View Single Post
  #4   Spotlight this post!  
Unread 22-01-2014, 23:23
Unsung FIRST Hero Woodie Flowers Award
Chris Fultz Chris Fultz is offline
My Other Car is a 500 HP Turbine
FRC #0234 (Cyber Blue)
Team Role: Engineer
 
Join Date: Jan 2002
Rookie Year: 1942
Location: Indianapolis, IN
Posts: 2,831
Chris Fultz has a reputation beyond reputeChris Fultz has a reputation beyond reputeChris Fultz has a reputation beyond reputeChris Fultz has a reputation beyond reputeChris Fultz has a reputation beyond reputeChris Fultz has a reputation beyond reputeChris Fultz has a reputation beyond reputeChris Fultz has a reputation beyond reputeChris Fultz has a reputation beyond reputeChris Fultz has a reputation beyond reputeChris Fultz has a reputation beyond repute
Re: Which gear ratio is ideal? deciding gear ratio

The JVN calculator can help you answer your question. Some basics of the math involved are below. (if anyone notices an error, please point it out!).

***

There are a few decisions you need to make, and then you can work backwards to select the gear ratios that will provide what you need.

What size wheels are you using? 4”, 6” 8” or something else?

Are you able to direct drive one wheel and then run belts / chains from it, or will you be using some speed reduction between the gearbox and the wheels.

And, how fast do you want the robot to travel in high and low speed?

I will make some assumptions and then you can calculate for what you are designing.

Let's say you want to use 6” wheels and you want a low speed of 9 feet per second and a high speed of 14 feet per second.

First determine the circumference of your wheels - 6” wheel = pi*D = 3.14*6 = 18.84”. So for every revolution of the wheel, your robot will travel 18.84”.

For the high speed of 14 feet per second:
18.84 inches per revolution / 12 inches per foot = 1.57 feet per revolution
14 feet per second = 14 x 60 = 840 feet per minute
840 feet per minute / 1.57 feet per revolution = 535 RPM (Revolutions per minute).

So the wheel needs to turn at 545 RPM.

If you are direct driving the wheel (connected to the gearbox output shaft or with a 1:1 sprocket set), then the reduction you need is (assuming you are using CIMs)

5310 RPM (CIM) / 545 RPM (Wheel) = 9.7:1 gearbox reduction.

If you are not able to direct drive the wheel, then you need to factor this reduction by the tooth counts of the sprockets (belt or chain) of your drive.

So if you have a 20 tooth sprocket on the gearbox and a 30 tooth sprocket on the wheel, then you are slowing down the RPMs going from the gearbox to the wheel, so that 1 revolution of the gearbox is only 0.66 revolutions of the wheel (20/30). So you need to reduce the gearbox reduction (so it has a higher RPM) by the ratio of 20/30. So instead of a 9.7:1 reduction you need a 9.7*.66 = 6.4:1 reduction.

You would do the same calculations for the low speed values, and then try to find a ratio combination that is close to the pair you need.

Remember the calculations are theoretical assuming everything is perfect in the system. Most people use a 95% efficiency number, so you can make this adjustment to your numbers to reflect this.
__________________
Chris Fultz
Cyber Blue - Team 234
2016 IRI Planning Committee
2016 IndyRAGE Planning Committee
2010 - Woodie Flowers Award - Championship