42 replies to this topic

[fordmustang1984]

I'm building a tractor, and am wondering how i should size a hydrualic motor to to drive it?

[hydriv]

Why not save yourself a ton of aggravation and just buy something like this.

Case Ingersoll 448 Tractor w/onan 18 HP engine - eBay (item 200494680454 end time Jul-18-10 14:42:49 PDT)

FYI, it comes with 18 hp, a 2 speed cast iron trans-axle, a full on hydraulic system that delivers 8.5 gallons of oil per minute at up to 2200 PSI that is air cooled and properly engineered. Whether you like it or not, this tractor will outperform anything that you might cobble together because it is a tried and true proven design with a history that dates back to 1962.

Sorry if this is not what you are looking for but you really have no idea what you are trying to do here and all I'm trying to do is to tell you that as gently as possible. This project of yours isn't as simple as you think it is or would like it to be. There are people called "hydraulic engineers" that are paid to design the type of system you apparently would like to have. They work with tables and formulas and every aspect of the system is sized to work with all the other components. There are good reasons why there are thousands of different pumps and motors out there. None of those products come cheap, either. So if you choose the wrong unit, then you just tossed several hundred dollars away because there is no return of such items. You buy it, you own it.

[fordmustang1984]

I see what your saying, and have tried to find one around here, but they are rare in this area. I already had the motor frame and axle, as well as scrap a basic comtrols like steering, so i figured i would go for it. Worst case is i end up with a really big gokart.

Also, the pump is 17 gpm.

[hydriv]

Look. I'm not trying to beat you up here or to make you out to be some sort of fool either. I understand what you are trying to do and I applaud your efforts to create something useful. However, the fact that you happen to have a 17 gpm pump kicking around is indicative of your lack of knowledge about hydraulics. In order to properly power that pump, you would need a motor that puts out at least 25 hp. Your hydraulic lines would all have to be 3/4" in diameter. You would need a reservior that would hold 15 gallons of oil.

I'm trying to make you understand that a hydraulic system is something that is engineered so that all of the components work together as a unit to accomplish a task. When you start out with "a pump" of a given size, then that pump dictates everything that follows or problems will ensue.

This sort of undertaking is fine if you have machine-shop facilities at your disposal and access to a wide variety of used parts to keep the costs down. I'm not going to tell you that this is an impossible project because it isn't. But if you do not have a source of cheap parts, then it can get real expensive, really fast.

[fordmustang1984]

I understand what your trying to say, and i have considered these things and thats why I am asking before i go and buy stuff. I'm not worried about any expenses or a lack of power. I have a 22 hp engine hooked up right now, and easy access to a 25 hp engine if need be. I know your being helpful and i'm glad theres someone who thinks of the issue i should watch out for.

That said, all i am trying to do is decide what size motor would work best with my pump. Then i'll start working on the plumbing and valves.

[hydriv]

The problem with a pump that big is that you need an even larger motor in order to get the sort of torque you need to propel the tractor with enough power that it will perform ground engaging work. If all you want is a parade tractor that will traverse fairly flat, paved surfaces, then it isn't as critical. Let me give you an idea as to what I'm talking about by using the Case garden tractor as an example.

Many of those GT's use a pump with an internal displacement of .61 cubic inches. The drive motor used on a 400 series tractor displaces 17 cubic inches. If you divide the motor displacement by the pump displacement you get a ratio of about 28 to 1.

If your pump produces 17 GPM @ 3600 RPM and 0 PSI, then it would have an internal displacement of about 1.1 cubic inches. If you apply that ratio of 28 to 1, then you would need a motor that displaces around 31 cubic inches. That's one big mother of a motor and it would be priced accordingly. Now what we have not talked about is the diameter of the tire rims on this axle and the gear ratio of the axle. You also have not stated whether this is just an axle or it's a trans-axle. And if it is a trans-axle, then you have to know what the final drive ratio is and what the various gear ratios are. Are you able to help me out with that information so I can help you out?

[fordmustang1984]

Yes, i can get you all of that info. The axle is out of a 1 ton truck, so i will ahve to measure the tires, but im positive the gear ratio is 4.10:1. It will also have a 3 speed tranny inbetween, and a chain drive to reduce it if need be. I will try and get the ratios for the tranny as best i can. Just let me know if there is any other information i should try and collect?

[hydriv]

If the axle is from a 1 ton, then the rims are likely 16 inch. The 4:10 ratio sounds quite plausible. If you are using an automotive type 3 speed transmission, then third gear will be one to one so essentially we are working with the 4:10 and the 16" rubber. Using a roller chain reduction between the two may be necessary if you intend to have three ranges by making use of the transmission gears. Take a tape measure and wrap it around the tire to get the circumference in inches. That way, we'll know how far the tire moves during each revolution.

How certain are you about the pump size? Do you have the specs for it? It would help if I knew for sure what the internal displacement of it is in cubic inches. Am I also to understand that the engines you intend to use are governed to run at 3600 RPM at full throttle?

[fordmustang1984]

Yes, the engines max out at 3600 rpm. I am extremely certain of the pump size. I used a formula from OCH to find the gpm and even found a conversion table for volumes to make sure everything was as exact as possible. I could find no model numbers on the pump, so i dont have the precise specs. I think i will be finding a 3 spd from a car now. The one i was going to use is out of a lawn tractor, so its not strong enough.

[hydriv]

According to my calculations, you could likely get by with this motor.

Motor, Hydraulic, 10.1 cu in/rev, 4 Bolt - General Purpose Hydraulic Motors - Hydraulic Motors - Hydraulics : Grainger Industrial Supply

[fordmustang1984]

When you say "get by" do you mean its slightly big, small, or too close to reallt tell? I'm just making sure. Thnaks for all the help. Its nice to have someone who can not only help figure things out, but also find things that should be taken into account when doing something.

[hydriv]

You have selected a one ton truck axle for this project. Most one ton trucks use 16" rims. Most of the tires used on those rims have an aspect ratio of 75 and a width of around 9". That combo means that a properly inflated tire will stand almost 30 inches off the pavement. It will also travel 93.865 inches every time it rotates on full revolution.

In order to figure out what motor to use, someone has to decide how fast this tractor is going to travel when in top gear and at wide open throttle. I felt that a "target speed" of 8 mph was reasonable. It's not crazy fast nor is it boringly slow. Thanks to the 3 speed transmission you intend to put in front of this rear axle, you will have two additional speed ranges that are much slower and will have much higher torque at the wheels.

If the target speed is 8 mph and there's 5280 feet in a mile, then (5280 x 8) the tractor must be capable of travelling 42,240 feet in one hour or (42,240 divided by 60) 704 feet each minute.

Since the tire rotates 93.865 inches per revolution, it will take (704 x 12 = 8448 inches divided by 93.865) 90 revolutions of the tire each minute to cover that distance of 704 feet. We now know what the target wheel speed is. 90 RPM

The rear end has a ratio of 4:10 to 1 so in order to get 90 RPM at the wheel, the input shaft of the rear end must spin at (90 x 4.1) 369 RPM. If you put a standard automobile 3 speed transmission in front of the rear end, 3rd gear is still going to be 1 to 1 so that means the hydraulic motor must still spin at 369 RPM.

You have this 17 GPM pump available and I'm presuming that it puts out 17 gpm when spinning at 3600 rpm. IF that is true, then the internal displacement of that pump must be 1.1 cubic inches. That means the pump will transfer 1.1 cubic inches of oil from the inlet side to the outlet side every time it completes on full revolution. If the pump spins at 3600 rpm, then (3600 x 1.1 equals 3960 cubic inches of oil divided by 231) the pump moves 17.142857 gallons of oil per minute. There are 231 cubic inches of oil in one US gallon

The motor I suggested is 10.1 cubic inches in displacement and it will handle up to 2300 PSI. However, the max rpm is limited to 353 rpm which is 16 rpm slower than our ideal speed. That drop translates to a wheel speed drop of 3.9 rpm and that means we won't hit our target speed of 8 mph.

A 10.1 cubic inch motor only needs (10.1 x 353 divided by 231) 15.434 gallons of oil per minute to achieve that speed. However.... we have not factored in anything for loss so far. If the efficiency of the system is 90 percent, then you have a 10 percent loss. That means, out of the 17 gallons that the pump SHOULD push, you will only see 15.3 effective gallons of oil. Some of that oil is lost in the pumping end and some is lost at the motor end. Both losses are due to necessary internal clearances inside both devices. I am not a hydraulics engineer. Specifying motors for an application is what they do for a living. However, you are free to compare what I wrote above to the components you have selected to verify my assumptions. There are often local hydraulics companies that have engineers on staff who could examine my work and tell you whether I am on target or way off base.

You should not rely upon my calculations alone because, as you can see, geroller motors are not an inexpensive item to buy and this is not an area where you want to play hit and miss.

In addition to the cost of the motor, you will also need a motor control valve that will allow you to govern both speed and direction. This valve must have an adjustable relief built into it so that the pump is protected at all times. Heat is a major enemy of both the hydraulic components and the oil that circulates through them. I urge you to select a reservoir tank that holds at least ten quarts of oil and also an oil cooler capable of flowing up to 20 gpm. Put an electric fan from a junked car in front of the cooler. Hoses will likely have to be 3/4" ID to easily handle 17 gpm and they should have a minimum working pressure of 3000 PSI. I would use straight grade 30 W motor oil in this system, not some foo foo hydraulic fluid that is like water when it heats up.

If you have a local scrap yard that gets old machinery brought to it, you might get lucky and find a suitable geroller motor but....know this. If you go with a larger internal displacement than 10.1, your ground speed will drop. If you find smaller motor, then the ground speed will increase but then you run the risk of "overspeeding" the motor thanks to the size of the pump.

If you choose the motor I selected, then you need to talk to Grainger's people first. I don't know if that motor is designed to handle "side loading" or not. Side loading is where you put a pulley or a sprocket on the motor's output shaft and then run a belt or chain to a pulley or sprocket that is fixed onto the input shaft of the transmission you intend to use. All of the force during motor rotation is being put on one side of the motor. The alternative is to put the shaft of the motor end to end with the shaft of the transmission and connect the two together using a LoveJoy style of coupling. Side loading is essentially non-existent with that set up.

Plumbing the system is simple. The output of the pump would go directly to the IN port of the motor control with built-in relief valve. The OUT port of that valve would go to the oil cooler which would get connected to the reservoir and it would feed the oil back to the pump. You would need ONE high pressure hose for the connection between the pump and the motor valve. After the motor valve, you can use low pressure return line to the cooler, to the reservior and to the pump.

The two "work ports" on the motor valve will need high pressure hoses to connect to the drive motor. Use Grade 8 bolts to secure the motor along with a bit of Lok tite. If you are using the truck rear end, then you have the ability to make use of the hydraulic brakes on that axle for stopping the tractor and an emergency brake to keep the tractor in place when unattended. You also have the ability to use individual master cylinders to control each wheel brake and this could come in handy for steering and for getting unstuck. Ag tires are available for 16 inch rims as are skid steer tires. I would suggest setting the relief valve at 2200 PSI. You will need a guage and some fittings to do this.

[olcowhand]

FM84....what is the displacement of the 2nd pump that you plan to use to power the hydraulics for the loader?

[hydriv]

Oooooh. I just love surprises. :confuse::confuse::confuse:

Am I now to understand that this home-made tractor is going to have a FEL on it?

[olcowhand]

Oooooh. I just love surprises. :confuse::confuse::confuse:

Am I now to understand that this home-made tractor is going to have a FEL on it?

I just had to throw that in.... there is a second pump for that. But there's a possibility that the one pump may be enough by using a flow divider. Yep, the plot is much thicker than you knew hydriv.