Eldron Posted September 4, 2014 Share Ok this spawned a many minute thought pattern in my head involving moments, newtons, force and distance... I probably need to apologise to a few people - I'm just not quite sure who.... Here's my take: It all comes down to travel of rotor through caliper/brake pad. The bigger the wheel the lower/shorter the amount of pad that moves through the caliper/pads.The bigger the rotor the higher/longer the amount of pad that moves through the caliper/pads. Given that the circumference calc (2*pi*r) is linear - the effect of wheel size and rotor size should be linear. So a 29" wheel/tyre (say 33" with 2" tyre) is 10% bigger than a 26" (30" with 2" tyre) then 10% LESS rotor will move through the caliper/pad resulting in 10% "worse" braking. Likewise a 180mm rotor will move 11.83% MORE rotor through the caliper/pad than a 160mm rotor.200mm rotor will move 11.11% MORE rotor through than 180mm rotor. Soooooo - moving from a 26" to 29" means you should move to the next rotor size up (160mm-180mm or 180mm-200mm). Interesting. If I'm right. Have at it mechanical engineers! Skott5 1 Link to comment Share on other sites More sharing options...
Robrider Posted September 4, 2014 Share Ok this spawned a many minute thought pattern in my head involving moments, newtons, force and distance... I probably need to apologise to a few people - I'm just not quite sure who.... Here's my take: It all comes down to travel of rotor through caliper/brake pad. The bigger the wheel the lower/shorter the amount of pad that moves through the caliper/pads.The bigger the rotor the higher/longer the amount of pad that moves through the caliper/pads. Given that the circumference calc (2*pi*r) is linear - the effect of wheel size and rotor size should be linear. So a 29" wheel/tyre (say 33" with 2" tyre) is 10% bigger than a 26" (30" with 2" tyre) then 10% LESS rotor will move through the caliper/pad resulting in 10% "worse" braking. Likewise a 180mm rotor will move 11.83% MORE rotor through the caliper/pad than a 160mm rotor.200mm rotor will move 11.11% MORE rotor through than 180mm rotor. Soooooo - moving from a 26" to 29" means you should move to the next rotor size up (160mm-180mm or 180mm-200mm). Interesting. If I'm right. Have at it mechanical engineers! Close enough. But you taking a long way around to get to: M = Fxd As someone else said, think about it as a lever. JXV and Jester6 2 Link to comment Share on other sites More sharing options...
Paul Ruinaard Posted September 4, 2014 Share Gots nothing to do with wheelsize...check your rotors and mineral oil. If they both pristine, up your rotor sizeWhat the man said. How heavy are you btw? I am riding 29er and not a small guy i went 180mm Shimano vented discs back and front and also braided hoses. My brakes are like a switch they are so sharp. Some bleeding and somebody that knows brakes (if you are in Jhb - Josh from Cyclists Workshop) Link to comment Share on other sites More sharing options...
Eldron Posted September 4, 2014 Share Close enough. But you taking a long way around to get to: M = Fxd As someone else said, think about it as a lever. Agreed - I just wanted to make it more practical. People tend to stop reading when you say pivot point, moment arm and Nm. Link to comment Share on other sites More sharing options...
openmind Posted September 4, 2014 Share Just from a science point of view.. Its about the application of force. Rotor pad apllies a force at a certain distance away from the hub. This creates a stopping moment (or torque if you would prefer) applied at the hub. The equation is M = Fxd. (d is the rotor radius) That same moment is then applied to the stopping power of the bike, using the same formula, except now the stopping force F is applied to the ground and the distance d is the radius of the wheel. if the rotor size and braking force are the same (ie one finger same pressure), then the increase in wheel size will decrease the stopping forceFinally, a reasoned response! I was slowly going mad reading all the pseudo-science! Glad to see someone paid attention in their std 8 science class Link to comment Share on other sites More sharing options...
zaslinger Posted September 4, 2014 Share Glad to see someone paid attention in their std 8 science class Best 3 years of our lives ;-) L'Bie 1 Link to comment Share on other sites More sharing options...
droo Posted September 4, 2014 Share Ok this spawned a many minute thought pattern in my head involving moments, newtons, force and distance... I probably need to apologise to a few people - I'm just not quite sure who.... Here's my take: It all comes down to travel of rotor through caliper/brake pad. The bigger the wheel the lower/shorter the amount of pad that moves through the caliper/pads.The bigger the rotor the higher/longer the amount of pad that moves through the caliper/pads. Given that the circumference calc (2*pi*r) is linear - the effect of wheel size and rotor size should be linear. So a 29" wheel/tyre (say 33" with 2" tyre) is 10% bigger than a 26" (30" with 2" tyre) then 10% LESS rotor will move through the caliper/pad resulting in 10% "worse" braking. Likewise a 180mm rotor will move 11.83% MORE rotor through the caliper/pad than a 160mm rotor.200mm rotor will move 11.11% MORE rotor through than 180mm rotor. Soooooo - moving from a 26" to 29" means you should move to the next rotor size up (160mm-180mm or 180mm-200mm). Interesting. If I'm right. Have at it mechanical engineers! Only one hole to poke in this - at a given speed, the 29" wheel will be turning slower. The leverage aspect is correct, it's just not the whole picture. If the wheels were rotating at the same angular velocity, your explanation would hold. Conservation of energy is the simplest way to explain this - K = Mv, and for a rider + bike at a given speed and assuming the 2 bikes weigh the same K will be constant. Therefore the force required to bring the two systems to a stop is the same regardless of wheel size. I could reduce this to a set of equations, but I have far too many bikes to get done before the weekend... Link to comment Share on other sites More sharing options...
Robrider Posted September 4, 2014 Share Only one hole to poke in this - at a given speed, the 29" wheel will be turning slower. The leverage aspect is correct, it's just not the whole picture. If the wheels were rotating at the same angular velocity, your explanation would hold. Conservation of energy is the simplest way to explain this - K = Mv, and for a rider + bike at a given speed and assuming the 2 bikes weigh the same K will be constant. Therefore the force required to bring the two systems to a stop is the same regardless of wheel size. I could reduce this to a set of equations, but I have far too many bikes to get done before the weekend... The same force will be required on the tyre that is in contact with the ground to bring two equal bodies to rest... yes. But the way that force is transferred to the brake it will not be the same. Its mechanical leverage. Brian Fantana and JXV 2 Link to comment Share on other sites More sharing options...
Eldron Posted September 4, 2014 Share Only one hole to poke in this - at a given speed, the 29" wheel will be turning slower. The leverage aspect is correct, it's just not the whole picture. If the wheels were rotating at the same angular velocity, your explanation would hold. Conservation of energy is the simplest way to explain this - K = Mv, and for a rider + bike at a given speed and assuming the 2 bikes weigh the same K will be constant. Therefore the force required to bring the two systems to a stop is the same regardless of wheel size. I could reduce this to a set of equations, but I have far too many bikes to get done before the weekend... That's exactly what my post is based on - the bigger the wheel the slower it turns so less rotor travels through the caliper/pads. Agreed on the conservation of energy rule but the lever effect/amount of rotor running through pad changes with wheel and rotor diameter. If the lever (displacement) changes then the force needs to change to keep the torque the same. Link to comment Share on other sites More sharing options...
droo Posted September 4, 2014 Share Eldron - I understand your reasoning, but here's mine: If the two systems are the same mass and travelling at the same speed they have the same Ek and as such require the same amount of power to be dissipated by the brake to bring them to a stop at the same rate. Assuming they have the same brake on the same size rotor, the force exerted must be the same since there are no material differences between the two systems. Robrider - if you're an engineer, show me the numbers and I'll believe you. JXV 1 Link to comment Share on other sites More sharing options...
Eldron Posted September 4, 2014 Share Eldron - I understand your reasoning, but here's mine: If the two systems are the same mass and travelling at the same speed they have the same Ek and as such require the same amount of power to be dissipated by the brake to bring them to a stop at the same rate. Assuming they have the same brake on the same size rotor, the force exerted must be the same since there are no material differences between the two systems. Robrider - if you're an engineer, show me the numbers and I'll believe you. My counter - counter I think Rob and I are effectively saying the same thing. My "amount of rotor through pads" is the same as Robrider's lever rule. For your conservation of energy/both bikes of equal weight stopping in an equal distance theory to be correct the amount of torque applied needs to be the same - if the lever/wheel size changes then the force has to change to keep the torque the same. If the force was being applied in the same position (for instance the handlebar) on both bikes then the force would be identical on both bikes. I think... Link to comment Share on other sites More sharing options...
droo Posted September 4, 2014 Share Looks like I'll have to do the sums then... Link to comment Share on other sites More sharing options...
Brian Fantana Posted September 4, 2014 Share Ok, I am not a geek, but I was sitting in a very boring seminar, and put pen to paper. My arrows are the wrong way round but the values won't change. Same energy required to stop the two systems, but different brake forces are required to achieve that. This is for ideal friction free environments, and it does not take the larger contact area of the larger diameter tyre into account. Edited September 4, 2014 by Brian Fantana zaslinger, Jester6 and Tumbleweed 3 Link to comment Share on other sites More sharing options...
LazyTrailRider Posted September 4, 2014 Share Er... I didOops, sorry! Link to comment Share on other sites More sharing options...
droo Posted September 4, 2014 Share Ok, I am not a geek, but... Riiiiight... I think the only thing that has been confirmed by this thread is that anyone who's still involved is a massive nerd Link to comment Share on other sites More sharing options...
jcza Posted September 4, 2014 Share Anoraks out in force Link to comment Share on other sites More sharing options...
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