I don’t quite get what mechanical advantage is. Is it different to the gear ratio, because it is also a ratio?
Please see image attached for more information.
Mechanical advantage is the ratio by which a machine multiplies the input force. For example, let’s say you want to lift a 600-pound object. You set up a machine (block and tackle, lever, inclined plane, etc. or combination of them) that allows you to lift this object by applying a force of 100 pounds. Your 100 pounds gets multiplied by 6 by the machine to lift the 600-pound object. That 6 is the mechanical advantage.
The general rule of thumb is that you sacrifice some sort of displacement (think distance) to gain force, though you can also do it the other way around, sacrificing force to gain some sort of displacement. Ideally the ratios match perfectly. For example, in the above case you would apply your 100 pounds of force over 6 times the distance you would want to lift the object under ideal conditions.
Even though mechanical advantage is defined with force, it still works with torque, the rotational analog to force. Similarly, because speed is based on distance, you can look at the displacement ratios as speed ratios, and we can use angular speed as well. This helps when we apply it to our motors since everything is rotation and we tend to care about the rate at which they’re turning. So you sacrifice speed to gain torque, or vice versa. The teeth on the gears (or similar), since they’re the same size, can be used as units for distance and so can be counted to determine these ratios.
Here is an example. Let’s say you attach your motor’s axle to a 12-tooth gear, and you mesh that gear with a 36-tooth gear on another axle. You would have to spin the 12-tooth gear 36/12=3 times (the gear ratio) to get the 36-tooth gear to spin one time. That means the second axle spins at 1/3 the angular speed of the motor’s axle. That is the speed sacrifice. In return, ideally you would get 3 times the torque provide directly by the motor. However, you lose some to increased friction and the like. So you’ll get a little less than 3 times the torque.
Be careful with the preceding post as it has a number of errors:
No. It’s the multiplication done to the force. The power supplied will decrease because some of the energy supplied goes into overcoming extra friction and the like. (Note: You can get some extra energy in some cases, like with a crowbar where its own weight can help you.)
Yes, ideally, this is correct. Really, that’s “ideal mechanical advantage,” not “mechanical advantage.”
No, it does not only refer to torque. Mechanical advantage specifically refers to force, but because of how the forces are relating it also applies to torque.
Two errors here: First, not the difference, the ratio. Division v. subtraction. Second, it does not describe the power at all.
Also, it’s kind of cool how power = torque * angular velocity, and since power provided by the motors is constant, if angular velocity doubles, then torque must halve, and vice versa.