Timeline for Hydraulic disc brakes - total friction as a function of rotor size
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 17, 2015 at 17:07 | comment | added | Daniel R Hicks | @ebrohman - For most materials, in the "medium" range of operation (where brakes would likely be operating), the "coefficient of friction" is a near-constant, meaning that sliding friction is directly proportional to caliper pressure. | |
| Apr 17, 2015 at 15:53 | comment | added | ebrohman | @andy256 that's what I was thinking, but it seems too simple. I'm looking to see that actual calculation. As maatnz said, the bigger rotor cools faster, has more air passing over it, etc. | |
| Apr 17, 2015 at 4:29 | comment | added | andy256 | IMHO this is correct. The rear disk is 7/8 the size of the front, so the same pressure el produce 7/8 of the torque. To get the same torque requires 8/7 of the pressure that is on the front to be applied to the back. | |
| Apr 17, 2015 at 1:23 | comment | added | Daniel R Hicks | @ebrohman - You can roughly assume that, above maybe 30% brake force, the force on the lever is proportional to the torque on the wheel. | |
| Apr 17, 2015 at 1:18 | comment | added | ebrohman | I know this, the larger rotor will create more torque on the wheel. I'd like to know what ratio of forces - front/rear lever pull - will cause the two different sized rotors to burn the speed equally. 160mm front, 140mm rear, or any ratio for that matter. You are going to have to pull the rear harder in this case. I'm asking how much harder. | |
| Apr 17, 2015 at 1:10 | history | answered | Daniel R Hicks | CC BY-SA 3.0 |