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Unread 30-09-2003, 13:36
Jnadke Jnadke is offline
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Quote:
Originally posted by Frank(Aflak)
edit: I looked over your post a second time, and it is true that at the diagonals you get sgrt(2) times the power of each individual set of wheels, but I think you failed to consider that you ALSO get sqrt(2) times the speed, so twice the power. You gotta look at both. And you need a driver who is aware of all of this.
Not exactly true, because velocity is a vector as well, and vectors can be divided into their x and y components. It's simple physics. If you deal with only one direction, then one side contributes sin 0 = 0 to travel and the other contributes sin 90 = 1 times the power, so 0+1 / 2 = 0.5 or 50% of the available power.

Now, if you travel diagonally, the angle of the vector with respect to the axis is 45 degrees. The vector can be split into its forward (y) and sideways (x) components. We have total Power (P), Power in the direction of travel (Py), Power in the sideways directions (Px), Motor power for one set of wheels, MS1, and motor power for another set of wheels, MS2. Py for the left set of diagonal wheels is MS1 sin 45 = P * 0.70. Py for the right side of diagonal wheels is MS2 sin 135 = P * 0.70. Now when you add these, you have to keep in mind that you're only taking into account the power for an individual set of wheels. When you do the multiplication and addition, you can see that you'll end up with LESS power than available.
Then there's the power in the x direction Px = MS1 * cos 45 = MS1 * 0.70 for one set of wheels. For the other set, you have Px = MS2 * cos 135 = MS2 * -0.70. You can see that the power in the x direction is lost because the motors can cancel eachother out.

Keep in mind that power is a function of both SPEED and TORQUE. The same equations can be used for both speed and torque individually. If you want, I can draw some force diagrams later to further prove my point.
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