Calculating hydraulic motor sizes

1. Oct 18, 2011

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 foot in one minute)

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” sprocket hooked by chain to rollers with 3 “ 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’t 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 an engine 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 for a backyard tinkerer 2. Oct 18, 2011 DaveC426913 Measure twice, cut once. 3. Oct 19, 2011 msn56 Sorry not following you 2 feet in 10 seconds = 12 feet in 60 seconds = 12 feet in one minute or is is that an ostentatious display of grammar is not indigenous to this site 4. Oct 19, 2011 DaveC426913 :Embarrassment: That was a display of my own stupidity. I actually read it wrong and thought you made a mistake. And didn't have anything else to add. Changing my own motto to "read twice, save once". 5. Oct 19, 2011 msn56 no problem - thought you were making fun of my grammar - in building we often say foot as a collection of single units even thought the proper term is feet now can anyone look at those calculations my back wont hold out through another firewood season ! LOL 6. Oct 20, 2011 MikeTheLost Could you explain a bit more what you are attempting in general and with what end result? My own experience with firewood is limited, but I recall cutting the big sections to the desired length and then splitting the sections to get manageable bits ready to toss in the fire place. Any lifting I would use a winch. 7. Oct 20, 2011 msn56 HI: Here is an example of what I wish to make: The above calculation would be for the motor that is moving the log forward. There would actually be several hydraulic motors and cylinders as follows MOtors; 1.) To move log forward 2.) To run chainsaw 3.) To run a conveyor to remove wood once split Cylinders 1.) To split the wood 2.) Arm to hold the wood while being split 3.) Possible to move wedge up and down 4.) To move chain saw in cut stroke although some people do this by hand THe motor to move the log forward would I believe require the most HP . I did calculations for the other two and if I am doing them right they would not need to be as powerful. The cylinder for the splitter is clearly the biggest demand there and I will either go with a 4 or 5 inch preferably at least a 24 " stroke. HOpe you can help I do have a seperate question on the chainsaw motor if we get to that MIke Last edited by a moderator: Sep 25, 2014 8. Oct 20, 2011 MikeTheLost The machine in the video seems to work, eyeball it and duplicate. My guess is that they used parts that were cheap and easy to find, and changed things when it didn't work well enough. I would approach the project in two steps, make it basically work, enhance after I had some firewood. Also I bet if you emailed them they would be happy to share any and all the details. Moving the log forward won't take a lot of power, its motion on rollers without any lifting. Look at the size of the motors used on each roller, pretty small, plus the whole thing is running off an average sized tractor. Doing the chainsaw portion manually looks like a good option for safety. I would make the chainsaw lever a two stage thing so when you pull the lever down it first clamps the log, then moves the blade of the chainsaw down. I'd leave the splitter higher up from the ground so the split firewood can drop off out of the way. I'd skip the conveyer dumper, doesn't all that wood need to be stacked? I'm also thinking this whole log splitting thing could be much more dramatic and dangerous looking. Maybe stand the uncut log upright and use a couple of electrodes to pass a large current down the center and exploding them? Or maybe a big arm, like the hammer amusement rides, that grabs a log and lifts it into the air and then smashes down on a splitter? 9. Oct 21, 2011 msn56 I am planning to take the conveyor and then dump it into bins - I will then be able to lift the bins with my tractor (so there is no stacking) . Then when i need fuel for my outdoor boiler i Just pick up a whole bin and move it - a lot less labor as far as dramatic here's an OSHA approved one (suprised he still has hands left!) lol!! still any idea on how do calculate the correct motor size ? Last edited by a moderator: Sep 25, 2014 10. Oct 21, 2011 MikeTheLost Make the splitter higher and/or push up a chute and dump into a bin, splitter should have plenty of force, and it removes a complicated system. I would not aim for a correct size motor, I would look for the cheapest surplus motor I was sure was big enough. You can see in the youtube link the roller motors are fairly small, the chainsaw around 5 to 10 hp, and the splitter is whatever size a splitter uses normally. Unless you want to, I would resist making a contraption, and focus on making firewood. Once you have the basic operation of cutting and splitting working, you can always add to it. Rule number one of making a thing, the first one you make may work, but primarily its an education for the second one. 11. Oct 24, 2011 potatoecannon I think there is a problem with your very first equation (HP required to move the log). The weight of the log (10 000#) is the force that would be required to LIFT the log at a constant speed. Initially, the force would need to be even higher, as you would need to accelerate the log from 0 speed. For the case of moving the log horizontally down a set of rollers, there will be 2 sets of forces the motor needs to supply: 1) Force to accelerate the log from 0 ft/min to 12 ft/min 2) Force to overcome the friction in the rollers This is not 10 000#, but instead is F=μ*N, where N is the Normal force (weight of log), and μ is the kinetic coefficient of friction of the roller bearings. Note that the initial force to start the log in motion will be higher (as the static coefficient of friction is always larger that the kinetic). As long as the motor can provide more torque than is necessary to overcome the static friction of the rollers, the log will accelerate and move down the conveyor. The more excess power you have, the faster it will accelerate up to the 12'/min the system is geared for. 12. Oct 24, 2011 msn56 Yes I absolutely agree there will be more force necessary initially to overcome the inertia of the "dead" weight and any resistance to movement. What that is I do not know but will factot hat into the equation. I have also figured out my calculations for torque reduction with the motor and understand that now. I will post a revision as well as explanation for anyone who may be interested 13. Oct 24, 2011 potatoecannon I think you may have misunderstood me. I think your motor is too big. While the log is moving horizontally, the only force required from the motor is to overcome the roller bearing friction. If μ≈0.002, then the force required to keep the log moving is only (10 000) * (0.002) = 20 lb. The rest of the force (9 980 lb, since you calculated the motor size of 3.6 HP based on 10 000 lb force at 12 ft/m) will be used to try to accelerate the log. Since V = a*t, and a is ≈ 1 gravity (from F=m*a), the log will try to accelerate at around 32.2 ft/s2. It will reach the 2 ft/s in about 0.06 seconds, assuming no slippage. The first roller may just spin against the log instead of pulling it (if the sliding friction between the roller and wood is less that the static friction of the log on the rest of surface it is on). 14. Oct 24, 2011 msn56 Yes I did misunderstand lol ! I shoudl listen first then type! I think iI see what you are saying - basically the most force would be required to break the log out of its inertia and into movement. This would be less than 10,000# because we are not lifting it but rather rolling it (?). Also once that initial inertial resistance is overcome the motor will still be producing that much force (power) into the log - however, we no longer need to overcome inertia and to only keep the log moving you need a lot less force. Therefore, the additional force will actually cause acceleration of the log 15. Oct 24, 2011 potatoecannon Exactly. And that would be fine, provided your motor is speed-controlled (ex. fixed volume supplied by a gear pump). If the motor is supplied by an on/off type valve connected to a pressure-regulated supply, it will overspeed quickly, and either just spin the roller against the bark, or fire off that log a lot quicker than expected. 16. Oct 24, 2011 MikeTheLost The motors will need to be a bit oversized due to the uneven surface of the logs, but I still think you should be looking at any convenient motor about the size other people seem to be using successfully. 17. Oct 26, 2011 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/-j...od-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

18. Oct 27, 2011

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?

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

Last edited by msn56; Yesterday at 04:17 PM.

19. Oct 27, 2011

MikeTheLost

Thats a lot to read, but I'll mention a few things.

Work = Energy
HP isn't changed by gearing (except for frictional losses), torque is changed by gearing.