hot rod golf cart

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May 31, 2013
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South Benfleet, Essex, England, UK
I am using 2000 rpm as a low reference point (this depends on the two speed selection) as 4500 rpm on the motor is probally dependent on the two speed selection.
this needs to be investigated. Will do when I get my new sprockets installed.

There is no jackshaft in my configuration, just the 13.5:1 differential ratio.
2000(rpm)/13.5(diff ratio)= 148.148 axle rpm. the fat tires are 87" in circuference so every miniute the tire rotates 148 rpm which equates to 148"/12=12.34 feet the cart moves per miniute at 2000 motor rpm.
hope this helps describes my configuration.
If the final axle RPM is 148RPM and the tyres are 87" circumference (roughly 7' 3") then each minute it moves 148 * 7.25 feet = 1073 feet per minute.
1073 feet per minute x 60 = 64,380 feet per hour. 1 mile = 5280 feet so the speed would be about 12.19 MPH. which is a nice speed to bimble along at. :)
 
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Apple Valley, California, USA
I apologize if I still don't get this correct.
But I am learning from this experience.
It may be I can eliminate my jackshaft and change to direct drive as well.
Hopefully going to a quiet belt drive .

Not sure if these two different calculators are helpful to 'verify' your figures or not.
Not quite so much math involved.


I in put your figures 2000 RPM - wheel diameter (87"/3.14) = 27.62) and 13.5 gear ratio
The result I get is 12.2 mph on one and 12.18 on the other.

An in-put of 3300 RPM was needed to achieve 20.1 mph.
 

Radical Brad

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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...

7301

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...

7302

Brad
 
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Joined
Apr 15, 2013
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Washington state
very descriptive thanks.
?? what is the 63360??
no explanation on what it is?
 
Joined
May 31, 2013
Messages
3,232
Location
South Benfleet, Essex, England, UK
very descriptive thanks.
?? what is the 63360??
no explanation on what it is?
Inches in a mile.
WS = 63360 x DS / 60 / (WD x 3.1416) is the clue. The wheel diameter was given in inches.
 
Last edited:
Joined
Apr 15, 2013
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Washington state
well so much for testing today.
the new sprockets and chain came last night.
The #25 chain is really small.
the 1/2" bore sprocket that mounts on the differential slips on the groved shaft. Thinking about putting a flat spot on the shaft but still need to contend with the #25 chain.
It is like jewelry chain almost.
Plan now is to order 2 #35 sprockets with 10 teeth then weld one to the orginal groved hub. Have a spare sprocket if something goes wrong.
Don't want to ruin the differential shaft.
I secured the steering wheel and worked on the brake mechnisim. no ereal sucess there.
back to the shop in he morning.
 
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Apr 15, 2013
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Washington state
cart is progressing at a slow pace. getting one thing done per day. taking my time to get everything just so. As well as waiting for parts;
 
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