Thoughts on converting lawn tractor snow blower to PT422

[KMA]

Thanks MossRoad.

That's what I was thinking originally too, but then I started trying to understand the effects of different pulley sizes on torque, etc. I think of the hydraulic motor sort of like a single gear "transmission". In other words, the power really comes from the tractor's PTO pump and the motor translates the hydraulic flow into a rotating shaft at a certain speed and torque. The smaller displacement motor will translate the power into a faster shaft speed with lower torque, and the larger displacement motor will result in a slower speed/higher torque connected to the same PTO circuit. Once I start changing the effective rotational speed by sizing up the pulleys, I'm left wondering what effects it will have on the larger motor's torque advantage...

For example, if we compare a 4 cyl car to a v8 in the same "gear", the v8 wins every time... but what if the 4 cyl car is in 1st gear and the v8 is in 4th? Changing the "gearing" (i.e. pulleys) we will affect the actual torque trying to achieve a certain auger shaft RPM.

That all being said, I would still choose the larger brush cutter motor and size up different pulleys since I believe I will have the most flexibility. For example, I could target a slightly slower auger speed which will decrease the max throwing distance of the snow, but in theory provide more torque to power through the heavier stuff. With the smaller motor I wouldn't have as many options. The only way the smaller motor wins out is if I decide to direct mount the motor using love joy style couplers. I understand there are downsides to this approach, but it would be a much simpler conversion with much less room for error.

 

[J_J]

KMA,

If you can determine the rpm of the hyd deck motor, I can compute the cu in of the hyd motor. That is if you are using the 8 GPM flow rate.

In order for the PT to obtain that 8 GPM, takes about 16 HP from the engine, and if it translate to the hyd motor, at 85 %, that means that the deck motor is developing about 13.6 HP.

You may need more or less HP for the snow blower.

 

[KMA]

If I remember correctly, the hyd motors on the finish mowing decks are about 3 cu in, and with 8 GPM's, that makes the motor turn at 616. You will have to bump up the speed by belt/pulley or a gear box setup.

In order to get direct driving speed using the 8 GPM's, you will need a hyd motor of .6 cu in for a speed of 3080 rpm.

Torque will be about 287 in lbs, using 3000 psi, and a .6 cu in hyd motor.

If the brush hog motor is larger in cu in, you will have more torque, but you will still need to bump up the speed even more as a 5 cu in hyd motor using 8 GPM's, is only 370 rpm's.

If you have a 6 in pulley on the impeller, in order to up the rpm to say 3000, rpm, you will need about a 5:1 ratio, so if using the 3 cu in motor, at 616 rpm, that ratio will provide about 3080 rpm's, however, the pulley will be huge.

Thanks J J.

If it really is a 3 cu in motor, it will be very difficult to get it to work. I can change the pulley out on the snowblower side (making it smaller), but even then I'm looking at ~10 in drive pulley according to the calculator. It's not impossible, but definitely makes it more difficult. Direct drive using the smaller finish mower motor would definitely be easier.

Does anyone know off hand what size those motors are? I think it would have to be .5 to .6 to get me in the ballpark for direct drive.

[edit]

Opps... didn't see your response until after posting mine. I'll have to get out to the shed tonight to find some markings on the finish mower motor (unless someone else knows them off hand). I was originally going to purchase a motor, before deciding to donate one from another attachment. If it turns out I need a slightly different size as I start testing, I'll buy one then. Thanks again.

KMA,

If you can determine the rpm of the hyd deck motor, I can compute the cu in of the hyd motor. That is if you are using the 8 GPM flow rate.

In order for the PT to obtain that 8 GPM, takes about 16 HP from the engine, and if it translate to the hyd motor, at 85 %, that means that the deck motor is developing about 13.6 HP.

You may need more or less HP for the snow blower.

 

[J_J]

What do you mean by what size?

If you want direct drive, you will need a .6 cu in hyd motor using the 8 GPM's, to get a rpm or 3080.

Do you have any idea of the speed the normal snow blower is turning?

 

[J_J]

Just looked at a skidsteer snow blower, and he is pushing 20 GPM through a 6.2 cu in motor for a rotor speed of 745. The velocity of the snow will be determined by the rotor size, 20 in, 22, 24, etc. Blade tip speed.

These are some representative flow rates and cu in motors.


Flow Rate,

gal./min ---cu. in./rev.
14-16------ 4.9
16-20------ 6.2
20-24------ 8.0
24 + -------9.6

 

[KMA]

What do you mean by what size?

If you want direct drive, you will need a .6 cu in hyd motor using the 8 GPM's, to get a rpm or 3080.

Do you have any idea of the speed the normal snow blower is turning?

J J,

By size I meant cu inches. I think .6 isn't quite ideal rpm-wise and .5 would be better in that regard, but I'm tempted go with something in between to balance rpm and torque in the directly coupled scenario. For example, maybe the .58 cu in Parker listed on Surplus Center's website for $166 (which should spin at just under 3200 rpm).

When I was picking up the snowblower, I measured the driver pulley on the guy's New Holland tractor to be 6", which is the same size as the pulley it connects to on the snowblower. I assumed based on the 1:1 ratio, the shaft on the snowblower would need to spin at the same speed as the shaft on the tractor's engine (3600 rpm at full throttle?). So my best guess is that a directly coupled hydraulic motor would also need to spin near that speed. A slightly slower speed might not be too terrible, but I wouldn't want to get much lower than 3200 or it probably won't be able to throw the snow far enough.

Thanks for your input, I really appreciate everyone's comments.

 

[KentT]

With a hydraulic motor, torque is basically the inverse of RPM...

Same hydraulic flow (GPM) would produce 2X torque at i/2 RPM -- of course this is an approximation, because of the inefficiencies of the hydraulic motors involved...

Same general principle holds true with pulleys, except it is circumference of the pulley vs torque. An engine turning 3600 RPM (typically design max for small gas engines) would produce twice the torque at the driven pulley, if you reduced the circumference of the drive pulley on the engine by 1/2.

Couple of important notes, IMO:

1. I'm talking pulley circumference in my comments above -- not diameter, a crucial distinction... make sure you do your ratio calculations based on circumference.

2. You have to "compromise" on pulley sizing. You don't want to use a real small pulley on a system that needs to transfer a lot of torque (i.e. power), because you need the larger "wrap" or footprint of the belt in the V-groove to provide the necessary friction to prevent it from slipping. In general, the larger the pulleys used on both the drive and the implement, the better they'll handle torque... 5" (diameter), 6" or even bigger works well. In contrast, smaller pulleys, such as 4" or so, are used on high-speed applications like mower spindles (aka quills) because they don't require a lot of torque, and the smaller pulley reduces friction.

For snowblowers 5" or 6" diameter pulleys work well, or perhaps as much as a 7" will work, if you can make the ratios for you....

Tillers, in comparison, often use 8" to 10" pulleys or perhaps even larger, because otherwise the belt will slip in hard soil.

Bottom line is if the belt is slipping you either need (a) more tension on the belt, or (b) you need to use bigger pulleys and a longer belt.

 

[J_J]

6 in pulley = 18.84 in circumference

4 : 1 ratio

24 in pulley = 76.36in circumference

6 in diameter is to 24 in diameter

as

18.84 in circumference is to 75.36 in circumference

8 GPM, 3 cu in hyd motor, = 616 rpm = Brush cutter hyd motor

616 rpm x 4 : 1 ratio = 2464 rpm

A 3 cu in hyd motor using 8 GPM will turn a 24 in pulley using a 4 : 1 ratio will turn a 6 in pulley, at 2464 rpm for the impeller speed.

Impractical, but could be done.

 

[KMA]

Right. Since pi (π) is the ratio of the circumference to diameter (c = π x d), its okay to use diameter in the discussion of pulley sizes (as long as we keep talking about ratios) like J J showed above. If the pulley used by the snowblower is 6" in diameter and a hydraulic motor's rpm is half of what is required, then a 12" drive pulley could be added on the motor to get the snowblower auger up to speed... essentially doubling the effective rpm of the motor (2:1 pulley ratio).

The challenge I currently have is three-fold: 1) I already have two motors to work with but the larger brush cutter motor rpm will likely make the drive pulley size impractical, and 2) adding pulleys to the system makes the design more complex and therefore more prone to failure and more difficult to build. (Additionally, a pulley system will increase the distance between the tractor and the snowblower, which will be more cumbersome to maneuver.) And 3), direct mounting with love joy style couplers won't tolerate severe jams of the auger and will force me to swap motors in order to fine tune auger speed. Also, I'm not sure I will get enough torque to keep the auger spinning fast enough to throw heavier snow with the smaller, faster motor. To summarize:

1) Bigger cu in motor has more torque, but spins very slow.
2) Pulley system is more complex, and might be impractically large when used with bigger motor.
3) Direct drive using smaller, faster motor is less tolerant of jams and might not have enough torque to throw heavier snow (especially with single stage blower)

Based on what I'm seeing, the direct coupling of a smaller motor simplifies the installation and allows the shortening of overall length by 6" or more. My gut is telling me I might be better off with a slightly larger motor than the one I'm borrowing from the PT finish mower (sacrifice max speed/throwing distance for a bit more torque), but I'm going to give it a try. Worst case I'll see where I'm at and buy a new motor if needed... at that point I'm hoping I'll know if another motor will help and what's the best size.

I'll post up pics once I get a chance to work on it. Hopefully it warms up a bit and I can start this weekend. Thanks again for all the input! :thumbsup:

 

[BobRip]

Right. Since pi (π) is the ratio of the circumference to diameter (c = π x d), its okay to use diameter in the discussion of pulley sizes (as long as we keep talking about ratios) like J J showed above. If the pulley used by the snowblower is 6" in diameter and a hydraulic motor's rpm is half of what is required, then a 12" drive pulley could be added on the motor to get the snowblower auger up to speed... essentially doubling the effective rpm of the motor (2:1 pulley ratio).

The challenge I currently have is three-fold: 1) I already have two motors to work with but the larger brush cutter motor rpm will likely make the drive pulley size impractical, and 2) adding pulleys to the system makes the design more complex and therefore more prone to failure and more difficult to build. (Additionally, a pulley systemF will increase the distance between the tractor and the snowblower, which will be more cumbersome to maneuver.) And 3), direct mounting with love joy style couplers won't tolerate severe jams of the auger and will force me to swap motors in order to fine tune auger speed. Also, I'm not sure I will get enough torque to keep the auger spinning fast enough to throw heavier snow with the smaller, faster motor. To summarize:

1) Bigger cu in motor has more torque, but spins very slow.
2) Pulley system is more complex, and might be impractically large when used with bigger motor.
3) Direct drive using smaller, faster motor is less tolerant of jams and might not have enough torque to throw heavier snow (especially with single stage blower)

Based on what I'm seeing, the direct coupling of a smaller motor simplifies the installation and allows the shortening of overall length by 6" or more. My gut is telling me I might be better off with a slightly larger motor than the one I'm borrowing from the PT finish mower (sacrifice max speed/throwing distance for a bit more torque), but I'm going to give it a try. Worst case I'll see where I'm at and buy a new motor if needed... at that point I'm hoping I'll know if another motor will help and what's the best size.

I'll post up pics once I get a chance to work on it. Hopefully it warms up a bit and I can start this weekend. Thanks again for all the input! :thumbsup:

Once you get the pulley size set to the desired impeller speed, the impeller will have the same torque regardless of which motor you selected. In other words speeding up the slow motor reduces the torque at the impeller. Also a belt has some losses, so you are loosing some horse power there.