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2 Speeds
Ok, so I was doing some math toady, on what the maximum ratio for the slow speed on a two speed transmission (with 4 CIMs) ought to be. I used a lot of math and am not very good at describing it, hints on how to do this are greatly appreciated. I started with these variables:
W (Force of Weight) = 700N (approx. a 70kg, 150lb robot)
S (Free Speed) = 88.5 rps
T (Stall Torque) = 9.7Nm
R (Reduction) = ????
Wr (Wheel Radius) = 0.0762m (6" wheel)
CoF = 1
And quickly derived these:
Ds (Drivetrain Free Speed) = S/R * Wr * 2 * Pi
Dp (Drivetrain Max Pushing Force) = T*R / Wr
Dt (Drivetrian Max Traction) = W * CoF (Or just "W" since CoF = 1)
Next I decided to determine the point at which max pushing force would be less than max traction. Having more traction prior to this point would be wasted.
Speed and pushing force can be placed on a graph with speed as the independant variable and max torque as the responding variable. The line representing the drivetrain will always be a linear relation it will meet with the y-axis (at stall torque) and the x-axis (at free speed). The three lines (x * y axes and the line), form a triangle. I decided to determine what fraction of the triangle was to the left (slower) than the point where max torque dipped below max traction. I began by figuring out what fraction of the triangle had torque above the line:
TgF (Torque Greater than Friction) = (Dp - Dt)/Dp
That much of the triangle would be left of the point. Next I found out what speed that point was at:
TgFS (TgF speed) = TgF * Ds
Now to expand:
TgFS = (Dp - Dt)/Dp * Ds
= ((T*R/Wr) - W)/(T*R/Wr) * (S/R*Wr*2*Pi)
= ((T*R/Wr) - W) * ((S/R*Wr*2*Pi)/(T*R/Wr))
= ((T*R/Wr) - W) * ((S*2*Pi*Wr^2)/(T*R^2))
I graphed this function with R as the manipulated variable and TgFS as the responding var, in order to find out at what reduction you could go the longest while still applying as much torque as traction will allow. It is my opinion that although it is perfectly justifiable to go with a lower gear reduction than this (for more speed), there is no justification for using a higher one, as all the extra torque will go towards spinning the wheels (or just be traction controlled out). When graphing, I found that the value of R that rendered the highest TgFS, would always result in Ds being twice TgFS. I have not yet figured out how to make a rigorous proof of this, if you could help here I would be very grateful.
Working off that assumption I created this equation which gave an equality that was only true in the optimal gear ratio:
2*TgFS = Ds
2(((T*R/Wr) - W) * ((S*2*Pi*Wr^2)/(T*R^2))) = (S/R*Wr*2*Pi)
2(((T*R/Wr) - W) * (2*S*Pi*Wr*Wr) / (T*R*R) = (S*Wr*2*Pi) / (R)
2(((T*R/Wr) - W) * (Wr)/(T*R) = 1
Now to isolate R, so that given any mix of input variable I can find the reduction at which TgFS is as high as possible.
2(T*R/Wr) - 2W = (T*R/Wr)
-2W = -(T*R/Wr)
(2W*Wr)/T = R
Up to now I have been using W instead of W*CoF since CoF in this example = 1. By substituting back, I get a more versatile equation:
(2*W*CoF*Wr) / T = R
Now filling in the the original values you get an optimal ratio for a large number of robots:
(2*700N*1*0.0762m) / 9.7Nm = R
1400N * 0.0762m / 9.7Nm = R
11 = R
Figuring a little more math on this (which I will spare you from), this gear must maintain traction control until the robot reaches 5.5ft/s (the TgFS point), which will take 0.25s and approximately 1 foot. This is the ratio which which will get to 5.5ft/s the fastest physically possible. Please note that this is in a hypothetical perfectly efficient drivetrain.
In my estimation there is no justifiable reason for having the low gear with a reduction greater than 11 in a drivetrain with 6" wheels. Is this accurate?
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