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 dont 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 = **.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 **.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