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Discussion Starter #121 (Edited)
If you want to proceed, can you explain;
From where does this magical load "more than 33.3%" come from?
I answered that too. You must have missed it, forum collapses long text. So here it is with your questions, answers in bold.

1. When traveling at constant speed:
Load on rear 50%, load on front 50%
Do you agree? Modern sportbikes are front biased 51-53%F to 49-47%R, but for the sake of simplicity 50/50 is fine.

2. Maximum front braking:
Load on rear 0%, load on front 100%
Do you agree? For stable braking, rear needs to be on the ground, so load won't be 100% on front. You don't see pro riders raising the rear at the end of every straight. So for realistic track riding disagree, from a theoretical physics perspective agree.

3. Maximum rear braking:
Load on rear 33.3%, load on front 66,7%
Do you agree? Your ratios are for stopping power, not load. Look at your 50/50 weight distribution above. Load on rear will be more than 33.3% with max rear braking.

In case number#3 was not clear. Assume bike weight 200kg, rider 100kg. With max rear braking, do you think the total load on rear wheel is only 33.3% of 300kg total mass?
 

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3. Maximum rear braking:
Load on rear 33.3%, load on front 66,7%
Do you agree? Your ratios are for stopping power, not load. Look at your 50/50 weight distribution above. Load on rear will be more than 33.3% with max rear braking.

In case number#3 was not clear. Assume bike weight 200kg, rider 100kg. With max rear braking, do you think the total load on rear wheel is only 33.3% of 300kg total mass?
Simplified model:
Wheelbase 1400 mm
CoG height 700 mm
Weight distribution 50% / 50%
Total mass 300 kg

(You should use Newtons, but lets keep this simpler)

Constant speed
Load R 150 kg
Load F 150 kg

Maximal rear braking, longitudinal acceleration/ deceleration 3.33 m/s^2
Load R: 150 - 3.33*(700/1400)*150 = 100 kg
Load F: 150 + 3.33*(700/1400)*150 = 200 kg

Load R 100/300 = 33.3%

Questions?
 

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Waiting on L74 to confirm. Plan to go with their underslug caliper/bracket (surprised it works with OEM and aftermarket wheels unlike Brembo's), and HEL thumb brake fork mounted.
IMA TBS was my 2nd choice after learning that it is not as compact as HEL one. And its installation is harder, requires moving OEM MC to thumb, IMA MC at foot lever.



I checked those sections, you are being vague. Either help show exactly the gaps in my knowledge, or there is no point. You wrote way more about puppies than anything helpful.
Here is the guy I picked up from the shelter recently. Maybe he will motivate you to be explicit and helpful on using the rear brake.

Are you against using it because it is not needed? Or due to extra time needed to adjust to it? Something else?

View attachment 214509
good looking pup.
 

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Discussion Starter #124
Simplified model:
Wheelbase 1400 mm
CoG height 700 mm
Weight distribution 50% / 50%
Total mass 300 kg

(You should use Newtons, but lets keep this simpler)

Constant speed
Load R 150 kg
Load F 150 kg

Maximal rear braking, longitudinal acceleration/ deceleration 3.33 m/s^2
Load R: 150 - 3.33*(700/1400)*150 = 100 kg
Load F: 150 + 3.33*(700/1400)*150 = 200 kg

Load R 100/300 = 33.3%

Questions?
Your math is wrong: (notice how I explain why you were wrong, I don't keep repeating it that you were wrong). I might be wrong too, but at least everyone can see the logic behind our answers.

Load F: 150 + 3.33*(700/1400)*150 = 400 kg, NOT 200kg. So it makes your calculation wrong on rear. Going by your calculation if 400kg is used for load on front, then load on rear would be 100/500 = 20%, not 33.3% as you mentioned (rear/total load on F + R). But I can see where you are going, I used the below to make it clear:

214698

214695

214697


wb = 1400
x = 700
h = 700

R = 300*(700/1400) + (300*(700/1400)* 3.33) = 650kg
F = 300*(700/1400) - (300*(700/1400)*3.33) = 350kg

Load on rear 650/1000 = 65%. Which goes back to what I said, load on rear would be greater than what you said. Questions?

Can we proceed now? How does the rear brake deceleration affect load transfer, increasing load on the front?
 

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How do you see this affecting mid corner and load transfer?
It goes against from what I observed from dyno video I posted. Does this mean the rear compresses under both acceleration or braking? I thought only under acceleration. It pulls down under braking.
[/QUOTE]

This does not show load transfer of course, but it does show the suspension compressing with use of the rear brake. Note that the bike is upside down in the video, so yes the suspension in that video is compressing with the rear brake.
 

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Your math is wrong: (notice how I explain why you were wrong, I don't keep repeating it that you were wrong). I might be wrong too, but at least everyone can see the logic behind our answers.

Load F: 150 + 3.33*(700/1400)*150 = 400 kg, NOT 200kg. So it makes your calculation wrong on rear. Going by your calculation if 400kg is used for load on front, then load on rear would be 100/500 = 20%, not 33.3% as you mentioned (rear/total load on F + R). But I can see where you are going, I used the below to make it clear:

wb = 1400
x = 700
h = 700

R = 300*(700/1400) + (300*(700/1400)* 3.33) = 650kg
F = 300*(700/1400) - (300*(700/1400)*3.33) = 350kg

Load on rear 650/1000 = 65%. Which goes back to what I said, load on rear would be greater than what you said. Questions?

Can we proceed now? How does the rear brake deceleration affect load transfer, increasing load on the front?
Yeah, there was a couple of errors in text in equation, because I tried to make it simpler and forgot to change it to the text.

So, corrected version:

Simplified model:
Wheelbase 1400 mm
CoG height 700 mm
Weight distribution 50% / 50%
Total mass 300 kg

(You should use Newtons, but lets keep this simpler and use kg, in that case 3.33 m/s^2 = 0.333 G)

Constant speed
Load R 150 kg
Load F 150 kg

Maximal rear braking, longitudinal acceleration/ deceleration 3.33 m/s^2
Load R: 150 - 0.333*(700/1400)*300 = 100 kg
Load F: 150 + 0.333*(700/1400)*300 = 200 kg

Load R 100/300 = 33.3%

Questions?
 

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Discussion Starter #128
Your math is still not adding up. Have you seen equations and results from my math? And from dyno and bike video where the rear compresses with rear brake? Making wheelbase longer.
Anyway, let's move on to my question how does the rear brake deceleration affect load transfer, increasing load on the front?
 

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Your math is still not adding up. Have you seen equations and results from my math?
You used erroneous equation.
For example, ask yourself, if total mass is 300 kg, how can load be more than 300 kg? That should have make you notice that there is something wrong in it.
And from dyno and bike video where the rear compresses with rear brake? Making wheelbase longer.
Yes, rear brake also causes other forces and things, not only load transfer.
Anyway, let's move on to my question how does the rear brake deceleration affect load transfer, increasing load on the front?
Rear and/or front brake > Deceleration > Load transfer > Increases load on front.

Questions?
 

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Discussion Starter #130
I was using the same accel multiplier you used. With .333 our results are the same but reversed.
Front load: You added, I subtracted.
Rear load: you subtracted, I added.

Final ratio 66% to 33%. Depending on acceleration or braking.

So since the rear shock compresses with rear brake, why do you think it increases load on the front? Look at dyno video again, see what happen when throttle is rolled off. Or watch a video of bike with rear brake only applied. Show how load on front is increased.

On corner entry, most load is on the front during hard braking phase, with braking reduced, replaced with lean angle load, what would adding rear brake in that point do to load transfer and bike steering (tighter line, no effect, wider, etc...)?
 

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I was using the same accel multiplier you used. With .333 our results are the same but reversed.
Front load: You added, I subtracted.
Rear load: you subtracted, I added.

Final ratio 66% to 33%. Depending on acceleration or braking.

So since the rear shock compresses with rear brake, why do you think it increases load on the front? Look at dyno video again, see what happen when throttle is rolled off. Or watch a video of bike with rear brake only applied. Show how load on front is increased.
Lets take this question first.

Rear brake also causes other forces and things, not only load transfer.

If load transfer would increase rear load when braking, you could brake as efficiently as with front brake.

 

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Discussion Starter #133
Rear brake also causes other forces and things, not only load transfer.

If load transfer would increase rear load when braking, you could brake as efficiently as with front brake.
Rear would never brake as efficient as the front due to inherit design, we all agree the rear has ~34% of braking force, front ~66%, so it cannot add much force to neither front nor rear.
How much load does the rear brake alone add on the front of bike? Non linked brake bike (front & rear brakes not linked).
 

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Rear would never brake as efficient as the front due to inherit design,
If load transfer would increase rear load when braking, you could brake as efficiently as with front brake and rear brake design would support that.
we all agree the rear has ~34% of braking force, front ~66%, so it cannot add much force to neither front nor rear.
Front has 100%.
How much load does the rear brake alone add on the front of bike? Non linked brake bike (front & rear brakes not linked).
Constant speed
Load R 150 kg
Load F 150 kg
Maximal rear braking, longitudinal acceleration/ deceleration 3.33 m/s^2
Load R: 150 - 0.333*(700/1400)*300 = 100 kg
Load F: 150 + 0.333*(700/1400)*300 = 200 kg

Load F 150 >> 200 = +33.3%
 

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Discussion Starter #137
If load transfer would increase rear load when braking, you could brake as efficiently as with front brake and rear brake design would support that.

Front has 100%.

Constant speed
Load R 150 kg
Load F 150 kg
Maximal rear braking, longitudinal acceleration/ deceleration 3.33 m/s^2
Load R: 150 - 0.333*(700/1400)*300 = 100 kg
Load F: 150 + 0.333*(700/1400)*300 = 200 kg

Load F 150 >> 200 = +33.3%
how would rear brake support such large load? It has a smaller rotor and 2 pistons.

What if you swap +/- Load R with Load F

Load R: 150 + 0.333*(700/1400)*300 = 200 kg
Load F: 150 - 0.333*(700/1400)*300 = 100 kg

So now Load F 150 >>100 = 0%
 

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how would rear brake support such large load? It has a smaller rotor and 2 pistons.

What if you swap +/- Load R with Load F

Load R: 150 + 0.333*(700/1400)*300 = 200 kg
Load F: 150 - 0.333*(700/1400)*300 = 100 kg

So now Load F 150 >>100 = 0%
Well yeah, if you want to decline Newtons first law aka inertia, then you can swap +/- how ever you want. I wouldn't.

Also check page 97 for info/ which way to put +/-.
"Load transfer during the braking "
"The dynamic load on the front wheel is equal to the sum of the static load and of the load transfer "
"while the dynamic load on the rear wheel is equal to the difference between the static load and the load transfer "
 

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Discussion Starter #139
@speedfinn How would the front have 100% of braking force if a motorcycle stops faster using front and rear brakes? Slower with just the front brake?

Can you answer the question about the rear compressing when rear brake applied?

Load F 150 >> 200 = +33.3%, load on front increased by 33.3%, should not it be 25%?
 

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@speedfinn How would the front have 100% of braking force if a motorcycle stops faster using front and rear brakes? Slower with just the front brake?
Typical track bike/600/1000 does not stop faster using front and rear brakes.
Typical track bike/600/1000 stops faster using front brake only.
Check msg #105:
Can you answer the question about the rear compressing when rear brake applied?
Rear brake also causes other forces and things, not only load transfer.
And if bike (mass) is not moving, then there is no load transfer at all, when braking, because there is no decelerating mass.
So other forces than load transfer are topmost.
 
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