Torque and gear ratio help please?

In summary: Thank you!In summary, this person is trying to figure out how to get the most out of a gear reduction system in order to lift a mass. They think that using formulas and graphs from the linked page, they could possibly calculate the load on the gear/pulley at a certain speed.
  • #1
Lokey
1
0
Hi, I'm new to this forum but I've read over probably a dozen threads so I figured that someone should probably be able to help me; please forgive me if this is the wrong area to post this question as well.

Dilemma:
So the problem right now is that I'm trying to use a gear reduction system to get the most out of some small electrical motors (those of an NXT robot) and I'd like to be able to calculate the amount of weight that can be pulled upward if I know everything about the gears and motor(s) hooked to the gear reduction system. Any ideas as to how I could calculate this?

Thoughts:
I imagine that given W=FD and P=W/T that I could possibly use these two formulas as well as the rpms of the motor and the size of the gears to go backwards and mathematically calculate weight? Just a thought though. I've been thinking through a half dozen designs all day so as I write this I'm fried lol, please forgive this post if it comes out half thought through.
Oh p.s. here's a link showing a lot of thorough graphs for rpm(s), torque vs. mechanical power, etc. http://www.philohome.com/nxtmotor/nxtmotor.htm

Thank you in advance if you even look at this lol
 
Engineering news on Phys.org
  • #2
Well i those two equations are not that far off.lol
Just a quick idea, you have your output speed in rpm.
1) using the speed-torque graph, find the torque at that speed.
2) using the torque, check the power-torque graph and find the mech power the motor is producing in Watts
3) Convert the rpm speed into a linear speed in m/s = (rpm x pi/30) x (radius of the lifting pulley or gear..whatever)
4) divide the power by the linear speed to find the load on the gear/pulley at that speed.
i.e
Tangential load (weight) in Newtons = Power/linear speed
Hope this helps
Anyone correct me if I'm wrong somewhere.I'm in a rush
 
  • #3
Welcome to PF!

Getting the most out of the motors depend a bit on what you mean exactly. If you want to be able to lift a mass as fast as possible you would want to go for maximum mechanical power, which the linked page mentions the motor will provide when loaded with a torque around 0.15 Nm. If you want to lift the mass as efficiently as possible (getting most mechanical work from your electrical source), diagrams on that page indicates that the torque should be a bit lower, around 0.10 Nm.

If your (frictionless) gear train has a reduction ratio of K (the output shaft rotates K times slower than the motor shaft, or equivalently, the output shaft has K times larger torque than the motor shaft), the output shaft is attached to a spool of radius R that winds the wire that lifts your mass M, and the motor shaft has torque N, then these relates as N*K = M*g*R or rearranged M = N*K / (g*R).

For example, if you attach an R = 0.05 m spool directly to the motor shaft (K=1) the maximum mass for N = 0.15 Nm motor torque is around 0.3 kg. If you instead put a K=10 gear train in, the mass increases 10-fold to around 3 kg. Likewise, if you put on a spool with half the radius you double the mass that can be lifted.

Furthermore, still assuming that the gear is frictionless (i.e. it transfers all power from the motor shaft to output shaft), you know that the lift power M*g*V must be equal to the mechanical power P provided by the motor, thus V = P/(M*g) = P*R / (N*K). So a change to K or R that allows you to lift double the mass will at the same time put the speed to half. Assuming P = 1.6 W, the last example above that gave M = 3 kg will give a speed of around 5 cm/s.

If the gear train has significant friction this will reduce the amount of torque on the output shaft that is available for lifting, thus reducing the maximum mass that can be lifted, but keeping the speed more or less the same.


If this did not help you feel free to elaborate your questions.
 

1. What is torque?

Torque is a measure of rotational or twisting force. It is the force that causes an object to rotate around an axis or pivot point.

2. How is torque related to gear ratio?

Torque and gear ratio are inversely proportional, meaning that as one increases, the other decreases. In a gear system, the torque is multiplied by the gear ratio, so a higher gear ratio will result in lower torque, and vice versa.

3. How do I calculate torque?

Torque is calculated by multiplying the force applied to an object by the distance from the pivot point to where the force is applied. The formula for torque is: torque = force x distance.

4. What is the importance of torque and gear ratio in a vehicle?

Torque and gear ratio play a crucial role in a vehicle's performance. Torque is needed to generate the power required to move the vehicle, while the gear ratio determines the speed and strength of the vehicle's movements. A higher gear ratio allows for faster speed, while a lower gear ratio provides more power for towing or climbing hills.

5. How can I increase torque and adjust gear ratio in my vehicle?

To increase torque, you can upgrade the engine or add aftermarket parts such as a supercharger or turbocharger. To adjust gear ratio, you can change the size of the gears in the transmission or differential, or install a different gear set. However, it is important to consult a professional mechanic before making any modifications to your vehicle.

Similar threads

  • Mechanical Engineering
Replies
2
Views
3K
Replies
2
Views
1K
Replies
9
Views
2K
  • Mechanical Engineering
Replies
14
Views
3K
  • Mechanical Engineering
Replies
1
Views
1K
  • Mechanical Engineering
Replies
3
Views
2K
  • Mechanical Engineering
Replies
8
Views
2K
  • Mechanical Engineering
Replies
2
Views
1K
  • Mechanical Engineering
Replies
3
Views
3K
  • Mechanical Engineering
Replies
10
Views
3K
Back
Top