PT AL-629 ( can it be done ? )

[Bob999]

But, the two wheel motors in a series circuit where the pump only puts out 1500 PSI (your propositition) would only be using 750 PSI each -- 750 PSI at the front motor, and 750 PSI at the rear motor... that's where you're misunderstanding lies.

I understand your point.

Perhaps you do not understand mine--Without specs or data there is nothing to anchor the analysis. What the calculators show is that doubling either pressure or flow doubles horsepower. But--so far as I know there is nothing to anchor either pressure or flow produced in the drive circuit. We both are just guessing.

That guessing can be resolved one of two ways--get specs or get measurements.

 

[KentT]

I understand your point.

Perhaps you do not understand mine--Without specs or data there is nothing to anchor the analysis. What the calculators show is that doubling either pressure or flow doubles horsepower. But--so far as I know there is nothing to anchor either pressure or flow produced in the drive circuit. We both are just guessing.

That guessing can be resolved one of two ways--get specs or get measurements.

But, we can reverse engineer an approximate gpm flow based upon the known specs of the White wheel motor and the speed the PT will travel. With 23" tires, it takes 14.61 RPM for the PT to go 1 MPH. It is rated for 8 MPH plus, and I've heard estimates of 10 MPH with these old wheel motors. 8 MPH would require 116 RPM. 8 MPH would require a flow of more than 6 gpm, at any rated PSI, according to the motor's published performance specs. 10 MPH would require 8 gpm at 500 - 750 PSI. So, though we can't reverse engineer a "hard number" we can get a range -- the tram pump is definitely capable of putting out between 12 - 16 gpm at 1000-1500 PSI (twice the PSI available to any wheel motor because of the series circuit.) We can anchor that much....

Meanwhile, we have the hard performance data to compare one wheel motor to another. For example, replacing the White RS series 12.5ci wheel motor with the Char-Lynn S series 18.2ci wheel motor will increase the torque by 61% at 4 gpm flow at 1000 PSI or 56.5% at 6 gpm at 1000 PSI. Comparing the performance data of the two motors, the least increase in torque is 39% -- from from 1809 PSI to 2897 PSI at 2 gpm and 1000 PSI. Even a 39% increase is significant....

And for those of us who need more torque, it is inviting. Even if my tram pump isn't operating at 100% efficiency, a 39% increase in the amount of torque currently available (whatever that may be) -- is still significant to me....

 

[Rivco]

I would love to post my honest thoughts about- where all this over analytical, redundant, MIT BS is headed with regards to increasing the wheel torque on the 180, 422, or 425, but I will restrain myself and just say- HAT'S OFF TO JOHARAL for taking the time to just upgrade to a larger displacement wheel motor that operates within the restaints of the OEM engine, pump etc. without rewriting the book on hydraulic's 101

 

[MossRoad]

But, the two wheel motors in a series circuit where the pump only puts out 1500 PSI (your propositition) would only be using 750 PSI each -- 750 PSI at the front motor, and 750 PSI at the rear motor... that's where you're misunderstanding lies. For a wheelmotor in a series circuit to produce the power that corresponds to 1500 PSI, the pump would have to output 3000 PSI.

You sure about that? Think of pressurizing a garden hose. Anywhere along the line the PSI is about the same until you open a valve at the other end. Then the pressure drops out.

So, if you drive your PT up against an imovable object and then step on the treadle, before the tires spin or a relief valve kicks in, I would think you would get the same PSI between the pump and the first motor and probably between the first and second motors, then dropping off dramatically after the second motor as it has free return to the pump. The PSI would not be divided in a series circuit, nor would the GPM.

 

[MossRoad]

I would love to post my honest thoughts about- where all this over analytical, redundant, MIT BS is headed with regards to increasing the wheel torque on the 180, 422, or 425, but I will restrain myself and just say- HAT'S OFF TO JOHARAL for taking the time to just upgrade to a larger displacement wheel motor that operates within the restaints of the OEM engine, pump etc. without rewriting the book on hydraulic's 101

Well, it never hurts to learn how something works inside and out. It is a great brain exercise and keeps us young.

HEY! I think I have just proved that owning a PT is the fountain of youth!

 

[KentT]

You sure about that? Think of pressurizing a garden hose. Anywhere along the line the PSI is about the same until you open a valve at the other end. Then the pressure drops out.

Yup, I'm sure. It helps to think of it from the perspective that it is the wheel motor that is causing the PSI, not the pump. The resistance of each wheel to turning adds 1/2 the pressure to the circuit (in a "100% efficient system" where you're not considering the resistance of hoses, fittings, etc.). The two parallel circuits on the PT have the same input pressure to the front wheel motor on each side. But, each wheel motor in that circuit operates at 1/2 the pressure that the circuit contains. The wheel motor in a series circuit operates on delta PSI -- the difference in pressure between its input and output ports. The wheel motor in front must be able to "contain" the full pressure, but because it has the back-pressure of the rear motor on its output port it can only produce the power of the input port pressure less the output port pressure.

 

[KentT]

... I would think you would get the same PSI between the pump and the first motor and probably between the first and second motors, then dropping off dramatically after the second motor as it has free return to the pump. The PSI would not be divided in a series circuit, nor would the GPM.

Think of it, MR... Why would the pressure drop off after the second wheel motor and not drop off after the first? Half of it drops at the first and the remaining half drops at the second... If the motor had the same pressure on on the input port and the output port, it wouldn't move at all.

It also helps to think of it as "load" on the system instead of pressure. Half the load comes from the front wheel motor and half from the rear one.

 

[MossRoad]

Well, if the first wheel motor had no resistance and the second one was locked against an imovable object, the first one would just be rolling with the flow and the second one would be providing all the resistance to the flow.

 

[KentT]

Well, if the first wheel motor had no resistance and the second one was locked against an imovable object, the first one would just be rolling with the flow and the second one would be providing all the resistance to the flow.

But, flow cannot take place except across a pressure differential... if the pressure was the same on both ports, no flow would be taking place. So, the first one wouldn't move at all.

Added: I wanted to add that if you are ever pushing against something with a PT bucket, for example, and the front tires start spinning faster than the rear because of down-pressure lifting them up, the difference in speed between the front and rear tires would be caused by the leakage past the rear wheel motors. If you'll notice, those tires still spin slowly when they do spin -- in relation to the treadlle position and throttle setting. But, if the wheels were plumbed parallel (the concept behind the Stray mod) the front wheels would get almost all the flow and they'd spin like mad...

 

[KentT]

I would love to post my honest thoughts about- where all this over analytical, redundant, MIT BS is headed with regards to increasing the wheel torque on the 180, 422, or 425, but I will restrain myself and just say- HAT'S OFF TO JOHARAL for taking the time to just upgrade to a larger displacement wheel motor that operates within the restaints of the OEM engine, pump etc. without rewriting the book on hydraulic's 101

Sorry if I'm redundant. I'm ****ytical by nature and a systems requirements ****yst by training, so I do tend to over-****yze things. My wife says that's only partially correct -- that I'm just ****...

That's how I learn, though, and how I apply that learning to new situations -- I try to understand the "theory of operation" behind something. In this case, this wheel motor upgrade will likely cost about 25% of what I originally paid for my PT and the 5 attachments that came with it. I could likely buy most any new attachment available (and perhaps two) for the price of this upgrade alone, by the time it's done, so it is significant -- at least to me...

 

[MossRoad]

But, flow cannot take place except across a pressure differential... if the pressure was the same on both ports, no flow would be taking place. So, the first one wouldn't move at all.

I think if you put the wheel up in the air on the first motor in the circuit, and locked the wheel on the second motor in the circuit, then applied the treadle, you would measure the pump's maximum PSI, let's assume 1500psi, at the output of the pump, 1500psi at the output of the first motor, and 0 psi at the output of the second motor, because nothing would get past the second motor, at least, not much, since it is locked.

Then, if you let the second motor turn slowly, you would see, say, 1450psi at the output of the pump, slightly less than 1450 at the output of the first wheel motor and only a few psi at the output of the second wheel motor, as there would be little resistance to the return side of the pump.

As long as there is no resistance on the wheel on the first motor, it will not cause much PSI drop in the circuit at all. Almost full pressure will get to the second wheel motor.

Think of a garden hose again. Run it to a T fitting with a sprinkler on it, then to another hose and a second sprinkler. Turn on the water and the first sprinkler in the line will spray a bit farther than the second one in the line. Plug the first sprinkler with your finger and the second one sprays harder and farther.

 

[Rivco]

Sorry if I'm redundant. I'm ****ytical by nature and a systems requirements ****yst by training, so I do tend to over-****yze things. My wife says that's only partially correct -- that I'm just ****...

That's how I learn, though, and how I apply that learning to new situations -- I try to understand the "theory of operation" behind something. In this case, this wheel motor upgrade will likely cost about 25% of what I originally paid for my PT and the 5 attachments that came with it. I could likely buy most any new attachment available (and perhaps two) for the price of this upgrade alone, by the time it's done, so it is significant -- at least to me...

Kent;
Sorry- no insult intended or implied!, I just think that sometimes (myself included ) we tend to over analyze problems trying to find a solution. I'm sure everyone including myself is applauding you for your efforts and are very appreciative, but to banter back and forth about flow rates, psi., wheel motor resistance, etc. can be overwhelming to someone ( with less knowledge ) thinking about attempting the same modification, and decides to shelf the whole idea because of what they are reading in these replies. That's why I stated in my earlier reply HATS OFF TO JOHARAL- he researched larger displacement wheel motors that would work for his needs and installed them without posting all the analytical inner workings on the PT hydraulic system. Bottom line- Let's get ur done and then post your performance results after the upgrade is completed. Again- thanks for all your efforts and research.

 

[KentT]

I think if you put the wheel up in the air on the first motor in the circuit, and locked the wheel on the second motor in the circuit, then applied the treadle, you would measure the pump's maximum PSI, let's assume 1500psi, at the output of the pump, 1500psi at the output of the first motor, and 0 psi at the output of the second motor, because nothing would get past the second motor, at least, not much, since it is locked.

Then, if you let the second motor turn slowly, you would see, say, 1450psi at the output of the pump, slightly less than 1450 at the output of the first wheel motor and only a few psi at the output of the second wheel motor, as there would be little resistance to the return side of the pump.

As long as there is no resistance on the wheel on the first motor, it will not cause much PSI drop in the circuit at all. Almost full pressure will get to the second wheel motor.

Think of a garden hose again. Run it to a T fitting with a sprinkler on it, then to another hose and a second sprinkler. Turn on the water and the first sprinkler in the line will spray a bit farther than the second one in the line. Plug the first sprinkler with your finger and the second one sprays harder and farther.

The pump doesn't cause PSI -- the wheel motors and resistance in the circuit cause the pressure... the pump causes flow that creates the pressure against the resistance of the wheel motors and lines. The motors serve somewhat as "pressure relief" valves.

What you'll truly see in your example is the front wheel turning fairly slowly (in relation to the throttle setting and treadle position) as leakage past the rear wheel motor occurs. This spinning will relieve some of the input pressure on the front wheel, so the difference between the input PSI and the output PSI of the front wheel will not be dramatic, but it would still be measurable. The only difference between the output PSI of the first wheel motor and the input of the second wheel motor will be the resistance in the lines and fittings, so it will be minimimal. Meanwhile the locked rear wheels are providing all the resistance that the pump can try to compensate for, and the pressure in the lines between the first and second pump will rise... so yes, your description is correct. That's one of the great features of the series-plumbed hydraulic drive system -- the power goes to the wheel that needs it automatically.

But, that is not the normal operation of the system. Allow both wheels to turn of their own accord (i.e. not locked) and you'll see the pressure on output side of the front wheel motor be essentially 1/2 of the pressure on the input side... the front wheel creates 1/2 the resistance (pressure) and the rear wheel creates the other 1/2. The turning of the front wheel relieves 1/2 the pressure and the turning of the rear wheel relieves the other 1/2...

BTW -- don't go too far with your garden hose and two sprinklers analogy because it is a parallel circuit -- not a serial one. In a series circuit the water coming out of the first sprinker would be the input for the second sprinkler, and that is certainly not the case.

 

[KentT]

Kent;
Bottom line- Let's get ur done and then post your performance results after the upgrade is completed. Again- thanks for all your efforts and research.

I plan to order the wheel motors early next week, likely the 22.7ci Char-Lynn ones, and then wait the 4-6 weeks for them to be built. I'll either have my friend install them or I'll install in late June when I'm down there on vacation. Either way, it'll be a while before I can report back.

One of the reasons why I'm being so **** is that these are special-order items that I can't return... I'm stuck with them if I order them.

 

[woodlandfarms]

What happened to the Whites? They seemed like an excellent deal...

 

[KentT]

What happened to the Whites? They seemed like an excellent deal...

They may be too long to fit in there readily -- they're 1.9" longer than stock. You have to pull the wheel motor inside the body (i.e. frame) of the PT far enough to clear the mounting box, and then slide it back out to mount it. That includes the 2.2" shaft plus the 2" of extra length.

I'll be talking be talking to my friend this weekend to see if he thinks he can get them in there without having to cut huge holes in the body to angle the motors and get them "threaded" up in there.

 

[woodlandfarms]

Wouldn't the length have the added benefit of stability? Also, it looks like on the 1850 that it is really simple to remove the engine (not that removing the engine is simple). But with 4 bolts you could have the engine loose enough just to push it one way or the other.

Not sure about your front situation, though....

 

[BobRip]

You sure about that? Think of pressurizing a garden hose. Anywhere along the line the PSI is about the same until you open a valve at the other end. Then the pressure drops out.

So, if you drive your PT up against an imovable object and then step on the treadle, before the tires spin or a relief valve kicks in, I would think you would get the same PSI between the pump and the first motor and probably between the first and second motors, then dropping off dramatically after the second motor as it has free return to the pump. The PSI would not be divided in a series circuit, nor would the GPM.

Moss, with the wheel motors stalled all you have for flow is leakage (or bypass) through them. If they are identical (like that's going to really happen) they can be modeled as two series resistors and half of the voltage (or pressure here) would be across each one. The total pressure across the two motors must be the total of what the pump is putting out.
With no leakage (that's not going to happen either), the first motor would have all of the pressure drop across it and none on the second. In reality the total pressure across the motors is going to be what the pump is putting out. How it balances between the two motors is a constantly changing thing. I hope this makes sense.

 

[MossRoad]

BTW -- don't go too far with your garden hose and two sprinklers analogy because it is a parallel circuit -- not a serial one. In a series circuit the water coming out of the first sprinker would be the input for the second sprinkler, and that is certainly not the case.

You know, I guess you're right.. The water would have to pass through the nozzle of the sprinkler and into the hose and onto the next device and that is definately not the case. Thanks for teachin' me a lesson.