2 Speeds

[lemiant]

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?

[Jared Russell]

Yep, your math is pretty much dead on for a theoretical system. For a real world design problem, though, I would also consider two (very important) additional factors:

1) Efficiency. You can expect to chop 10-40% off the top of your total available power due to transmission inefficiencies. The exact number has to do with the number of reduction stages, types of reductions, tolerances, alignment, and the like.

2) Current draw. If you gear so that your motors are just barely traction-limited at stall in low gear, you won't be able to push (or resist pushing) all day - you will be drawing ~120A per motor and will eventually pop your breakers. It won't be an instantaneous thing, but eventually they will trip (temporarily disabling the motor while it cools). If you design for 40A per motor or below, you will be able to push all day without popping a breaker (in theory). Whether this matters or not depends on your application.

So there are definitely justifiable reasons for going lower.

Originally Posted by lemiant

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.

What you discovered is indeed true. It has to do with the physics of DC motors. The angular speed of the motor decreases linearly with load torque. The total output power of the motor is the product of angular speed (rad/sec) and torque (N*m). (1 N*m/s = 1 Watt). You can always trade off between speed and torque with gearing on the output, but you often** want to make sure that the torque on the motor at the input to the gearing is as close as possible to the "max power" point. As it happens, max power is maximized when the motor is running at 1/2 free speed and 1/2 stall torque (on your triangle graph, imagine graphing x*y for each point on the hypotenuse-that is your power curve).

** often but not always, as the most EFFICIENT point on the power curve (in terms of transforming electrical power into mechanical power) is NOT at the max power point. Different applications will have different design criteria.

[lemiant]

In this case the robot would be capable of double the traction of the tires when not moving (it is capable of exactly friction force when moving at half speed), so if you traction controlled the robot during a pushing match it would draw 60 amps, is that ok?

[Hawiian Cadder]

in my experience the 40 amp breakers are not the first thing to go, in a 4 cim drivetrain, 40*4 =160, so shouldn't the 120 amp main breaker pop first? in testing our drive this year we were having a little fun dragging some team members around the floor, and we never had any of the 40 amp breakers pop, the 120 amp breaker got hot enough after 5 min to pop.

[Chris is me]

Originally Posted by Hawiian Cadder

in my experience

40*4 =160, so shouldn't the 120 amp main breaker pop first?

Please don't speak from experience if you're actually just guessing, and you aren't sure what you're talking about.

in testing our drive this year we were having a little fun dragging some team members around the floor, and we never had any of the 40 amp breakers pop, the 120 amp breaker got hot enough after 5 min to pop.

Note how it took 5 minutes to pop. Not 2:15.

Generally, the 120 amp breaker has such a large safety margin that it shouldn't be popping, even if you push in high gear. If it is, it may be faulty.

[Trent B]

Just going to drop this tidbit here from the spec sheet located here which is about the 120 amp Breaker and is also in the PNG

Based on their tolerances the breaker should be able to run at 240amps for 10-40 seconds without tripping or 160 amps for somewhere around a minute of time.

The 40 amp snap actions are much more prone to tripping, as low as 6 seconds at 135% current as seen here (Snap action breakers)

[Vikesrock]

Originally Posted by Ether

The graph in that link and the graph in the PNG file you attached seem to disagree wildly.

The attached file is for the 120amp breaker. The link is for the 40A snap action breakers.

[Ether]

Originally Posted by Vikesrock

The attached file is for the 120amp breaker. The link is for the 40A snap action breakers.

You are quick :-) You grabbed it after I hit the delete key. Check the timestamps.

[Trent B]

I edited my post to try and make it a little more clear, I linked to both spec sheets and took a png shot off of the 120amp spec sheet.