J
Joe Riel
Guest
Diablo Scott <[email protected]> writes:
> Paul Kopit wrote:
>
> The third is the most interesting to me. In Jobst's proposed test
> he'll ride a measured mile on his bike with the computer that has
> perfect calibration. I asked him not to concentrate too much on
> maintaining a perfectly straight line but to ride "normally". In
> normal riding we swerve around potholes and debris, we take different
> lines around corners, we vary our position relative to the edge of the
> road, we get out of the saddle and stretch or sprint - there are a
> whole host of things we do that make the path of our front wheels
> different (longer) than the straight line distance or whatever path
> GPS uses in its measurement. Since Jobst is certain of his
> calibration data, at the end of his test he'll know exactly how much
> he's deviated from a straight line by how much his computer differs
> from 1.00 miles.
It's easy to estimate an expected error.
Assume that the rider's follows a sinusoid with amplitude A and
period (length) L. The ratio that distance along the sine wave
to the period is greater one can be well approximated by
(1) dL/L ~ 1/4*(A/L)^2
The exact value is an elliptic integral.
Assume that L is equal to the development of the gear train.
(2) L = (F/R)*Dw*pi
= (52/16)(27*inch)(3.14)
~ 260 inch
The amplitude, A, is half the side to side deviation. Six inches
seems a reasonable upper bound for any reasonably competent rider.
(3) A = (6 inch)/2 = 3 inch.
Plugging into (1) gives
(4) dL/L = 1/4*(3/260)^2 = 130e-6 = 0.013%
For a 1 mile course that corresponds to
(5) dL = (130e-6)(5280 ft)(12 in/ft) = 8.4 inch
Good luck distinguishing the two.
Joe Riel
> Paul Kopit wrote:
>
> The third is the most interesting to me. In Jobst's proposed test
> he'll ride a measured mile on his bike with the computer that has
> perfect calibration. I asked him not to concentrate too much on
> maintaining a perfectly straight line but to ride "normally". In
> normal riding we swerve around potholes and debris, we take different
> lines around corners, we vary our position relative to the edge of the
> road, we get out of the saddle and stretch or sprint - there are a
> whole host of things we do that make the path of our front wheels
> different (longer) than the straight line distance or whatever path
> GPS uses in its measurement. Since Jobst is certain of his
> calibration data, at the end of his test he'll know exactly how much
> he's deviated from a straight line by how much his computer differs
> from 1.00 miles.
It's easy to estimate an expected error.
Assume that the rider's follows a sinusoid with amplitude A and
period (length) L. The ratio that distance along the sine wave
to the period is greater one can be well approximated by
(1) dL/L ~ 1/4*(A/L)^2
The exact value is an elliptic integral.
Assume that L is equal to the development of the gear train.
(2) L = (F/R)*Dw*pi
= (52/16)(27*inch)(3.14)
~ 260 inch
The amplitude, A, is half the side to side deviation. Six inches
seems a reasonable upper bound for any reasonably competent rider.
(3) A = (6 inch)/2 = 3 inch.
Plugging into (1) gives
(4) dL/L = 1/4*(3/260)^2 = 130e-6 = 0.013%
For a 1 mile course that corresponds to
(5) dL = (130e-6)(5280 ft)(12 in/ft) = 8.4 inch
Good luck distinguishing the two.
Joe Riel