SpeedAccelerationAndClimb

From Simreal

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If you want to calculate what motor to use for your robot's wheel-based drive system, you need to take three things into consideration: the robotís top speed, how quickly you want it to be able to accelerate, and what kind of hills you want it to be able to climb.

Speed

A moving object has kinetic energy (KE), so when you take a still object and make it move you add energy. The same applies if you have a moving object and want to make it move faster; you need to add more energy. These calculations are more natural in SI units where KE is in joules, mass m is in kilograms and velocity v is in meters per second. I, however, tend to think in imperial units. So a conversion is in order before we get started: assume a 50 pound robot traveling 10 feet per second is described in SI units as:

mkg = mlb / 2.20462

mkg = 50 / 2.20462 = 22.68

vm/s = vft/s / 3.28084

vm/s = 10 / 3.28084 = 3.048

The formula for kinetic energy is:

KE = ( m / 2 ) v2

So this 23 kg robot moving at 3 m/s has a kinetic energy of:

KE = ( m / 2 ) v2

KE = 11.5 * 9 = 103.5

Over a hundred joules (the relevent unit for kinetic energy). (Thanks to Soeren for finding a mistake in the calculation).

Acceleration

To accelerate the robot to this velocity, we need to pump 1,150 joules of energy into it. But our electronics donít work in joules, they work in amps, volts, and watts. Both mechanically and electrically speaking, one watt of power is equal to one joule of energy expended across one second of time. This relationship helps us bridge the gap between mechanical energy and electrical energy.

To determine the power our motor needs to accelerate itself to a 1,150 joule potential, we need to pick a time frame for the acceleration -- say, T=10 seconds -- and a voltage, such as the common E=12 volts.

We can state the equation as:

E * I' * T = J

12 * I * 10 = 1,150

120 I = 1,150

I = 9.58

So this kicky acceleration will require almost 10 amps of motor power, which is not too bad.

Climb

An elevated object has potential energy (PE), so to drive our robot up an incline requires adding energy to it. The amount of energy depends on the mass of the robot, the height of the ramp, and the (constant) accelerating force of gravity. Say we want to drive our robot up a 1 meter ramp:

PE = m * g * h

PE = 28 * 9.80 * 1 = 225.4

By climbing this ramp, weíve added 225 joules to the robotís potential energy. If we want to climb the ramp in 5 seconds, we can again determine the current needed:

E * I * T = J

12 * I * 5 = 225

60 I = 225

I = 3.75

Piece of cake! But there is one other consideration...

Climbing a gradient means adding a percentage of the robotís weight to the rolling resistance. The steeper the hill, the more the robotís weight pushes back. This force F varies with the sine of the angle Θ of the hill:

F = m * sin(Θ)

Given a 10 degree hill (quite steep), the 23 kg robot is essentially lifting:

F = 23 * sin(10)

F = 23 * 0.174 = 4

About 4 kilograms of weight. Converting kilograms to pounds (or newtons), and given your wheel diameter, you can calculate the torque needed to ascend this hill. This exercise is left to you, the reader (donít you just hate that?)