We have a Andy Mark Super shifter with 9.6:1 and 24:1 gear ratios. Question: What is the best speed to shift from 24:1 to 9.6:1?
The ratio of gears n = 24/9.6 = 2.5
My thought is to switch when the power in first gear drops lower than the power available in second gear.
We can derive this point easily using the torque-speed curves for the two gears and setting them equal and solving for speed.
First gear torque speed w1 = w1_max*(1- torq/torq1_max)
First gear pwr_1 = w1*torq
Second gear torque speed w2 = w2_max*(1 – torq/torq2_max)
Second gear power pwr_2 = w2*torq
Setting pwr_1 = pwr_2 we can solve for torq and then for w_1 where the pwr curves cross and we should shift.
This leads to w_1 = w_2 since torq cancels out.
w1_max*(1- torq/torq1_max) = w2_max*(1-torq/torq2_max)
Solving for torq
torq = (w2_max/w1_max-1)/(w2_max/w1_max/torq2_max -1/torq1_max)
And plugging into first gear speed equation gives
w1 _equal_pwr_point = w1_max*(1 – ( w2_max/w1_max-1)/(w2_max/w1_max*torq1_max/torq2max -1)
We can now make the following substitutions:
w2_wmax/ w1_max = n
torq1_max/torq2_max = n*eff1/eff2
where eff1 and eff2 are the efficiencies associated with each gear.
So w1_equal_pwr_ point = w1_max*( 1 -(n-1)/(n^2*eff1/eff2 -1))
For spur gears we assume .95 for each 5:1 ratio so gear 25:1 for gear 1 would have eff1 = .95^2 and 9:1 ratio for gear 2 we assume just .95 .
Thats it: if n= 2.5 then the best switching point is
w1_max*(1-(2.5-1)/(2.5^2*.95^2/.95-1)) = w1_max* .696
or .7 w1_max
Also, the Torq_equal_pwr_point = .3 *Torq1_max.
Using encoders to measure speed we would use a shift logic as follows:
If (w_encoder > .7 w1_max)
then gear = gear2 // shift to high speed gear
else gear = gear1 // shift to low gear
w1_max is calculated from w_free of Cim motors = 5310 rpm * V_bat/12/gear1_ratio. Since gear1_ratio = 24 then:
w1_max 12 v = 221 rpm (7.7 fps with 8 in wheels)
w1_max 10.5v = 194 rpm (6.8 fps with 8 in wheels)
So it would be best to read the bat voltage and take this into account when shifting.