## Hydraulic Motor calculations

ipt>
Hello:

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)

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