AHAHAHAHHAHA, I CAN'T BELIEVE THIS THREAD IS STILL GOING, SORT IT OUT YOU BUNCH OF **** LOSERS!!!!!
Bike Forum
Why are 180mm rotors more powerful than 160mm?
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Posted 10 months ago #
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Don't forget that a bigger disc is heavier, completely negating any stopping advantage it may offer....
DrP
(you know I'm playing, right...?!)
Posted 10 months ago # -
Double arse, I think I've convinced myself of your theory now pdw.
Definitions:
1) r = disc radius
2) Ff = frictional force = Force applied by pads x Coeef Friction
3) SL = Swept length = circumf of disc (or effective circumf of disc which might be half way across the friction surface) for one rev is 2 x pi x r
4) T = torque = Ff x r
5) Wf = Work done by Friction = Ff x Sl
6) Av = angular velocity - for 1 rev is 2pi/t
7) P= Power = T x Av
8,) Wt = P x time = work done by torque.Wf = Ff x 2 x pi x r
Wt = Ff x r x (2 x pi)/t x tThe t's cancel and so work done by friction = work done by torque.
Funny how as soon as you sit down and write it out, it all makes sense..
Doh.The friction on the disc x the dia is what makes the torque.The work is that torque (or friction) applied for a period of time ( which equates to a distance travelled over the disc)
Posted 10 months ago # -
That was a good bout, well played everybody, well played. Good to get the brain working once in a while.
(high five pdw)
Posted 10 months ago # -
In future I'm saying anything until I've proved the maths to myself..
Posted 10 months ago # -
bigger disks are more powerful because they cost more. if they weren't more powerful nobody would bother paying the extra would they?
Posted 10 months ago # -
how do you guys find helmets big enough to fit all them brains in?
Posted 10 months ago # -
i use a modified dustbin
Posted 10 months ago # -
I used to brake my bmx with my trainers, would a bigger shoe-size have provided more torque i wonder?
Posted 10 months ago # -
I did the same sunnrider - did you use sintered or organic compound soles?
Posted 10 months ago # -
I think I need a smaller lid after that..
In fact I have realised that the swept length thing is a massive red herring. As assuming everything else is equal the increase in torque will quite obviously reduce the angular displacement to stop the bike by a factor which will make the swept length exactly equal for any size disc.
Posted 10 months ago # -
As assuming everything else is equal the increase in torque will quite obviously reduce the angular displacement to stop the bike by a factor which will make the swept length exactly equal for any size disc.
But to sweep a given area on a disc with a greater circumference means the wheel hasn't rotated as much...
Posted 10 months ago # -
I don't understand your statement? Can you explain it a bit more?
edit actually i think i do.
Thats exactly what I mean - "reduced angualr dispalcemnt" means "hasn't rotated as much"
Posted 10 months ago # -
Yes I think I just reversed your statement, probably because in true STW style I hadn't read it properly.
Posted 10 months ago # -
Yes. The pad on a bigger rotor covers more distance on each wheel rotation, so for a given force it produces more heat, meaning you must have slowed down more
But if you have stopped in a shorter distance with the bigger rotor then does that not negate the difference in distance travelled by the rotor?
Posted 10 months ago # -
But if you have stopped in a shorter distance with the bigger rotor then does that not negate the difference in distance travelled by the rotor?
edit. I did agree but I've gone fuzzy headed. Bollocks, I need to think about this more..
pdw whats your answer to this?
Posted 10 months ago # -
its angular delta will remain the same as the wheel
Posted 10 months ago # -
composite pro, I had hoped that was implicit in what I said before..
Posted 10 months ago # -
Bigger distance with a smaller rotor, or shorter distance with a bigger rotor, I think for a given force at the calipers, the pads will have covered the same distance over the disc, as you've lost the same amount of kinetic energy either way.
Posted 10 months ago # -
this is the same conclusion that I came to above and the way you have written it makes it seem sensible. If its shorter distance (or less revolutions) to come to a stop this means you come to a stop faster, this is exercising me slightly..
Posted 10 months ago # -
If it takes one rev to stop and its doing 1 rev a second then the equation I listed above works fine, the time t which divides through the 2 pi radians is equal to the time t for the work done.
Wt = Ff x r x (2 x pi)/t x t
So for a 180 disc the term r goes up and hence I had assumed the 2 pi is multiplied the ratio of the radii (80/90) = 0.88, but actually I think both the number of radians goes down and the time goes down so its probably not the same distance travelled..
Am I making sense?
I think this is possibly only solved by integration.Posted 10 months ago # -
the revolution (or rotation) is the same on either sized rotor (160 or 180), unless I am missing something drastic, which is quite possible.
Posted 10 months ago # -
it's summink to do with physics innit..?
Posted 10 months ago # -
Heechee. If a larger disc has more torque then it will stop your bike faster. Agreed? So if it stops faster the total number of revolutions from when you apply the lever till when it stops is less. Agreed?
Posted 10 months ago # -
I agree with the larger diameter increases torque. Not so sure the revolution or rotation is any different. I am no expert though, and would take no offense at being proven wrong.
Posted 10 months ago # -
YAWN!!!
Posted 10 months ago # -
Easy explanation is leverage applied to the disc verses the wheel size
Put massive 19" wheels on your chav corsa the brakes feel rubbish because the big wheel has more leverage against the disc go back to a smaller wheel 15" and the leverage on the disc is less so the brakes feel more powerful
Swap this around to fitting a bigger disc for the same size wheel the leverage is the same at the wheel but the bigger disc doesn't work as hard for the same braking effort at the caliper
Effectivly giving you more available power only to the point of how much grip you have before you lock up
Also bigger discs get rid of the heat better giving better braking as everthing heats up the disc can keep cooler as thereis more material to disapaite the heat
Didn't read all four pages so I may have missed the point
Posted 10 months ago # -
Toys - your equation shows that work done in a revolution is proportional to disc radius. Or to look at it another way, it's proportional to the swept distance i.e. The distance that the pad covers around the circumference of the disc. To stop a bike, you need to do the same amount of work however you do it, so whatever the disc size, the pads need to sweep the same distance. Which means more revolutions of a smaller disc i.e. A longer stopping distance.
Posted 10 months ago # -
pdw yeah I understand that, but look at the time taken to stop. In the work done equation there is a time term t, if you reduce the number of rotations by using a bigger disc, then it takes less time (or increase by using a smaller disc.)
All I am saying is that at first I thought I could equate the work done equations for a 160 and a 180 disc to find the reduced number of revolutions, (which is obviously the ratio between the disc radii) but actually you have two unknowns, the change in revolutions and the change in time, so my crap maths doesnt work. Needs integration, or possibly an approximation method. Its obviously too dull for anyone else but it has just interested me a bit..
Posted 10 months ago # -
Jebus! That's 5 min of my life I won't get back, and I was only skimming too!
They stop you quicker, cos they're bigger.
Night night.Posted 10 months ago # -
HOW DID THIS GET TO FOUR PAGES?
Only skimmed the junk but there's more made-up 'science' here than in a creationist's textbook.
Posted 10 months ago # -
LOCK THIS THREAD PLEASE, IT IS DOING MY HEAD IN!!
Posted 10 months ago # -
pdw yeah I understand that, but look at the time taken to stop. In the work done equation there is a time term t, if you reduce the number of rotations by using a bigger disc, then it takes less time (or increase by using a smaller disc.)
I'm sure no one will thank me for resurrecting this thread but...
There shouldn't be a time term in the work done equation. In the equation you wrote, the time terms cancel out. It really is just force * distance, which in the case of a disc is force * radius * 2 * pi * number of revolutions.
Posted 10 months ago #
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