Determining Hydraulic motor size

[msn56]

Hello:

I am new at this. I am not an engineer . I want to build a firewood processor. I read several books on hydraulics and am now trying to design it.

I have a question on calculating motor requirements . If someone could review my calculations I would sincerely appreciate it .

I want to be able to move a log 10,000# 2 feet in 10 seconds ( = 12 feet in one minute). This is where the log would be brought up so the chain saw could cut it

I want to figure out what size motor I need to do that. I am ff a model calculation in a book

The log will be moved by a motor with a 3 in sprocket hooked by chain to rollers with 3 in sprockets

1.) Horsepower Required

HP = F x L / t x 33,000 → 10,000 * 12 * 60 / 60 * 30,000 = 3.6 HP

2.) Number of sprocket turns needed to move 12 feet in 60 seconds

n = d / ∏ * diameter → 12? (3.14*3/12) = 15.2 turns

3.) RPM required to achieve 15.2 turns of the sprockets

N = n/t → 15.2 * 60 / 60 = 15.2 rpm

4.) Torque Required

HP = T1 * N1 / 5252 → T1 = 5252* 3.6 HP / 15.2 → 1243 lbft ft

First question if the motor had a smaller sprocket than the rollers than I would need more turns of the sprocket on the motor and I could reduce the torque required Correct?

5.) Torque Reduction

Ok SO now this is where I get confused in the book I am looking at there is a torque reduction N tm/ Nt → rpm motor / req rpm -

I guess I don稚 understand this
Its going to require the same force to move the log no matter what the rpm of the motor is so why does this get reduced ?

IN my instance if I chose a motor with 1000 rpm then the torque reduction would be

1000/ 15.2 → 65.7 reduction so that would be 1243/65.7 = 18.9 lbft-ft
this seems like a tremendous reduction

6.) Motor Flow rate required

assuming 2000 psi

Qm = p * Q / 1714 where p = pressure and Q = HP required

Qm = 2000 * 3.6 / 1714 = 4.2 gal / min

7.) Displacement of the motor

V = Q/N * 231 = 4.2 / 1000 * 231 = 0.97 in (3)

So to move a 10000 # log 2 feet in 10 seconds (12 in one minute) I would need

1.) 3.6 HP
2.) Output torque 18.9 lbft-ft
3.) Flow rate at 3000 psi of 6.3 gal / min
4.) Motor displacement of 0.97 cubic inches

Ok so now pulling out my handy dandy surplus center flier I note a hydralic motor that is $119 Its 4.5 cubic inch Char-Lynn 1044 in- lbs ( 87 ft- lbs) with an rpm of 760 at 15 gpm

I would have to recalculate torque reduction to 760 / 15.2 = 50

1243 / 50 = 24 ft lbs .

so that would still meet the requirements? Now to look a this little munchkin motor and to think its going to move a 10000 log is hard for me to imagine.

Are these calculations correct?

Also I have bee trying to research what motor I would use for the chain saw. The chain saw must go at about 9000 rpm at 5 HP. It seems the faster the motor the lower the HP. Is it possible to get a motor that goes that fast wiht that HP or would I have to mechanically increase the speed and if so that would seem way too fast to use a chain and sprocket ?

I really appreciate any help as does my back ! lol

 

[oldnslo]

For the chain saw finding motors that will operate at 9,000 RPM is a challenge. I think Rexroth makes some that are in the 8,000 - 9,000 RPM range. Look at their (A)A2FM series.

On your motor calcs for moving the log the one thing that is difficult to calculate is the inertia factors. I.e. getting 10,000 lbs moving and then stopping it.

Another thing to consider is the efficiency of the Char-Lynn motor. For the model you mentioned I would guess 50 - 60% efficiency. You may want to go the Eaton website and do some looking at the catalog specs on these motors.

Roy

 

[SPYDERLK]

I would just use a bigger drive sprocket on the chain saw to reduce drive rpm requirement. What is the hyd delivery and pressure capabilty of your prime mover?
larry

 

[oldnslo]

MSN56

For moving the log I would look at keeping the motor RPM's lower by using a larger displacement motor. These motors are called LSHT (Low Speed High Torque) and are designed for just that. This should help reduce the start-up time and stop time by reducing over running torque that would be created by a gear reduction system that turns into a gear increasing system during stopping. Flow input should remain similar for the same load.

On the Chain saw part I agree with SPYDERLK on using a larger drive sprocket. What you need to do here is figure torque at what ever chain speed you require. Then start bag figuring motor size. Personally I would probably look real hard and a axial piston style motor for this application since the speed will vary less with load changes due to having an efficiency around 95% and they are rated to higher RPM"s than gear or vane motors. NOTE: Using a piston style motor will require a case drain line that goes DIRECTLY to tank with less than 10 PSI back pressure.

Roy

 

[J_J]

You seem to be talking hyd motors, and using a chainsaw to run something. What about the hyd pump?

You will lose some efficiency from the chainsaw to the pump, and then another 85% to the hyd motor.

To pump 2 GPM at 3000 psi, will require about 4 HP starting out, and then you will lose some HP at the pump, and the hyd motor.

 

[msn56]

Hi Guys!

Thanks for all of the great input. I ma sure I am going to have more questions when it comes to making up the circuit. IN the meantime I have done a lot of reading and have gotten the calculations down especially the torque reduction. I agree that using a high torque low rpm motor makes the most sense otherwise I would need to do significant mechanical reduction of the rpm which would not be efficient. For the time being I am using the Char lynnn motor for the calculation as that's what I did the math for but the math would work for any motor. I will look at a high torque low rpm motor later and post a calc for that as well

This may be helpful to someone else so I am going to post it here


Determining Hydraulic Motor Size:

I recently posted some calculation in determining how to size a hydraulic motor and I had some confusion in my mind over the torque reduction . I think I have this figured out at this point and would like to post it for anyone else that may want to figure out how to calculate what size hydraulic motor you need.

First just some basic definitions which I found were very useful in helping me understand what we are talking about.

Energy = Force

This is like pulling on a 100# weight that has been nailed to the floor you are expending a lot of energy (although you are not really accomplishing anything) sort of like our government spending a lot of money but not really accomplishing anything lol!)

WORK is force applied over a distance W= F * L

SO if that weight were unscrewed and you lifted it 1 foot now you have accomplished something. Work is force that moves something

So if you lift 100# a foot you have done 100 ft lbs of work

And finally we have Power

Power is force and distance (work) over time
Lift 100# 1 foot in 1 minute

So

Energy = Force

Work = Force x Distance

Power = Force x distance / time

Now note that distance/time may be feet per second or feet per minute or rpm (revolutions per minute) or gpm (gallons per minute)

OK so say we have that 10,000 log and we wish to move it 2 feet per second or twelve feet per minute.

(Note : The actual force needed may not be 10K lbs due to resistance inertia etc (Remember call of the North and the dog that has to "break the load out" )etc. The log could be 15,000 etc and its going to have bark that gets stuck and irregular surfaces etc so you can plug in what numbers you think are correct

lets just say that we will have to push 10,000 # 12 feet in one minute.

HP (Horse power) = F * L / (time * 33,000)

So 10,000* 12 *60 /(60 * 30,000) = 3.6 HP

All right so we need at least 3.6 HP (if the motor were 100% efficient) - but most are not say you have an 85% efficient motor so then you would need 3.6 / 0.85 = 4.2 HP motor.

How much torque would need to apply to move that log assuming that you had a 1" shaft with 1/2 " nubs welded to it?

(see this pic https://lh3.googleusercontent.com/-...AAAAGc/4Q1qSxRuMbw/s1600/firewood-processor-l )

Torque is a force that is simply acting in a circular fashion (rotational force)
We have all felt torque. WE know that when we turn a bolt with a wrench the force we put into that wrench is torque (ie torque wrench)

Motors also make TORQUE

It is important to note that the torque rating of the motor is the torque at the motors shaft. (ie if you put a sprocket on it the torque would change see explanation below)

We know that Power = force * distance / time

SO for a motor the distance / time part is the same as rpm ie the revolution of the shaft determines the distanced covered (circumference of the circle) over time

ie if you have a motor with a one inch shaft and its rpm is 200 then the distance covered in one minute is PI * D * rpm = 3.14 *1 * 200 = 628 inches in one minute

HP = torque *rpm / 5252

SO in this instance 4.2 HP = Torque * rpm / 5252
Torque = HP * 5252/ rpm

So to determine what hp we need we need to figure out what the rpm is


To determine the rpm we have to know what size bar will be moving the log. If you note in the pic its a 1" bar with nubs welded to it. Lets say that its diameter is 2"

OK so how many times would the 2" bar need to turn to move the log 12 feet in one minute

C = Pi * D ---> 2 * 3.14 = 6.28 inches per turn

144 inches (12 feet) / 6.28 = 23 times

SO that bar in the picture would have to turn 23 times in one minute to move the log 12 feet in one minute (assuming that each time it turned it turned the log that much)

OK and what torque would that bar moving at an rpm of 23 need to apply to move that 10,000 lbs ?

We know that torque = 5252 * hp / rpm

Torque= 5252 * 4.2 / 23 = 959 ft lbs - seems like a lot

ok now here is where you have to look at the motor you have chosen and this is where the torque reduction (amplification ) comes in

Say I picked a motor that develops 87 ft lbs of torque at 750 rpm

HP = Torque * rpm /5252 87 (750)/5252 = 12.4 HP that should be plenty
according to the above need of 4.2 HP

Yet it says it only has 87 ft lbs of torque and my calculation says I need 959!

True but that motor is turning at 750 rpm

If I hooked it up using 1:1 gear ratio it would be spinning way too fast so I have to slow it down - to do that we of course would have to use some sprockets and the reduction in gearing would have to be 750 / 23 = 32 this would therefore need about a 32 :1 reduction in rpm - this would be like putting a 1 inch sprocket on the motor and a 32 inch sprocket on the hub - that motor would now have to spin around 32 times to just move that small 1" bar with the welded on nubs to turn around once ! its it like taking a 32 inch bar and using it as a lever you get tremendous force over a short distance and that is what the torque reduction is all about -or looking at in the other direction the torque magnification of the motor. SO the motors torque would be increased by a factor of 32 in this situation

and that holds true with what we know about power

we know power = force * distance/ time

we know HP stays the same and we know time stays the same so if distance changes (the motors 1 " shaft goes around 32 times while the nubbed shaft only goes around once ) then the force must change to keep the equation balanced

Thus the torque of the motor get significantly amplified (by 32 times)

87 * 32 = 2784 ft lbs well above the required 969 ft lbs

I always try to relate these things to life experiences - say you have a set of pulleys and you are trying to lift 1000# you would pull the rope quite a distance to move the 1000# a foot or so - you can pull the rope because the force is low but the distance long - the 1000# weight can be moved because there is a lot of force acting over a short distance.

Next what size pump would I need for this motor ?

Well looking at the ad in surplus center for a motor it says 15 gpm it also says 1800 psi 760 rpm and a torque of 87 ft lbs.

Lets see if that 15 gpm number makes sense

HP = Torque * rpm /5252 --> 87 (750)/5252 = 12.4 HP
so that is the HP of this motor at a flow rate of 15 gpm

Flow rare required Qm = p*Q /1714 where p = pressure in system and Q is the HP required

Qm = 1800*12.4/1714 = 13 gpm but if the pumo is only 85% efficient (which is usual) then 13/.85 = 15.3 gpm that sounds about right!

OK but how about for what we are using the motor for

well we only need 4.2 hp

Qm = 1800*4.2 /1714 = 4.4 gpm

So for our need we would only need 4.4 gpm

Well seeing that I am not mass producing these - it looks to me that the $119 motor would work just fine wow a lot cheaper than gas one !!

but of course we have to buy a pump and the gas motor to run the pump - but we can run multiple motors from the one gas one

Now as was pointed out a low rpm high torque motor would be the best choice and I will post a calculation for that.

PS In reality you will note that fergusons wood processor the shaft is spinning much faster than 23 rom as it doesnt always move the log (ie slips a lot ) but at least this gives you an idea as to what size motor would be useful


OK so that is how you calculate the motor size you need

1.) First determine the hp needed to move a weight x feet in one minute
2.) Next determine what size shaft you will be using so you can determine the number of rpms needed
3.) From that determine the torque you would need at that shaft
4.) Compare the rpms of the motor to that of the shaft that is moving the log and do a torque reduction.
5.) IS the torque of the motor greater than what you come up with in #4 ? if so then the motor probably is strong enough

1.) HP = f * d / (time * 33,000) = 10,000 * 12 * 60 / (60 * 33,000) = 3.6 hp
Say motor only has 85% efficiency - that means 3.5 / 0.85 = 4.2 HP
SO minimum is 4.2 HP

2.) OK the shaft is 2" circumference that will be moving the log
# of rpm to make 12 feet in one minute = 3.14 *2 = 6.28 inches per revolution
144/ 6.28 = 23 revolutions

3.) Torque needed at shaft to move 10000#
HP = T * rpm / 5252 T = HP *5252 / rpm = 4.2 *5252 / 24 = 919 ft lbs

4.) Motor torque can be reduced by 750 / 23 = 32
919/32 = 28.7 ft lbs

the motor puts out 87 ft lbs therefore it should be able to handle this situation !


(As a side point I was thinking a lot about torque and its measurement lets put a 6 sprocket on that motors shaft - is the torque at the outer edge of the sprocket the same as the shaft ? Well according to the formula HP = Torque * rpm / 5252 . The rpm has not changed and the HP is the same so the instantaneous torque should be the same correct ?

Wrong you cant - why because we know that work = F * distance and that torque reading is specific to the size of the motors shaft ( no one seems to tell you this anywhere .) Again use your real life experiences to think about this. Put a huge sprocket on the little motor you could probably hold it with your hand yet you wouldnt dare do that with the shaft of the motor.

Why? Because you are covering so much more distance with a big sprocket the force must be smaller to keep the HP constant ) SO if the shaft twIsts one inch it actually goes through about 1/6 of its rotations. If you had a ten foot sprocket on the motOr one inch of movement would not represent 1/600th of A ROTATION. THEREFORE YOU HAVE A WHOLE LOT LESS FORCE PRESENT

I hope this helps anyone else who may have struggled with these calculations


:thumbsup::confused2:

 

[oldnslo]

MSN,
torque is always stated as a unit of measure in distance. I.e. in-lbs, ft-lbs, nM etc. The torque a motor produces is always stated in one of these units of measure so stating that torque chages with distance is a true statement.

Roy

 

[muddstopper]

Hello:

I am new at this. I am not an engineer . I want to build a firewood processor. I read several books on hydraulics and am now trying to design it.

I have a question on calculating motor requirements . If someone could review my calculations I would sincerely appreciate it .

I want to be able to move a log 10,000# 2 feet in 10 seconds ( = 12 feet in one minute). This is where the log would be brought up so the chain saw could cut it

I want to figure out what size motor I need to do that. I am ff a model calculation in a book

The log will be moved by a motor with a 3 in sprocket hooked by chain to rollers with 3 in sprockets

1.) Horsepower Required

HP = F x L / t x 33,000 → 10,000 * 12 * 60 / 60 * 30,000 = 3.6 HP

2.) Number of sprocket turns needed to move 12 feet in 60 seconds

n = d / ∏ * diameter → 12? (3.14*3/12) = 15.2 turns

3.) RPM required to achieve 15.2 turns of the sprockets

N = n/t → 15.2 * 60 / 60 = 15.2 rpm

4.) Torque Required

HP = T1 * N1 / 5252 → T1 = 5252* 3.6 HP / 15.2 → 1243 lbft ft

First question if the motor had a smaller sprocket than the rollers than I would need more turns of the sprocket on the motor and I could reduce the torque required Correct?

5.) Torque Reduction

Ok SO now this is where I get confused in the book I am looking at there is a torque reduction N tm/ Nt → rpm motor / req rpm -

I guess I don稚 understand this
Its going to require the same force to move the log no matter what the rpm of the motor is so why does this get reduced ?

IN my instance if I chose a motor with 1000 rpm then the torque reduction would be

1000/ 15.2 → 65.7 reduction so that would be 1243/65.7 = 18.9 lbft-ft
this seems like a tremendous reduction

6.) Motor Flow rate required

assuming 2000 psi

Qm = p * Q / 1714 where p = pressure and Q = HP required

Qm = 2000 * 3.6 / 1714 = 4.2 gal / min

7.) Displacement of the motor

V = Q/N * 231 = 4.2 / 1000 * 231 = 0.97 in (3)

So to move a 10000 # log 2 feet in 10 seconds (12 in one minute) I would need

1.) 3.6 HP
2.) Output torque 18.9 lbft-ft
3.) Flow rate at 3000 psi of 6.3 gal / min
4.) Motor displacement of 0.97 cubic inches

Ok so now pulling out my handy dandy surplus center flier I note a hydralic motor that is $119 Its 4.5 cubic inch Char-Lynn 1044 in- lbs ( 87 ft- lbs) with an rpm of 760 at 15 gpm

I would have to recalculate torque reduction to 760 / 15.2 = 50

1243 / 50 = 24 ft lbs .

so that would still meet the requirements? Now to look a this little munchkin motor and to think its going to move a 10000 log is hard for me to imagine.

Are these calculations correct?

Also I have bee trying to research what motor I would use for the chain saw. The chain saw must go at about 9000 rpm at 5 HP. It seems the faster the motor the lower the HP. Is it possible to get a motor that goes that fast wiht that HP or would I have to mechanically increase the speed and if so that would seem way too fast to use a chain and sprocket ?

I really appreciate any help as does my back ! lol

I really think you have over thought this whole build process, making it much more complicated than it actually is.

First, what kind of wood are your processing that would weigh in at 10,000 lbs per log, thats some pretty big wood to be cutting up into fire wood. Still instead of adding more hydraulics and cost to your project, why not let the splitter cylinder provide the force to advance your log foward for your cutoff saw. All it would take would be to attach a long metal strap to your push plate on you splitting cyl, weld a few teeth to the strap and whenever the splitter cyl, runs out to split a stick of wood, it would pull the log to be cut, into the saw. Attach a clamp to your frame to hold the log in place whenever the splitter retracts and you are ready to cut your next block of wood. This would be a ton cheaper than buying another hydraulic motor, ( not to mention all the hoses and drive chains and sprockets), and making the rollers to move the log. I can invision a limitswitch and a electric over hydraulic control valve to automaticly clamp the log whenever it reaches the desired cutting position, Anyways??????

You are also overthinking your hydraulic motor for your chainsaw. To find out what requirements for cutting power you need, just look up the specs for a husky 3120 chainsaw. Its 8.5 hp, uses a .404 chain and has a chain cutting speed of around 4200 feet per min. The chainsaw turns 9600 rpms. In this situation, the rpms used are meaninless, what you are looking for is the FPM the chain is actually moving. It takes around 8.5 hp to move that chain at 4200fpm. The Huskey 3120 chainsaw has a very small dia sprocket, You can buy many different sprocket sizes to get your chain speed to where it needs to be, choose a hydraulic motor that will give you around 8.5hp turning a sprocket that is sized to turn your motor rpms into the proper fpm for the chain. A 3000 rpm or even a 5000rpm hydraulic motor is much cheaper and easier to find that the bent axis motors used on timber harvester equipment. They also run at much lower PSI's. You can build your complete hydraulic saw using new parts for around $500 with the slower speed motors, whereas you will pay $800-and way upwards, depending on size for a bentaxis hydraulic motor by itself. (hint, I have looked at several factory processors that use the old Greesen, now Parker, gear motors for their saw motors). Think around 1cuin displacement, 15gpm, 2500psi and you will be very close to what you need.
Check out Surplus Center - 0.700 cu in PARKER MGG20030-BB1B3 HYD MOTOR
Or, Surplus Center - 1 cu in CROSS HYD MOTOR 40MH10DACSC

 

[msn56]

Hey Mudstopper I like that idea :thumbsup: use the cylinder to move the log - could just use rollers on pillow bearings . I am trying to envision this. I could weld a rod hinge system on the plate of the splitter cylinder then have two arms that come up and connect via a horizontal piece that has teeth in it to clamp down on the log. This would need a hydraulic cylinder to push that horizontal var down and hold it there and that holding cylinder would need to move with the log

 

[muddstopper]

Hey Mudstopper I like that idea :thumbsup: use the cylinder to move the log - could just use rollers on pillow bearings . I am trying to envision this. I could weld a rod hinge system on the plate of the splitter cylinder then have two arms that come up and connect via a horizontal piece that has teeth in it to clamp down on the log. This would need a hydraulic cylinder to push that horizontal var down and hold it there and that holding cylinder would need to move with the log

I am trying to envision the clamp you are suggesting, but I cant see any need for it to move with the log. Not sure I can describe what I had in mine. I'll just throw out some made up measurements that can surely vary according to actual needs.

Suppose your trough that holds your log before being sawed, has a bottom width of 8 inches and is 16ft long. You could purchase a 8"X??"X16ft metal band. This band would lay in the bottom of your log trough, maybe on rollers as you suggested. One end would be attached to the push plate on your splitter cylinder. On top of the metal band material, you could weld a series of teeth, (large key stock could work). These teeth should be big enough to grap the log and pull it forward, yet shaped (quartermoon) so that as the metal band slides back under the log, so they cant grap the log to pull it back when splitter cylinder is retracted. As for the clamp to hold the log, this should be designed to grap the log and hold it in place to be sawn when the proper cut lenght is reached. I would suggest maybe make the clamp a couple of vee shaped fixed plate or stough bar material that actually will lift the log slightly as it holds it in place.

To take this one step further, suppose that you use a electric control valve to supply power to the hydraulic cylinder that clamps the log. If you did this, you could use a limit switch to be tripped when the log reachs a pe-set or pre-determined measurement for the wood to be sawn. As the log is advanced by the splitter cylinder, the logend would trip the limitswitch, activating the clamp cylinder, grapping and holding the log the same lenght for every cut. This is the reason I suggest that the clamp lift the log slightly when it clamps the log, this would ensure the log wont be pulled back as the splitter cylinder is retracted.

I can see (in my head) my design plainly, I think it will work, I just dont know if I described it well enought that you can figure out what I mean.