Help with my 8inch?

CHEWY

CHEWY

Eats Squid
Second reply to this: I think your ratios are wrong. The leverage ratio is not the ratio of the axle-caliper vs caliper-rim distance, because if the radius of the rim is 12" and you have a 6" DIAMETER rotor, your ratio is only 3:1 (3 being the rim, 1 being the caliper). However, it should be 4:1.
Braking ratios in relation to the rim involve the radius of the braking surface divided by the radius of the WHEEL (including tyre). The rim is roughly 11" radius from the axle, whereas the tyre is 13" from the axle (on a 26" wheel). Using your force X etc:
- The force required by a rim brake to provide X braking force at the tyre is the inverse of 11/13 "X", which is 13/11 "X"
- The force required by an 8" rotor to provide X braking force at the tyre is the inverse of 4/13 "x", which is 13/4 "x".
- The force required by a 6" disc rotor to provide X braking force at the tyre is the inverse of 3/13 "x", which is 13/3 "x"

Let X = 1, for the sake of simplicity:
13/11 x = 1.18
13/4 x = 3.25
13/3 x = 4.33
Click to expand...
always wondered why the hell they taught us algebra. it does have a use... wow
 
F

frank-oi

Likes Dirt
ok don't roll backward fast and jam on the brakes .that also helps stop the tabs snaping off .seen it happen.

now everyone can go back and stick there noses into there phiysics books again and figure out something else to confuse me with. gezz.
 
S.

S.

ex offender
CHEWY said:
Second reply to this: I think your ratios are wrong. The leverage ratio is not the ratio of the axle-caliper vs caliper-rim distance, because if the radius of the rim is 12" and you have a 6" DIAMETER rotor, your ratio is only 3:1 (3 being the rim, 1 being the caliper). However, it should be 4:1.
Braking ratios in relation to the rim involve the radius of the braking surface divided by the radius of the WHEEL (including tyre). The rim is roughly 11" radius from the axle, whereas the tyre is 13" from the axle (on a 26" wheel). Using your force X etc:
- The force required by a rim brake to provide X braking force at the tyre is the inverse of 11/13 "X", which is 13/11 "X"
- The force required by an 8" rotor to provide X braking force at the tyre is the inverse of 4/13 "x", which is 13/4 "x".
- The force required by a 6" disc rotor to provide X braking force at the tyre is the inverse of 3/13 "x", which is 13/3 "x"

Let X = 1, for the sake of simplicity:
13/11 x = 1.18
13/4 x = 3.25
13/3 x = 4.33
Click to expand...
always wondered why the hell they taught us algebra. it does have a use... wow
Click to expand...
This is pretty simple stuff if you think about it a bit, it doesn't involve any tricky formulas or that. Supposedly this is the stuff that a 14 yr old kid should be able to do... it's just ratios.
 
S.

S.

ex offender
belly_up said:
you can work the load as being the tyre or the caliper, as long as you're consistent it will still work out ok, just a conceptual difference.
And you're right about my formula not being right, i musn't have been thinking that day. but i think it still worked out ok because i was just calculating the percentage difference between 6" and 8" and rim (and using the same formula for each), not trying to work out actual load values. then i would have been stuffed.
Click to expand...
Yeah true, load and force are usually interchangeable unless you have a multi-pivot lever like a suspension linkage... and even then it doesn't matter much. It's just a way of distinguishing between two opposing forces by using different names for each.
 
Scott

Scott

bAdmin
Staff member
Mountain bikers shouldn't have to think that much.. there should be a law against it!
 
K

kalem

Likes Bikes and Dirt
Steve you're a smart young lad aren't you! Well all i have to say is i'm getting a 203mm avid rotor for my bike and i don't care what anyone says!!!!!
 
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