Quote:
Originally Posted by Lil' Lavery
Before we get carried away, let's examine this statement.
Let's start with the simple equation F=ma. Based on this, when you reduce your weight (and consequently mass) and keep your force the same, your acceleration should increase (which agrees with the bolded statement). However, when examined further, we find that the force does not always remain constant in the FRC world when your mass changes.
The force we're speaking about in this application is the force exerted by the drive wheels. This is limited by friction between the wheel surface and the playing surface. The frictional force is calculated as Ff=μN, where μ is the coefficient of friction and N is the normal force. Subbing this back into the first equation, we now have:
μN=ma
In most situations, the normal force is going to be a function of the mass, usually just N=mg when on a flat surface. Putting this back into the previous equation we have:
μmg=ma
Which simplifies to:
μg=a
In other words, in friction limited drivetrains (ie drive systems that have enough power to "spin out" their wheels), acceleration is not governed by the mass of the robot, as the mass plays into both sides of the equation.
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Good point, and well worded.
However, consider that the driving force (i.e. output torque of the motors) decreases with speed. At some speed the motors will lack the torque required to spin the wheels, barring extreme cases. At that point, the situation will become the F=ma situation we all know and love and the lighter robot will be able to accelerate faster.
Another interesting tidbit is that the wheels' coefficient of friction may not be constant, but rather vary with contact pressure. Assuming that the coefficient of friction is constant the heavy and light robots will accelerate at the same rate in a friction-limited case. However, if the tread/playing surface is sensitive to contact pressure then the lighter robot will have the advantage, all else being equal.
Also note that a drive-train does not require
power to break traction, but rather requires
torque, specifically torque on the drive wheels.