On 10 Jul 2003 00:30:37 +0950, TCG <
[email protected]> wrote:
>Greetings...We have a discussion here and need some educated advice...
>
>If you shave 1 gram off of each cycling shoe (total 2g), how much energy will you save based on 1
>hour of riding at cadence of 60rpm using 172.5 or 175 crank arms...and how much less mass have you
>moved over that same hour.
>
>It's been slow at the office lately...
>
Work Done (W) = Force (F) x Distance Moved (D)
but
F= mass (m) x acceleration (a)
Substituting,
W = maD - or at least you probably are, keeping up that cadence for an hour!
Simplifying greatly,
Ignore friction (sigma), gravity (g), and the fact that the direction of linear acceleration is
always changing:
Linear acceleration is always towards the centre of the crank. Assume this is constant. For a radius
of 172.5mm:
Linear Velocity (V) = Radius (r) x Angular Velocity (Omega)
Omega = 60rpm = 120 Pi Radians per minute = 2 Pi Radians per second
Substituting
V = 0.1725 x 2 x Pi metres per second = 1.083
Acceleration = rate of change of distance with respect to time
or
a = v^2 / r
so
a = 1.083*1.083 / 0.1725 = 6.81 m/s^2
So,
W = maD
W = 0.002 x 6.81 x D
Linear Distance moved in one hour (D) = circumference x rpm x 60
Circumference = pi X Diameter (d)
So
W = 0.002 x 6.81 x pi x 2 x .1725 x 60 x 60
W = 53 Joules
So, the energy needed to move (and thus saved, if you don't have to move it) 2g through one hour of
revolutions at 60 rpm on 172.5mm cranks is 53 Joules - the eqivalent of moving a mass of 53kg (the
mass of a small adult) through a distance of one metre in one second. That's also how much less mass
you've moved over the hour.
Comparing like for like, if two identical riders, of mass 53kg, save for the missing 2g, rode
identical bikes in ideal conditions at the same cadence over the same route, one would finish the
course in 1 hour - the lighter rider would finish the course one second faster.
You can substitute the values for the other sized crank yourself...
--
MatSav