’s also simple with just the front brake or just the rear. If you do just the rear, then yes you get 63% on the rear wheel, which consequently accounts for 63% of your max braking force/0.315 g if μ is 0.5. So I’m assuming you calculated 0.32 and accidentally used 0.23 g. If the same weight distribution is used for front brake only then you get 30% of the weight on the front wheel. So given that weight distribution I’d 100% go with rear only over front only if μ is 0.5.
You've gone further than the example I've presented. I showed back only braking but you've gone to front and rear together. Which was never on the...
You've gone further than the example I've presented. I showed back only braking but you've gone to front and rear together. Which was never on the table.
I'm trying to keep this simple. Static weight distribution, then rear only and front only. Keeping it simple so we can check the physics matches.
But we agree on 0.34g to hold a bike on a 20 degree slope. So if the back wheel can't do 0.34g the bike will keep accelerating.
It is simple with both brakes. It’s also simple with just the front brake or just the rear. If you do just the rear, then yes...
It is simple with both brakes. It’s also simple with just the front brake or just the rear. If you do just the rear, then yes you get 63% on the rear wheel, which consequently accounts for 63% of your max braking force/0.315 g if μ is 0.5. So I’m assuming you calculated 0.32 and accidentally used 0.23 g. If the same weight distribution is used for front brake only then you get 30% of the weight on the front wheel. So given that weight distribution I’d 100% go with rear only over front only if μ is 0.5.
Just reorganised this to make it clearer about flat vs slope.
So we agree that 20 degrees gives you 0.34g needed. But from there you're doing it very differently. No I haven't made a typo with 0.23g.
A 0.34G force couple would result in a forward weight shift moment of acceleration * mass * COG height. That is resisted at the wheels by extra weight on the front and less on the back. Ratio of COG/Wheelbase. Which is conveniently 0.5 (0.65/1.299).
Weight shift = acceleration*mass/2 = 0.34*86/2 =14.6kg. Call it 15.
Flat weight distribution was 22/64kg
Back only braking on the flat gives new weight distribution becomes, 49kg rear, 37kg front. But do we have enough traction for that?
0.34G needs 29kg of traction. 49kg on the rear with a friction coefficient of 0.5 only gives us 24.5kg traction.
So the rear brake can't produce enough traction on the flat to hold us on a 20 degree slope. It can't produce 0.34g. It's closer to 0.28g, On the flat.
On a slope the weight moves further forward simply because the COG is above ground. This reduces the rear weight and traction further. I'll get into that later.
I dunno about you but when Im riding steeps my goal isn’t to come to a complete stop. Usually its just to manage speed and braking zones are chosen based on flatter areas or higher traction surfaces.
Just reorganised this to make it clearer about flat vs slope.So we agree that 20 degrees gives you 0.34g needed. But from there you're doing it...
Just reorganised this to make it clearer about flat vs slope.
So we agree that 20 degrees gives you 0.34g needed. But from there you're doing it very differently. No I haven't made a typo with 0.23g.
A 0.34G force couple would result in a forward weight shift moment of acceleration * mass * COG height. That is resisted at the wheels by extra weight on the front and less on the back. Ratio of COG/Wheelbase. Which is conveniently 0.5 (0.65/1.299).
Weight shift = acceleration*mass/2 = 0.34*86/2 =14.6kg. Call it 15.
Flat weight distribution was 22/64kg
Back only braking on the flat gives new weight distribution becomes, 49kg rear, 37kg front. But do we have enough traction for that?
0.34G needs 29kg of traction. 49kg on the rear with a friction coefficient of 0.5 only gives us 24.5kg traction.
So the rear brake can't produce enough traction on the flat to hold us on a 20 degree slope. It can't produce 0.34g. It's closer to 0.28g, On the flat.
On a slope the weight moves further forward simply because the COG is above ground. This reduces the rear weight and traction further. I'll get into that later.
Sorry but you’re lost in the sauce. Rear wheel normal force multiplied by coefficient of friction is always going to be equal to max braking force from the rear wheel and that is your braking force applied at the CG. There are zero iterations required to solve for front and rear wheel normal forces regardless of whether it’s front and rear, rear only, or front only. Weight distribution percentages do not change with slope angle in any of those scenarios. There is no getting into that later. It just doesn’t happen. It is clear that m, g, and cos(theta) all cancel out. Braking force available decreases with slope angel which is why slope angle factors out of weight distribution. This is all there is to it for rear brake only:
Just reorganised this to make it clearer about flat vs slope.So we agree that 20 degrees gives you 0.34g needed. But from there you're doing it...
Just reorganised this to make it clearer about flat vs slope.
So we agree that 20 degrees gives you 0.34g needed. But from there you're doing it very differently. No I haven't made a typo with 0.23g.
A 0.34G force couple would result in a forward weight shift moment of acceleration * mass * COG height. That is resisted at the wheels by extra weight on the front and less on the back. Ratio of COG/Wheelbase. Which is conveniently 0.5 (0.65/1.299).
Weight shift = acceleration*mass/2 = 0.34*86/2 =14.6kg. Call it 15.
Flat weight distribution was 22/64kg
Back only braking on the flat gives new weight distribution becomes, 49kg rear, 37kg front. But do we have enough traction for that?
0.34G needs 29kg of traction. 49kg on the rear with a friction coefficient of 0.5 only gives us 24.5kg traction.
So the rear brake can't produce enough traction on the flat to hold us on a 20 degree slope. It can't produce 0.34g. It's closer to 0.28g, On the flat.
On a slope the weight moves further forward simply because the COG is above ground. This reduces the rear weight and traction further. I'll get into that later.
Sorry but you’re lost in the sauce. Rear wheel normal force multiplied by coefficient of friction is always going to be equal to max braking force...
Sorry but you’re lost in the sauce. Rear wheel normal force multiplied by coefficient of friction is always going to be equal to max braking force from the rear wheel and that is your braking force applied at the CG. There are zero iterations required to solve for front and rear wheel normal forces regardless of whether it’s front and rear, rear only, or front only. Weight distribution percentages do not change with slope angle in any of those scenarios. There is no getting into that later. It just doesn’t happen. It is clear that m, g, and cos(theta) all cancel out. Braking force available decreases with slope angel which is why slope angle factors out of weight distribution. This is all there is to it for rear brake only:
You're correct for a static situation. But braking is not static, it's dynamic and you need to address the weight shift.
I'm walking you through flat braking. Ignore slope. The only slope introduced was to quantify the braking acceleration needed and We agree on that.
Here it is again. Tell me which part you disagree with.
Start with 64/22kg F/R weight distribution on the flat. COG 650, WB 1299, COF 0.5. Haul on the back brake, weight shifts forwards. Max back braking can potentially give you 32kg braking force (0.37g). But a 32kg braking force causes a forward weight shift, moving 16kg from your back wheel to front wheel and reducing rear traction. First iteration would have 48kg on the back wheel and 38 on the front. 48kg reduces your braking traction to 24kg which reduces max deceleration to 0.28g.
Iteration further, results converge on a balanced dynamic situation. Final result is an almost 13kg weight shift, 51/35kg R/F weight distribution with 2.9m/s^2 (0.3g) braking rear only on the flat.
I did just find and fix 2 errors in my spreadsheet just now. They almost cancelled out but threw previous results a bit.
You're correct for a static situation. But braking is not static, it's dynamic and you need to address the weight shift. I'm walking you through flat...
You're correct for a static situation. But braking is not static, it's dynamic and you need to address the weight shift.
I'm walking you through flat braking. Ignore slope. The only slope introduced was to quantify the braking acceleration needed and We agree on that.
Here it is again. Tell me which part you disagree with.
Start with 64/22kg F/R weight distribution on the flat. COG 650, WB 1299, COF 0.5. Haul on the back brake, weight shifts forwards. Max back braking can potentially give you 32kg braking force (0.37g). But a 32kg braking force causes a forward weight shift, moving 16kg from your back wheel to front wheel and reducing rear traction. First iteration would have 48kg on the back wheel and 38 on the front. 48kg reduces your braking traction to 24kg which reduces max deceleration to 0.28g.
Iteration further, results converge on a balanced dynamic situation. Final result is an almost 13kg weight shift, 51/35kg R/F weight distribution with 2.9m/s^2 (0.3g) braking rear only on the flat.
I did just find and fix 2 errors in my spreadsheet just now. They almost cancelled out but threw previous results a bit.
I guess if I had to pick something the very first thing would be that we were talking about a 78% rear wheel bias but then suddenly it changed. You mentioned d'Alembert’s principle. That is exactly what this evaluation is. It’s solving for a new weight distribution given an initial weight distribution and friction and assuming constant deceleration is possible. The end result is exactly when deceleration available is deceleration used. It is not solving for slope angle you can hold the bike on, whether or not the bike is accelerating or decelerating, or anything like that. Purely is it at a steady state for weight distribution.
I guess if I had to pick something the very first thing would be that we were talking about a 78% rear wheel bias but then...
I guess if I had to pick something the very first thing would be that we were talking about a 78% rear wheel bias but then suddenly it changed. You mentioned d'Alembert’s principle. That is exactly what this evaluation is. It’s solving for a new weight distribution given an initial weight distribution and friction and assuming constant deceleration is possible. The end result is exactly when deceleration available is deceleration used. It is not solving for slope angle you can hold the bike on, whether or not the bike is accelerating or decelerating, or anything like that. Purely is it at a steady state for weight distribution.
I got interrupted editing the post above and just finished, feel free to check it again.
Ignore all slopes. Flat braking 64/22kg (74% rear).
Final stable deceleration is 0.3g with the back brake only, giving you 51/35kg weight distribution on the flat.
I got interrupted editing the post above and just finished, feel free to check it again.Ignore all slopes. Flat braking 64/22kg (74% rear).Final stable deceleration is...
I got interrupted editing the post above and just finished, feel free to check it again.
Ignore all slopes. Flat braking 64/22kg (74% rear).
Final stable deceleration is 0.3g with the back brake only, giving you 51/35kg weight distribution on the flat.
Which is exactly what that equation spits out… it doesn’t require any iteration and, again, does not depend on slope angle.
God this thread is wounding af
I measured to approximately that point based on body position relative to a table. I’m not very tall. There is certainly some error in guessing CG height, but it’s +/- not inherently low.
There are some inconsistencies in your numbers. For max braking with both brakes with μ of 0.5 the weight should be 53/47 rear to front. Additionally, the g’s that can be attributed to the rear wheel are always going to be equal to the fraction of weight on the rear wheel times μ. So if you’ve got 63% of your weight on the rear wheel in this scenario it should be accounting for 0.315 of the 0.5 g’s available. 53% and 0.265 of the 0.5 is coming from the rear wheel. At a 20 degree angle it is correct that you need 0.34 g’s to hold you so just the rear brake won’t. And yes you can get all 0.5 g’s from the front wheel, but we don’t ride the front wheel down steep things unless we want to go OTB. As @TEAMROBOT said, front brake only would be Russian roulette. Either completely fine or spectacular crash. I could redo the calculations for just rear brake only and it should give a slightly higher amount of braking available from the rear, but will always be less than what front only could do in real world scenarios.
I like both those things, I left engineering because I found my name on a drawing for a part that I'd never seen before, doing a job it wasn't fit for, in a critical part of the structure of a personnel lift. And had seen the same lack of care taken in documentation/safety across multiple companies that I worked for, could have happened at any of them... I wasn't about that.
Not having wounding conversations with munters is just a positive by-product.
You've gone further than the example I've presented. I showed back only braking but you've gone to front and rear together. Which was never on the table.
I'm trying to keep this simple. Static weight distribution, then rear only and front only. Keeping it simple so we can check the physics matches.
But we agree on 0.34g to hold a bike on a 20 degree slope. So if the back wheel can't do 0.34g the bike will keep accelerating.
That's the sort of thing you leave a company for. But not normally a whole industry.
What about the guy who put wood chips in his pellet bbq though
Good comedy is always appreciated
See: "multiple companies". Bit of a theme that I was sick of. thanks for your insight though.
It is simple with both brakes. It’s also simple with just the front brake or just the rear. If you do just the rear, then yes you get 63% on the rear wheel, which consequently accounts for 63% of your max braking force/0.315 g if μ is 0.5. So I’m assuming you calculated 0.32 and accidentally used 0.23 g. If the same weight distribution is used for front brake only then you get 30% of the weight on the front wheel. So given that weight distribution I’d 100% go with rear only over front only if μ is 0.5.
Just reorganised this to make it clearer about flat vs slope.
So we agree that 20 degrees gives you 0.34g needed. But from there you're doing it very differently. No I haven't made a typo with 0.23g.
A 0.34G force couple would result in a forward weight shift moment of acceleration * mass * COG height.
That is resisted at the wheels by extra weight on the front and less on the back. Ratio of COG/Wheelbase. Which is conveniently 0.5 (0.65/1.299).
Weight shift = acceleration*mass/2
= 0.34*86/2
=14.6kg. Call it 15.
Flat weight distribution was 22/64kg
Back only braking on the flat gives new weight distribution becomes, 49kg rear, 37kg front.
But do we have enough traction for that?
0.34G needs 29kg of traction.
49kg on the rear with a friction coefficient of 0.5 only gives us 24.5kg traction.
So the rear brake can't produce enough traction on the flat to hold us on a 20 degree slope. It can't produce 0.34g. It's closer to 0.28g,
On the flat.
On a slope the weight moves further forward simply because the COG is above ground. This reduces the rear weight and traction further. I'll get into that later.
I dunno about you but when Im riding steeps my goal isn’t to come to a complete stop. Usually its just to manage speed and braking zones are chosen based on flatter areas or higher traction surfaces.
Sorry but you’re lost in the sauce. Rear wheel normal force multiplied by coefficient of friction is always going to be equal to max braking force from the rear wheel and that is your braking force applied at the CG. There are zero iterations required to solve for front and rear wheel normal forces regardless of whether it’s front and rear, rear only, or front only. Weight distribution percentages do not change with slope angle in any of those scenarios. There is no getting into that later. It just doesn’t happen. It is clear that m, g, and cos(theta) all cancel out. Braking force available decreases with slope angel which is why slope angle factors out of weight distribution. This is all there is to it for rear brake only:
You're correct for a static situation. But braking is not static, it's dynamic and you need to address the weight shift.
I'm walking you through flat braking. Ignore slope. The only slope introduced was to quantify the braking acceleration needed and We agree on that.
Here it is again. Tell me which part you disagree with.
Start with 64/22kg F/R weight distribution on the flat. COG 650, WB 1299, COF 0.5.
Haul on the back brake, weight shifts forwards.
Max back braking can potentially give you 32kg braking force (0.37g).
But a 32kg braking force causes a forward weight shift, moving 16kg from your back wheel to front wheel and reducing rear traction.
First iteration would have 48kg on the back wheel and 38 on the front.
48kg reduces your braking traction to 24kg which reduces max deceleration to 0.28g.
Iteration further, results converge on a balanced dynamic situation.
Final result is an almost 13kg weight shift, 51/35kg R/F weight distribution with 2.9m/s^2 (0.3g) braking rear only on the flat.
I did just find and fix 2 errors in my spreadsheet just now. They almost cancelled out but threw previous results a bit.
I guess if I had to pick something the very first thing would be that we were talking about a 78% rear wheel bias but then suddenly it changed. You mentioned d'Alembert’s principle. That is exactly what this evaluation is. It’s solving for a new weight distribution given an initial weight distribution and friction and assuming constant deceleration is possible. The end result is exactly when deceleration available is deceleration used. It is not solving for slope angle you can hold the bike on, whether or not the bike is accelerating or decelerating, or anything like that. Purely is it at a steady state for weight distribution.
I got interrupted editing the post above and just finished, feel free to check it again.
Ignore all slopes. Flat braking 64/22kg (74% rear).
Final stable deceleration is 0.3g with the back brake only, giving you 51/35kg weight distribution on the flat.
Which is exactly what that equation spits out… it doesn’t require any iteration and, again, does not depend on slope angle.
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