Homemade Ground-Drive PTO Forecart
by Ken Gies of Fort Plain, NY
Editor’s Note: Genius pure genius – and elegant to boot. Here’s another clear bit of evidence that our future with the working animals is as bright as we choose to make it. Thanks and kudos to you Ken. LRM
I begin with apologies to Basil Scarberry. I can in no way equal his craftsmanship as seen in his ground drive pto forecart found in Winter 2005 SFJ, on page 30. My intention is to discuss finding a good ratio for the pto shaft. I also want you to avoid some of the mistakes I made. So let’s look at my rather crude differential based cart. It has worked well for me since 2003.
As we start, consider a few things when building a pto cart. Are big drive tires necessary? Is a lot of weight needed? What is cheapest? What is least complicated? Are replacement parts still available? What do you want to power with the cart? Will it take more than two horses? Will the pto need to run in slippery conditions? Hydraulics? Brakes? Lights? Will ground clearance matter? Does the width matter? Imagine the cart in use. Try to see it working where you normally go and where you almost never go. Will it be safe and easy to mount or dismount? Can you access the controls of the implement conveniently? Is it easy to hook and unhook? Where is the balance point? I’m sure you will think of other details as you daydream about it.
I built my cart for two Haflinger ponies, but the tongue extends for full sized horses. I use three ponies on it occasionally, but still prefer using two. It is fifty-four inches wide with thirteen-inch highway tires on it. It has about eight inches of clearance underneath. Hay snags when I triple up windrows and straddle them, but I manage. It has no brakes, lights or hydraulics, but then again, I have yet to need them. It is easy to unhook. It runs my tedder nicely. I also use it to pull my manure spreader, hay rake, and sometimes a wagon.
When you find a good differential, figure out what size of tires you will put on it. In order to get the pto speed right, the tires you will use should be on it. Many people think that bigger tires will give them more torque at the output shaft. There may be some gain because of traction increase, but the drive ratio will have a far greater effect on the performance of the cart. A good example is a #9 mower. The wheels are not that big, about 32” tall, but they still run the cutter bar adequately. I have seen some mowers with seven-foot bars doing a fine job. Normal 235-75R-15 light truck tires are between 26 and 28 inches tall. I think that for most applications this size is adequate. Differentials in this size range are not too heavy or wide, are readily available, and are repairable if a bearing ever goes. It is easy to rig up brakes too. I also know that tractor-style traction tires are available for 14, 15 and 16 inch rims should you really need them. If you intend to run a baler, you need more horses and an increase in traction through weight and improved/bigger tires. Which direction is the pinion shaft turning? Will it need to face front or back to turn the jackshaft in the right direction? Once all this is determined, then you can figure out the drive ratio to get the pto up to speed.
Just a note on balers; some brands of balers need to run at a certain number of plunger strokes per minute. I worked as an ag mechanic for a year or so and “fixed” several balers that were having plunger problems and breaking shear bolts because the machine was operated at the wrong rpm. Check the baler manual before you get too far into this project if baling is part of your plan.
I am not a mathematician. It took me a long time to figure out the final speed of the pto shaft. The number I wanted to find was how many revolutions the pto shaft needed to make over a given distance to get the equivalent of 540 rpm at horse speeds. Here is my attempt at the math. First, I took the speed that a good team moves. (This will be the basis for all calculations. If the ground speed differed significantly, a new value would have to be computed using similar messy math.) The average speed that many people use for horses is 2.5 miles per hour. The distance in a mile is 5,280 feet. Multiply the two numbers: 2.5 x 5,280 = 13,200 feet per hour. I reduced the feet per hour to feet per minute by dividing by 60: 13,200 / 60 = 220 feet per minute. Now I had common time units. I divided the revolutions per minute by the feet per minute to get how many revolutions per foot I needed at 2.5 mph to get 540 rpm: 540 rpm / 220 fpm = 2.45 revolutions per foot. I used 245 revolutions per 100 feet for figuring because a 100 ft tape is easy to obtain and it is not an extreme distance to push a differential. Remember that number, 245.
When I first got my differential, I had two of my kids push it while I counted the number of revolutions the pinion shaft made in 100 feet. My differential pinion shaft turns 64 times per 100 feet. In order to get it up to the equivalent of 540 rpm I should increase the rpm four times: 245 / 64 = 3.8.