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Discussion Starter · #1 · (Edited)
One thing that concerns me is the possible loss of traction (and braking action) if one of the rear wheels is off the ground or on a slippery surface with a hydrostatic drive and non-locking differential. That's one thing I like about Kubotas. So I'm wondering if hydraulic motors could be attached to the front axle to drive the front wheels via hydraulic lines?

There may be some problem synchronizing speeds, but if the drive is basically torque-regulated, it might work OK. I found some motors on eBay that might work, for about $100 each:
http://www.ebay.com/itm/EATON-HYDRA...780?pt=LH_DefaultDomain_0&hash=item519ad93f64

Eaton Char-Lynn S Series Orbital Motor Model 103-1029

Specification English/Metric

Displacement 11.4 in3/r
Max Speed @ Cont. Flow 304
Flow Continuous 15 GPM / 57 LPM
Flow Intermittent** 20 GPM 76 LPM
Torque Continuous 2950 in lb 333 Nm
Torque Intermittent** 3533 in lb 399 Nm
Min. Starting Torque @ Continuous 263 in lb 2670 Nm
Min. Starting Torque @ Intermittent**329 in lb 2910 Nm
Pressure @ Continuous 1850 PSI 128 Bar
Pressure @ Intermittent** 2250 PSI 155 Bar
Max Recommended Operating Temp. 180° F 82° C
**Intermittent Ratings 10% of every minute

Mechanical

2 Bolt Flange Mount.
Two .540” Diameter thru holes on 4.187” Bolt Circle.
3.250” Diameter Pilot
1.00” Diameter Shaft with Woodruff Key 1.625” Long
½ inch NPTF Ports
Port A Pressurized CW Rotation
Port B Pressurized CCW Rotation

WEIGHT: 17 LBS

I don't know if the torque is in the right range for this application. I figure that 2950 in-lb is about 245 ft-lb, which would exert about 490 lb of force at the surface of a 12" diameter tire, so driving both wheels would provide almost a half-ton of force. And 304 RPM on a 12" tire is about 10.8 MPH.

This might be useful for something like a FEL or dozer blade which puts a lot of weight on the front axle.

Anybody ever tried this? Are my calculations right? :pcwiz:

:thanku:
 

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A lot of expense and work for something that is of questionable need.

The secret for dealing with the potential of loss of traction because a rear wheel lifted of the ground is ...........................................don't let the rear wheels get off the ground.

I have a 2wd MF1655 with a FEL with a 210 lb bucket which will carry 1000 lb in the bucket. The back wheels stay on the ground with that load!

Load the rear tires with RimGuard, add wheel weights and 2-link tire chains, and a weight box either on the 3PH or on the rear of the tractor. My tractor weighs over 2400 lb with me in the saddle, a 250 lb 5' backblade on the 3PH............. and no wheel weights.

PStechPaul, you are coming up with a lot of suppositions about theoretical problems that might occur with a tractor that you haven't even tried yet. Try it and give it a chance. It just might surprise you with what it will do.

If you really want to go through the trials and tribulations of adding that type of front wheel drive as an exercise and a method of getting rid of an excess of cash, we'll try to help. Hydraulic motors are cheap. Hoses, fittings, and the additional components needed to make such a system work well in conjunction with the original rear wheel drive are not. And I'm rather doubtful that it can be made to work well.

If you are really looking for a 4wd tractor, go buy one. It'll be about the same price and a lot less frustrating
 

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John Deere already does this on the X700 Series of GTs. If you want to do this, just replicate their system. Or better yet, buy an X728 (gas) or X748 (diesel). You can even get All-Wheel-Steer along with the AWD, if you purchase an X729 or X749.
 

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Discussion Starter · #4 ·
I just got my tractor a couple hours ago. I have some guys working on my property today, and they helped load the mower deck onto the trailer and he drove it up a small hill to a more level access road. Several times the rear wheels slipped in the mud, even with chains, but eventually he got more of a running start and was able to do it. He said the tractor seemed to run OK, and didn't think the steering was too bad. I'll play around with it a bit more later, when they have left. It may have been more difficult because the front blade was still attached, and the trailer with the mower deck added more dead weight to pull. Also the trailer tires need air, as does one of the tractor's rear tires. But it seemed to run fine, and I didn't see any smoke, so I think the engine is pretty good. Probably a good deal for $1000 delivered.

The next hurdle is to take the tractor up to the top of the hill via the access road, which is partly washed out with deep ruts. I may need to throw some rocks and old bricks and concrete chunks into the ruts and then maybe lay down some stone in the worst sections. The lower part of the road is the steepest and most damaged, so if I get past that I should be OK. But then I need to consider any problems I might have coming back down, especially with a trailer load of firewood. That's where having four wheel drive, or at least four wheel brakes, is an advantage. Maybe I should call my friend who has the skid steer loader to fix the road before I attempt it.

As you may have noticed, I love to design things and think "outside the box", rather than just take the easiest and most obvious route to solve a problem. Your comments and suggestions are most welcome, and have caused me to think more clearly about what I need to do. I have entertained many ideas over the years, but most of them have been "pipe dreams" and I never completed most of these projects. Maybe my next project should be a tractor shed to protect my investment, and maybe even make it big enough to do some work in it. At least I think I should cover it with a good tarp, and maybe drive the wheels onto some cement blocks to keep it off the muddy ground.

I might still go a bit further with the idea of using hydraulic motors, and if nothing else they could be used as a winch or as a rotary power source for an implement such as a crane. But mostly I think I should get comfortable with my tractor and fix some of the little things that need TLC, such as a detached seat spring, and replacing the battery with a newer, more powerful one, and getting a block heater for easier cold weather starting (if we actually get any - I'm sitting outside now!).

:thanku:
 

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I don't know if the torque is in the right range for this application. I figure that 2950 in-lb is about 245 ft-lb, which would exert about 490 lb of force at the surface of a 12" diameter tire, so driving both wheels would provide almost a half-ton of force. :thanku:
I think you should be decreasing the torque by half not doubling it ? You loose torque the further away from the centerline of the axle. Think having to gear down your pumpkins for larger mud tires.
 

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Discussion Starter · #6 ·
Well, a 12" diameter tire has a radius of 6", so 245 ft-lb * 0.5 ft = 490 lb. And that is on each wheel, so both wheels should push 980 lb. That seems like more than enough. I could probably use an even larger tire/wheel, or a smaller motor, and still have it be effective.

I just took my tractor for a "spin", and I went down and back up the little hill on the service road with no major problem, although the drive wheels did lose traction at a couple of places. It seems like this tractor frame is rigid, so on uneven ground I was usually on three wheels. At least it felt that way. Not like my Broadmoor which has a mid-frame pivot so all wheels stay in contact.

:fing32:
 

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The front axle should be providing the pivot to keep all the wheels on the ground, however if its very uneven terrain there might not be enough travel to always keep the wheels in contact. What kind of tractor do you have?
 

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Almost 20 years ago I went to a JD dealer and priced a complete rear axle from a 4 wheel drive riding mower with a front deck and rear steer. I don't remember if it was a 900 or 1100 series. The price was $2,000.
 

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As you may have noticed, I love to design things and think "outside the box", rather than just take the easiest and most obvious route to solve a problem. Your comments and suggestions are most welcome, and have caused me to think more clearly about what I need to do. I have entertained many ideas over the years, but most of them have been "pipe dreams" and I never completed most of these projects. Maybe my next project should be a tractor shed to protect my investment, and maybe even make it big enough to do some work in it. At least I think I should cover it with a good tarp, and maybe drive the wheels onto some cement blocks to keep it off the muddy ground.
You have lots of company with those thoughts "outside the box" and "pipe dreams". Being a millwright in heavy industry for 30 years sorta chased some of them out of my head, but some still remain, plus a bunch of new ones collected from the job.

A tractor shed for storing and working on your "new" tractor would be a great idea.


Well, a 12" diameter tire has a radius of 6", so 245 ft-lb * 0.5 ft = 490 lb. And that is on each wheel, so both wheels should push 980 lb. That seems like more than enough. I could probably use an even larger tire/wheel, or a smaller motor, and still have it be effective.

:fing32:
The problem with theoretical torque numbers is that you need traction to make use of them. I can pretty much break traction with my 2400 lb tractor at will and the rear end is rated at 1250 ft lb of torque. Obviously 2/3 throttle on a 20 hp Onan is not going to generate anywhere near that amount of torque, even with the gear reduction involved.

Your formula should read: 245 ft-lb/ 0.5 ft = 490 lb.
 

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Well, a 12" diameter tire has a radius of 6", so 245 ft-lb * 0.5 ft = 490 lb. And that is on each wheel, so both wheels should push 980 lb. That seems like more than enough. I could probably use an even larger tire/wheel, or a smaller motor, and still have it be effective.

I just took my tractor for a "spin", and I went down and back up the little hill on the service road with no major problem, although the drive wheels did lose traction at a couple of places. It seems like this tractor frame is rigid, so on uneven ground I was usually on three wheels. At least it felt that way. Not like my Broadmoor which has a mid-frame pivot so all wheels stay in contact.

:fing32:
Your equation is correct if you were tightening a bolt with a one foot long handle and the force on the handle is 245 lbs then the force at the nut would be 490 ft lbs.

Think of putting that same handle on your motor shaft and spinning the motor . If your motor is creating 245ft lbs max at the splindle then the one foot bar would require only 122.5 pounds of force to stop it and a 2 foot bar would require only 61.25 lbs and so on. The force will always be less the further you extend the moment arm from the pivot point. In this case the point where the power is being supplied from is the center not the end of the moment arm.

You will never get any more power from that motor than what it is listed for unless your use something to multiply the torque like a transmission etc...

I am not trying to be argumentative or stir something up I am just trying to help you figure out your project.
 

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Discussion Starter · #12 ·
According to the Wiki http://en.wikipedia.org/wiki/Pound-foot_(torque):

"A pound-foot (lb·ft or lbf·ft) is a unit of torque (a vector). One pound-foot is the torque created by one pound force acting at a perpendicular distance of one foot from a pivot point."

Apparently lb-ft is the preferred designation for torque, with ft-lb as a unit of energy, but ft-lb seems to be most commonly used for torque. Watt-hours, calories, and BTUs are usually used for energy.

Here is a more detailed example:
http://www.ehow.com/how_12022516_18-footpounds-torque-wrench.html

and, yes, my formula should have been 245 ft-lb/ 0.5 ft = 490 lb, and I also realize that traction will reduce the actual push or pull exerted on the axle. That is a function of the weight on the axle and the break-away friction of the tire on the ground. It is probably in the same ballpark as the weight on the wheel, under ideal conditions. For a tractor, operating on mud, ice, or snow, it could obviously be much less. And for a track vehicle, I think it could approach the maximum available torque, independent of weight. And it also depends on the angle of the tractor on an incline. You can have enough torque to lift the entire weight of the tractor, but you casn't climb a vertical wall (unless you have a special mechanism and a ladder) :)

The equations for a wheel depend on the radius, and not the diameter, so there is the factor of 2 which may be confusing. But thanks for analyzing this. It is important to post correct information, since others may look at a thread and use the information. It's good to discuss issues like this. I learned a lot about power, speed, torque, energy, traction, slope, and efficiency, when I was working out a design for my proposed rail system, and the hybrid converter package for cars: http://www.pstech-inc.com/SHAMPAC.htm

And I made a spreadsheet for calculating various battery sources and gear ratios and motor speeds and such: http://www.pstech-inc.com/VehiclePower.xls

I'm an electronics engineer so my mechanical skills are not extensive, especially when it comes to hydraulics, so I need to read the links supplied. But I also work at a place www.etiinc.com where they calibrate all sorts of things, including torque wrenches, although I'm not much involved with that.
 

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According to the Wiki http://en.wikipedia.org/wiki/Pound-foot_(torque):

"A pound-foot (lb·ft or lbf·ft) is a unit of torque (a vector). One pound-foot is the torque created by one pound force acting at a perpendicular distance of one foot from a pivot point."

Apparently lb-ft is the preferred designation for torque, with ft-lb as a unit of energy, but ft-lb seems to be most commonly used for torque. Watt-hours, calories, and BTUs are usually used for energy.

Here is a more detailed example:
http://www.ehow.com/how_12022516_18-footpounds-torque-wrench.html

and, yes, my formula should have been 245 ft-lb/ 0.5 ft = 490 lb, and I also realize that traction will reduce the actual push or pull exerted on the axle. That is a function of the weight on the axle and the break-away friction of the tire on the ground. It is probably in the same ballpark as the weight on the wheel, under ideal conditions. For a tractor, operating on mud, ice, or snow, it could obviously be much less. And for a track vehicle, I think it could approach the maximum available torque, independent of weight. And it also depends on the angle of the tractor on an incline. You can have enough torque to lift the entire weight of the tractor, but you casn't climb a vertical wall (unless you have a special mechanism and a ladder) :)

The equations for a wheel depend on the radius, and not the diameter, so there is the factor of 2 which may be confusing. But thanks for analyzing this. It is important to post correct information, since others may look at a thread and use the information. It's good to discuss issues like this. I learned a lot about power, speed, torque, energy, traction, slope, and efficiency, when I was working out a design for my proposed rail system, and the hybrid converter package for cars: http://www.pstech-inc.com/SHAMPAC.htm

And I made a spreadsheet for calculating various battery sources and gear ratios and motor speeds and such: http://www.pstech-inc.com/VehiclePower.xls

I'm an electronics engineer so my mechanical skills are not extensive, especially when it comes to hydraulics, so I need to read the links supplied. But I also work at a place www.etiinc.com where they calibrate all sorts of things, including torque wrenches, although I'm not much involved with that.
I also love trying to figure this stuff out, my background is in building, but since my accident I have been trying too learn welding and hydraulics with the help of this forum.

Your equation is correct, but not for calculating the max force that the tire exerts against the ground.

"A pound-foot (lb·ft or lbf·ft) is a unit of torque (a vector). One pound-foot is the torque created by one pound force acting at a perpendicular distance of one foot from a pivot point."

You are not applying a force of one pound one foot away , your force is at the pivot and you need to calculate the force one foot away.

It is force lbs x distance ft = torque ft.lbs , you should not be dividing in this case. You will have 122.5 lb ft of torque available for tractive forces.
 

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It is probably in the same ballpark as the weight on the wheel, under ideal conditions. For a tractor, operating on mud, ice, or snow, it could obviously be much less. And for a track vehicle, I think it could approach the maximum available torque, independent of weight. And it also depends on the angle of the tractor on an incline. You can have enough torque to lift the entire weight of the tractor, but you casn't climb a vertical wall (unless you have a special mechanism and a ladder) :)

The equations for a wheel depend on the radius, and not the diameter, so there is the factor of 2 which may be confusing. But thanks for analyzing this. It is important to post correct information, since others may look at a thread and use the information. It's good to discuss issues like this.
Not even close under ideal normal operating conditions. Only if you wrap a cable around the tire and use it like a hoist to lift a load will you achieve "ideal" conditions. Normal operation depends on the coefficient of friction between ground and tire for actual torque generation.

The Sundstrand transmission in my 2400 lb tractor will generate 160 in lb (13.3 lb-ft) of torque @1500 psi and 3500 rpm. There is approximately a 31:1 reduction between the hydro's motor and the axle with its 26" tall tire for a maximum torque value at the tires of 413 +/- lb-ft of torque in high range. Double that (826 lb-ft) for low range. At approximately 2400 rpm with 2-link chains on the turf tires, the coefficient of friction on grass is easily exceeded and the chains will chew a pair of 12" wide holes down to the depth of the rear axle housing. Without chains on rough concrete takes a bit more torque, but it will still spin the tires at full throttle. The loaded tires on this tractor weighed 200 lb each and it is for sure that the weight of the tractor on those tires exceeded 426 lb.

The Sundstrand Series 15 hydro was the premium hydro in its day and I believe it is comparable to the premium hydros in the top of the line heavy GTs today. I haven't seen the specs on the newer hydros to make a real comparison.

I, also, make my posts for the uninitiated tractor owners sitting on the sidelines as guests, as well as for those members who have more experience. That's one reason for my excessively long explanations on some threads. Discussion is good!
 

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I also love trying to figure this stuff out, my background is in building, but since my accident I have been trying too learn welding and hydraulics with the help of this forum.

Your equation is correct, but not for calculating the max force that the tire exerts against the ground.

"A pound-foot (lb·ft or lbf·ft) is a unit of torque (a vector). One pound-foot is the torque created by one pound force acting at a perpendicular distance of one foot from a pivot point."

You are not applying a force of one pound one foot away , your force is at the pivot and you need to calculate the force one foot away.

It is force lbs x distance ft = torque ft.lbs , you should not be dividing in this case. You will have 122.5 lb ft of torque available for tractive forces.
Chris, set your torque wrench to 50 ft-lb and tighten a 1/2" bolt, then move your hand half way down the shank and pull to loosen it. It will take twice the effort (lb of force) to loosen it compared to the effort to tighten it. Leverage is the name of the game, half the leverage generates or requires twice the force for an equal result.
 

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TUDOR;2050538 The Sundstrand transmission in my 2400 lb tractor will generate 160 in lb (13.3 lb-ft) of torque @1500 psi and 3500 rpm. There is approximately a 31:1 reduction between the hydro's motor and the axle with its 26" tall tire for a maximum torque value at the tires of 413 +/- lb-ft of torque in high rangeGTs today. I haven't seen the specs on the newer hydros to make a real comparison. [/QUOTE said:
So if you increased your tire size the torque available at the wheels would go down not up , correct.

So how can the measured torque value increase, the longer the moment arm gets ?
 

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Discussion Starter · #17 ·
Ah, this is getting interesting! The amount of force exerted on a tractor at the axle is also related to acceleration, and here is a good reference: http://physics.info/acceleration/

It shows that an ordinary car can accelerate at 0.3-0.5 G, and the maximum for a traction vehicle (as opposed to jet or fan powered) is 1.7 G for an F-1 race car. This means that such a vehicle will exert about 1/2 its weight in forward force for acceleration. Weight is really just the acceleration of mass due to gravity.

Now the important thing for me is really the ability of the tractor to pull a certain weight up a certain grade. I do not care about accelerating but just moving at a fixed speed, so all of the force will be just the gravitational acceleration. Slope is described as percent grade http://en.wikipedia.org/wiki/Grade_(slope), is 100 * (rise/run).

On a 15% grade, which is what I estimate my access drive to be, the gravitational force in the direction of motion of a 2000 pound tractor will be 2000*15/100, or 300 lb. So even in this extreme instance, and assuming all of the tractor weight is on the drive wheels, each wheel only needs to exert 150 lb of force. So for 24" diameter drive wheels, 150 lb-ft torque is all that is needed, and since such conditions often cause the tires to slip, it's easy to see how a tractor with 413 lb-ft torque can spin the tires even on flat ground when attempting to accelerate.

The other factor about traction is the reduced effective weight, or pressure, of the wheels on the ground surface as the slope increases. But on the usual grades of less than 20%, most of the weight is still on the point of traction, so it is mostly insignificant.
 

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So if you increased your tire size the torque available at the wheels would go down not up , correct.

So how can the measured torque value increase, the longer the moment arm gets ?
My head hurts from trying to dredge up the lessons from high school daze back in prehistoric times.

Torque is theoretically measured at the center of a shaft. The force required to create that torque, or resulting from that torque, is proportional to the length of the lever attached to that shaft.

eg.
100 lb-ft torque/ 1 ft = 100 lb force
100 lb-ft torque/ 2 ft = 50 lb force
100 lb-ft torque/ 0.5 ft = 200 lb force

Perhaps if you think of it as a lever lifting a weight. A longer lever requires less force to lift a given weight than a short lever. Same result, different approach. Is it easier to lift the back of your tractor with a 2' lever or a 10' lever? The load (torque) is the same.

Larger diameter tires have less force available at the ground/tread interface than smaller diameter tires.
 

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Ah, this is getting interesting! The amount of force exerted on a tractor at the axle is also related to acceleration, and here is a good reference: http://physics.info/acceleration/

It shows that an ordinary car can accelerate at 0.3-0.5 G, and the maximum for a traction vehicle (as opposed to jet or fan powered) is 1.7 G for an F-1 race car. This means that such a vehicle will exert about 1/2 its weight in forward force for acceleration. Weight is really just the acceleration of mass due to gravity.

Now the important thing for me is really the ability of the tractor to pull a certain weight up a certain grade. I do not care about accelerating but just moving at a fixed speed, so all of the force will be just the gravitational acceleration. Slope is described as percent grade http://en.wikipedia.org/wiki/Grade_(slope), is 100 * (rise/run).

On a 15% grade, which is what I estimate my access drive to be, the gravitational force in the direction of motion of a 2000 pound tractor will be 2000*15/100, or 300 lb. So even in this extreme instance, and assuming all of the tractor weight is on the drive wheels, each wheel only needs to exert 150 lb of force. So for 24" diameter drive wheels, 150 lb-ft torque is all that is needed, and since such conditions often cause the tires to slip, it's easy to see how a tractor with 413 lb-ft torque can spin the tires even on flat ground when attempting to accelerate.

The other factor about traction is the reduced effective weight, or pressure, of the wheels on the ground surface as the slope increases. But on the usual grades of less than 20%, most of the weight is still on the point of traction, so it is mostly insignificant.
A couple of minor comments for you to ponder.

Here's a guote taken from a recent thread. The first time this was posted, several months ago, it included a link to the original article. The newer thread is linked below the quote. It is very interesting.

In order to exceed 300 mph in 4.5 seconds, dragsters must accelerate an average of over 4G's. In order to reach 200 mph well before half-track, the launch acceleration approaches 8G's.

http://www.mytractorforum.com/showthread.php?t=225280

While it is true that the effective weight on the rear drive wheels is reduced when going up a slope with a 2 wheel tractor, there is also a weight transfer from the front wheels of a 4 wheel tractor (or any vehicle) back to the rear wheels which results in slightly increased actual load on the rear wheels on relatively low slopes. As the slope becomes greater, this effect is reduced to the point where the front wheels lift off the ground and the load at the tread/ ground interface begins to reduce drastically. How high the center of gravity is above ground will have a marked effect on the rate of weight transfer.
 

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Discussion Starter · #20 ·
Good points! I found the following
http://hypertextbook.com/facts/2007/AnamAhmed.shtml (Acceleration of dragster)
http://en.wikipedia.org/wiki/Top_Fuel
http://www.automobilemag.com/features/columns/0403_top_fuel_dragsters/index.html (Probably the article you mentioned)
http://www.rlengines.com/tech/Acceleration.pdf (same article in PDF)

I did not know that such vehicles could reach such levels of acceleration. Those must be some amazingly "sticky" tires!

It's also a good point about the shift in the weight on the rear tires when going uphill. So it may actually help when I'm going up the service road with an empty trailer, which also adds some weight to the taction wheels. But my concern is going down the same hill, fully loaded, on uneven ground. Without front wheel braking or trailer brakes, there could be some instances where one of the drive wheels is off the ground. At that point, as I understand it, the hydrostatic drive no longer functions as a brake, and the wheel that has lost ground contact will spin backwards. Hopefully the foot brake engages friction brakes in each drive wheel. Otherwise things may get out of control.

That's what scared me with my dozer, which had poor brakes. But there was no way both tracks could leave the ground, and the engine compression braking was sufficient as long as the clutch did not slip. And I could also drop the blade in an emergency, as long as I was not going too fast and hit something really solid.

Maybe I should consider installing front brakes? Or maybe trailer brakes? Probably the usual on-road trailer brakes would do the job. I don't know how likely it might be for the trailer to jacknife, but I could see it happening if the tractor slipped at an angle and the road surface was slippery.

I have also considered adding powered wheels to the trailer, controlled by the force on the tongue. It would only add enough forward force to keep the pulling force less than, say, 50 lb, and would engage braking when a pushing force was sensed.

Maybe I think/worry too much!
 
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