Here are my calcs from the YardMule.

It has a jackshaft and a differential, so maybe this can help?

My resulting speed was bang on.

All based on this configuration...

The goal of this exercise is to create a jackshaft gear reduction system that will convert the high rotational speed of your motor shaft into a slow but powerful speed at your drive wheels. You will be increasing torque (power) in exchange for speed.

I will use two letter short codes for all of the values so that the formulas used to calculate them are easy to follow. These numbers you already know and are required for calculations:

**(DS)** Desired maximum vehicle speed in miles per hour
**(WD)** Diameter of your drive tires in inches
**(GR)** Total differential gear box reduction
**(MR)** Maximum RPM of your drive motor
**(MT)** Number of teeth on your motor shaft sprocket

These numbers you do not have yet and will be working out using the above numbers:

**(WS)** Required wheel RPM for desired speed

**(PS)** Required pinion RPM for desired speed

**(JR)** Required jackshaft gear reduction for desired speed.

The value for

**DS** (desired vehicle speed) is your choice. For reference, the average brisk walking speed is about 3 to 4 miles per hour, which is what I chose for my top speed. A slower speed will directly translate into towing or hill climbing power. I am very pleased with my top speed and almost endless power. You could certainly create a much faster vehicle, but as you double the top speed, you cut the drive torque in half. I will use my 4 MPH as an example.

Let’s begin by calculating

**(WS)** Required wheel RPM for desired speed.

To determine how fast your wheels need to turn

**(WS)** in order to reach your desired top speed, use this formula:

**WS** = 63360 x

**DS** / 60 / (

**WD** x 3.1416)

So, for my desired 4 MPH top speed, using my 32 inch tires, my calculation would be:

**WS** = 63360 x

**4** / 60 / (

**32** x 3.1416)

The result of that calculation is that

**WS = 42.017**, or rounded to

**42 RPM.**
So, my wheels will need to be turning at

**42 rotations per** minute to achieve a maximum speed of

**4 miles per hour**. Once you have the value of

**WS**, write it down.

Ok, now that you have

**WS**, we can easily calculate

**PS**, the RPM speed that the sprocket mounted on your differential’s pinion flange needs to turn at in order to bring your vehicle up to

**DS**, your desired maximum speed.

**PS** =

**WS** x

**GR**
This one if somewhat obvious because the gear reduction inside the differential gear box forces the pinion shaft to spin that many times more in order to turn the wheels around once. So, for my desired top speed of

**4 MPH**, using my differential with its

**3.55** total gear reduction as figured earlier by spinning the wheels, the calculation would be:

**PS** =

**42** x

**3.55**
My answer for

**PS** (required pinion speed) is

**149.1**, or rounded off to

**149 RPM**. That means that to travel at my desired 4 miles per hour top speed, my differential sprocket has to turn at 149 rotations per minute.

At this point, you have a lot of the math done, and you can see that further gear reduction will be required as most motors turn at speeds well over 2000 RPM. If my motor were to turn the differential pinion directly at 2000 RPM, then my Yard Mule would be moving at a top speed of 54 miles per hour, which would not be safe for this kind of vehicle!

The last bit of math you need to do will determine the required gear reduction of your jackshaft assembly

**(JR)** in order to trade your motor’s speed for torque. This calculation is completely based on the ratio of teeth from one sprocket to the other, which creates the gear reduction. You will need to count or calculate all of the teeth on all of your sprockets. Let’s begin by naming them according to where they are on the vehicle.

You can see that sprockets

**S1** and

**S1** form a pair as well as sprockets

**S3** and

**S4**. You already have sprockets

**S4** (differential) and

**S1** (motor). You only need to calculate the number of teeth for sprockets

**S2** and

**S3** in order to add the required jack shaft gear reduction

**(JS).**
Gear reduction through a jackshaft is multiplied. The resulting total reduction is a multiple of the pairs, which means

**(S1:S2) x (S3:S4)** is the total reduction from motor to differential.

Derived gear reduction through a small to large sprocket pair is calculated by dividing the larger number of teeth by the smaller number of teeth. Working on the first pair

**(S1:S2)**, I have

**15** teeth for

**S1**, and

**60** teeth for

**S2**. This is a gear reduction of

**60/15**, which equals a total reduction of

**4 **times.

The second sprocket pair

**(S3:S4)** have

**15** teeth for

**S3**, and

**42** teeth for

**S4.** The resulting gear reduction is

**42/15**, which equals a reduction of

**2.8** times.

Now, we multiply both calculated reductions together to determine the value for JR

**,** which is the total jackshaft reduction. This value will be

**4 x 2.8**, making

**JR = 11.2**.

Now we know that the motor’s maximum speed

**(MR)** will be divided by

**JR,** which in my case will be

**11.2** before it reaches the differentials drive pinion.

If you divide my motor’s maximum speed of

**MR** by

**JR**, that equals

**1670 / 11.2 = 149**.

This should be very close to the calculated

**PS** (pinion speed) value, which it is.

One final tip on choosing sprockets for your jackshaft is to make room for a larger sprocket for

**S2** to slow your vehicle down if required. If you don’t add that clearance, it will be impossible to adjust the gear ratio for slower speed later on if required.

Here is a different view of the motor and diff...

Brad