Help with my 8inch?

we've gone through this before, firstly, 8 inch rotors place more stress on parts than a 6 inch rotor, thats why they provide stronger braking (equal and oppsite forces etc etc) so from what i've gathered it's a bad idea to run a 8 inch rotor on a standard 9 mm drop out single crown fork the reason for this is that there is too much twisting force from a 8 inch rotor with normal drop outs, get a mate to hold a wheel and spin in slowly, grab the spokes on one side and try to stop it, the wheel will twist in one direction, this twist makes the brake tabs pull/push at the wrong angle (not stright) and they can tear off or can twist the fork enough to snap weaker archs (like found on the EXR), with triple clamp forks the aid of the top crown reists some of this twishing force so you could run a 8 inch with say a JrT with standard drop outs, or on a single crown fork with bolt on axels the 20 mm axel helps stop some of the twisting force so it would be safe to use also. part of this was told to me buy the guys at dirt works in reference to Fox Vanillas which are a fair bit tougher than maz EXRs.
 
My personal opinion is unless you are DH racing, there is no need for them. Seeing as it is on a exr, I would say stick with a 6inch
 
if your riding hard enough to require a 8 inch rotor your probably pushing the limits of the EXR as turley suggested, but on the flip side, it's your bike, do what you want with it.
 
Ty` said:
we've gone through this before, firstly, 8 inch rotors place more stress on parts than a 6 inch rotor, thats why they provide stronger braking (equal and oppsite forces etc etc) so from what i've gathered it's a bad idea to run a 8 inch rotor on a standard 9 mm drop out single crown fork the reason for this is that there is too much twisting force from a 8 inch rotor with normal drop outs, get a mate to hold a wheel and spin in slowly, grab the spokes on one side and try to stop it, the wheel will twist in one direction, this twist makes the brake tabs pull/push at the wrong angle (not stright) and they can tear off or can twist the fork enough to snap weaker archs (like found on the EXR), with triple clamp forks the aid of the top crown reists some of this twishing force so you could run a 8 inch with say a JrT with standard drop outs, or on a single crown fork with bolt on axels the 20 mm axel helps stop some of the twisting force so it would be safe to use also. part of this was told to me buy the guys at dirt works in reference to Fox Vanillas which are a fair bit tougher than maz EXRs.
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I tend to disagree with that.... remember that the rotor is bolted to the hub, not the spokes. Unless you managed to flex/twist your entire hub, the wheel isn't going to twist because of that.... what COULD cause it to twist however, is the leverage (pivoting at the caliper) from the wheel causing the fork leg to flex backwards. But in reality, I've NEVER felt any twist in my forks due to braking - I think it's 99% BS really.

More dangerous than that is the possibility of tearing your wheel out of your dropouts. This can happen when the wheel, pivoting at the caliper (if the caliper grabs the rotor and doesn't let it go) and being forced at the bottom of the wheel (contact patch with the ground), is pulled with more force than your QR can withstand, and tears your axle out of the drops and forces it into the down tube. Don't believe me? Try this simple and relatively safe test: turn your bike upside down, take your QR skewer out of the axle, and have the wheel sitting on the dropouts, only held in by its own weight. Spin the wheel fast, and suddenly jam on the front brakes. The wheel will try to leap out of the dropouts. Alternatively, lock the brakes fairly gently (with the wheel stopped) and push backwards on the wheel (while still in the same setup as before - bike upside down, skewer out etc). As you will see, the bike uses the rotor/caliper interface as a pivot, and the axle comes out of the dropouts.

However, given that all forks nowadays have lawyer tabs (those little rings sticking out around the edge of your dropouts which force you to unscrew your QR skewer slightly before you can take your wheel out of the drops), you'd have to break your QR off before the wheel could come out (assuming it's done up in the first place).

Another thing: basic leverage ratios (in this case, wheel contact patch to caliper distance versus axle to caliper distance) tell us that a higher leverage ratio will be more likely to pull the wheel out of the dropouts. In other words, the bigger the relative difference between contact patch to caliper distance and axle to caliper distance, the more likely the axle is to be pulled out of the dropouts. Theoretically, this would mean that an 8" rotor is actually LESS likely to pull the wheel out of the dropouts than a 6" rotor, except for the fact that the tangent of the radius from caliper to axle is closer to vertical with an 8" rotor (in other words, the angle from contact patch to axle to caliper is closer to 90 degrees).

Keep in mind that you also have more leverage on an 8" adaptor than a 6" one, and if the fork disc tabs aren't designed to take the stresses (in relation to the amount of force and its direction) of an 8" rotor, they may break off. However, I would be reasonably confident that the tabs are pretty damn strong and can hold up to 8" rotors.

All in all, there is likely to be little or no difference to safety or anything when using 8" rotors instead of 6". The only thing likely to cause a difference is that you might be tempted to go harder with a stronger brake... causing you to run your fork into something you wouldn't otherwise ;). Which ultimately asks the question: do you really need an 8" rotor on a cross-country-intended fork (even though I know Kona spec them on jump bikes)?

Wow, that took forever. Sorry for the long-winded post!
 
for the benefit of the other farkin members (cause you all want to hear what i've got to say :D ) this is what i posted in reply over on TRC:

The load is being applied from the rim, pivoting around the axle, with the braking force in between, making it a class 3 lever (http://www.professorbeaker.com/lever_fact.html). The closer the you move the braking force towards the rim, the less force you need to apply to offset a given load. Thus why V-brakes don't need to apply anywhere near as much force to the rim as a disk brake caliper does to it's disk, to achieve similar stopping power. The leverage ratio is determined by the ratio of the pivot to the caliper, and the caliper to the rim. so a 6 inch rotor has a leverage ratio of 10/3, an 8 inch rotor has a ratio of 9/4, and a rim brake has a leverage ratio of 0 ie, the braking force is acting directly on the opposing force, without any lever.
To provide a given rate of deceleration from a given speed, a force, "x" is required. a rim brake needs to apply "x" amount of force to slow the bike, an 8 inch rotor needs to apply 9/4 * "x" = 2.25 "x", and a 6 inch rotor needs to apply 10/3 "x" = 3.33 "x".
So the reason that DH bikes use an 8 inch rotor is that, assuming the caliper is the same as the 6in version (which it normally is), you can effectivly apply 48.1% more stopping power to the tyre.
As for the brake tabs, because the caliper is mounted 1in further out, there's another lever effect happening, and by my quick mental calculations, you can put about 33% more force through the mounting tabs, although as chester pointed out, as long as the bike is going forward the tabs should only really be in compression anyway.
So thats what i think anyway... :D
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Ty` said:
socket, point taken
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I wrote a bloody million word essay and you have a one line reply? :twisted:

Bastard :lol:

Belly_up: Exactly. I hadn't bothered to work out the ratios though.
 
belly_up said:
for the benefit of the other farkin members (cause you all want to hear what i've got to say :D ) this is what i posted in reply over on TRC:

The load is being applied from the rim, pivoting around the axle, with the braking force in between, making it a class 3 lever (http://www.professorbeaker.com/lever_fact.html). The closer the you move the braking force towards the rim, the less force you need to apply to offset a given load. Thus why V-brakes don't need to apply anywhere near as much force to the rim as a disk brake caliper does to it's disk, to achieve similar stopping power. The leverage ratio is determined by the ratio of the pivot to the caliper, and the caliper to the rim. so a 6 inch rotor has a leverage ratio of 10/3, an 8 inch rotor has a ratio of 9/4, and a rim brake has a leverage ratio of 0 ie, the braking force is acting directly on the opposing force, without any lever.
To provide a given rate of deceleration from a given speed, a force, "x" is required. a rim brake needs to apply "x" amount of force to slow the bike, an 8 inch rotor needs to apply 9/4 * "x" = 2.25 "x", and a 6 inch rotor needs to apply 10/3 "x" = 3.33 "x".
So the reason that DH bikes use an 8 inch rotor is that, assuming the caliper is the same as the 6in version (which it normally is), you can effectivly apply 48.1% more stopping power to the tyre.
As for the brake tabs, because the caliper is mounted 1in further out, there's another lever effect happening, and by my quick mental calculations, you can put about 33% more force through the mounting tabs, although as chester pointed out, as long as the bike is going forward the tabs should only really be in compression anyway.
So thats what i think anyway... :D
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Actually, now that I think about it, where did you get the ratios 9/4 and 10/3? There should be a constant in there somewhere (the distance from axle to rim).... I'm assuming that the 3 (in the ratio 10/3) represents the 3" radius of a 6" rotor, and the 4 (in the ratio 9/4) represents the 4" radius of an 8" rotor.

Edit: Another thing, the axle isn't really the ultimate point of reference for forces in regards to twisting and stuff, because the fork is mounted to the frame at the steerer tube, not the axle... ie when you're braking hard, it's putting force into the steerer/crown area, using the fork and wheel as a 2ft long lever. If the fork isn't designed to be super strong at that interface, heavy braking could induce the notorious creaking problem in the crown, or if you hit a sudden, sharp bump (like a kerb) under hard braking, you might put more force on the steerer than it can handle, resulting in considerable amounts of pain...
 
ok now i think after all this debateing that an 8inch rotor just..... just won't do it :(
thanks guys for your help, i wouldnt have thought of ANY of that shit u guys were saying........
so i guess i'll have to get rid of this 8inch hayes rotor :(
 
peachy said:
ok now i think after all this debateing that an 8inch rotor just..... just won't do it :(
thanks guys for your help, i wouldnt have thought of ANY of that shit u guys were saying........
so i guess i'll have to get rid of this 8inch hayes rotor :(
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NO! Don't ever get rid of rotors if they're useable! Keep it as a spare, one day you or a mate will need it, and trust me you won't get anything like retail cost if you sell it. They are about the best spare part to have around, even if they don't fit your current bike. Seriously, you'll get some use out of it somehow...
 
And just in case there's anyone else out there that hasn't been beaten into submission by the 8 vs 6 inch rotor discussion... remember this...
6 inch rotors are almost always less expensive than 8, and are less likely to hit obstacles and thereby getting bent/warped/dented/etc....
 
Yea,so ummm, Im not quite as 'technicaly' minded as you guys but what are you trying to say?

8 inch yes? no? same as 6 inch

cheers :)
 
Insane said:
Yea,so ummm, Im not quite as 'technicaly' advanced as you guys but what are you trying to say?

8 inch yes? no? same as 6 inch

cheers :)
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Realistically, an 8" rotor isn't going to cause much problem. Except that you're more likely to get in over your head (ie going too fast) and damage the fork like that. You'd have to be pretty unlucky though.
 
belly_up said:
for the benefit of the other farkin members (cause you all want to hear what i've got to say :D ) this is what i posted in reply over on TRC:

The load is being applied from the rim, pivoting around the axle, with the braking force in between, making it a class 3 lever (http://www.professorbeaker.com/lever_fact.html). The closer the you move the braking force towards the rim, the less force you need to apply to offset a given load. Thus why V-brakes don't need to apply anywhere near as much force to the rim as a disk brake caliper does to it's disk, to achieve similar stopping power. The leverage ratio is determined by the ratio of the pivot to the caliper, and the caliper to the rim. so a 6 inch rotor has a leverage ratio of 10/3, an 8 inch rotor has a ratio of 9/4, and a rim brake has a leverage ratio of 0 ie, the braking force is acting directly on the opposing force, without any lever.
To provide a given rate of deceleration from a given speed, a force, "x" is required. a rim brake needs to apply "x" amount of force to slow the bike, an 8 inch rotor needs to apply 9/4 * "x" = 2.25 "x", and a 6 inch rotor needs to apply 10/3 "x" = 3.33 "x".
So the reason that DH bikes use an 8 inch rotor is that, assuming the caliper is the same as the 6in version (which it normally is), you can effectivly apply 48.1% more stopping power to the tyre.
As for the brake tabs, because the caliper is mounted 1in further out, there's another lever effect happening, and by my quick mental calculations, you can put about 33% more force through the mounting tabs, although as chester pointed out, as long as the bike is going forward the tabs should only really be in compression anyway.
So thats what i think anyway... :D
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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

As you can see from these figures, a 6" rotor requires (4.33/1.18 = 3.67) times as much force at the caliper to provide the same actual braking force as a rim brake, and an 8" rotor requires (3.25/1.18 = 2.75) times as much force at the caliper to provide the same actual braking force as a rim brake. Therefore, an 8" rotor provides 33.33% more braking power at the tyre (for a given braking power at the caliper) than a 6" rotor.

As far as adaptors goes: If you're using a Hayes post-mount adaptor on an IS fork tab, you have roughly 1" of leverage to begin with. Add one inch, and you've doubled the leverage on the tabs (that's debatable though, due to the angle and position that the caliper is attached at, shear forces vs leverage forces etc). Let us assume that for the sake of realistic argument, we have gone from 1" of ideal leverage via the tabs, to 1.8" of ideal leverage. However, the 8" rotor puts 33% LESS force into the caliper than a 6" rotor (because torque = force x length of lever, if the torque is constant and the length of the lever increases, the amount of force must decrease in proportion to the increase in lever length). So 0.67 x 1.8 = 1.206 times the amount of force being applied to the tabs. Not much extra really.

Edit: fixed a mistake
 
i'd forgotten my high school physics, so i went and looked up some sites, and they said for that class of lever ie:
pivot------load--------force
that the leverage ratio was equal to the force->load distance over the load->pivot distance.
And yeah, i didn't include the tyre in my calculations, but our numbers seem to be fairly similar, bearing that in mind.
 
belly_up said:
i'd forgotten my high school physics, so i went and looked up some sites, and they said for that class of lever ie:
pivot------load--------force
that the leverage ratio was equal to the force->load distance over the load->pivot distance.
And yeah, i didn't include the tyre in my calculations, but our numbers seem to be fairly similar, bearing that in mind.
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Hmm, no I don't think that way of calculating it is right. You can't have two variables in calculations like that. 1:1 is when you're applying force at the load (ie no lever), but if you worked it out the way you suggested, 1:1 would suggest that you are applying force at a distance halfway between the pivot and the load.

I think this is a different kind of lever anyway, technically the load is the tyre, not the caliper...
 
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.
 
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