Original Posted By: Oscar Zuniga
Jack wrote:>I have 21" motorcycle rims.>With the tall wheels, I find my brakes are adequate>for stopping and to hold it for a runup, but they can't>hold it against full power.Right. The typical 6.00x6 tire is going to have an outsidediameter of about 17" (except for Mike Cuy's, since heinflates his tires by the worm's-eye-view eyeball methodrather than a tire inflation gauge ;o) and the outside diameterof a motorcycle tire on a 21" rim is going to be about 26".Torque being proportional to the distance from the point ofapplication of the force, for any given forward thrust, themotorcycle wheel/tire setup with the same Clevelands as the6.00x6s will require roughly 1.5 times the braking powerto hold (13/8.5 = 1.53). Correct my math???Oscar ZunigaAir Camper NX41CCSan Antonio, TXmailto: taildrags(at)hotmail.comwebsite at http://www.flysquirrel.net ________________________________________________________________________________
Pietenpol-List: brake holding power
Re: Pietenpol-List: brake holding power
Original Posted By: Jim Markle
More torque, right on that one, and the same kinetic energy to dissipate. That's another criteria to know.To find the kinetic energy in foot-pounds, multiply the weight in pounds times the landing speed, in mph, squared by .01671. That gives the kinetic energy per wheel.For example, a 1,320 lb LSA landing at 40 mph would have to get rid of about 35,300 foot pounds per wheel. Of course some of that goes into ground friction, some into air drag, it's not all into the brakes.David PauleThe typical 6.00x6 tire is going to have an outsidediameter of about 17" (except for Mike Cuy's, since heinflates his tires by the worm's-eye-view eyeball methodrather than a tire inflation gauge ;o) and the outside diameterof a motorcycle tire on a 21" rim is going to be about 26".Torque being proportional to the distance from the point ofapplication of the force, for any given forward thrust, themotorcycle wheel/tire setup with the same Clevelands as the6.00x6s will require roughly 1.5 times the braking powerto hold (13/8.5 = 1.53). Correct my math???Oscar ZunigaAir Camper NX41CCSan Antonio, TXmailto: taildrags(at)hotmail.comwebsite at http://www.flysquirrel.net________________________________________________________________________________Date: Tue, 13 Apr 2010 08:59:51 -0600 (GMT-06:00)
More torque, right on that one, and the same kinetic energy to dissipate. That's another criteria to know.To find the kinetic energy in foot-pounds, multiply the weight in pounds times the landing speed, in mph, squared by .01671. That gives the kinetic energy per wheel.For example, a 1,320 lb LSA landing at 40 mph would have to get rid of about 35,300 foot pounds per wheel. Of course some of that goes into ground friction, some into air drag, it's not all into the brakes.David PauleThe typical 6.00x6 tire is going to have an outsidediameter of about 17" (except for Mike Cuy's, since heinflates his tires by the worm's-eye-view eyeball methodrather than a tire inflation gauge ;o) and the outside diameterof a motorcycle tire on a 21" rim is going to be about 26".Torque being proportional to the distance from the point ofapplication of the force, for any given forward thrust, themotorcycle wheel/tire setup with the same Clevelands as the6.00x6s will require roughly 1.5 times the braking powerto hold (13/8.5 = 1.53). Correct my math???Oscar ZunigaAir Camper NX41CCSan Antonio, TXmailto: taildrags(at)hotmail.comwebsite at http://www.flysquirrel.net________________________________________________________________________________Date: Tue, 13 Apr 2010 08:59:51 -0600 (GMT-06:00)