|Feb3-12, 07:17 PM||#1|
Torque calculations for logging winch
Hi Guys :
I have been reading about torque and gears and not being an engineer or physicist would like to know if my thinking is correct in the following example
Ok so Say I have a hydraulic motor and it generates 3000 ft lbs of torque and it is going to be turned at 650 RPM (tractor PTO) please see pic to explain this
Coming out of that motor is a 1 " shaft. Therefore the torque at the 1" shaft is 3000. Ok if I put a 2 " pulley on that shaft its torque would be 1/2 of that or 1500.
NOw I will connect the 2" pulley to a ten inch one by chain. THat 10 inch pulley also has a one inch shaft So now the torque of the the one inch shaft on that the 10" pulley woudl be 5 x as much or 7500 . If I put a 2" gear on the end of that one inch shaft I again lose half my torque and am down to 3750.
ONce again I connect that 2" gear to a 10 inch one and that would give me 5 x as much torque - if I had a one inch shaft coming out of it 0r 3750 *5 = 18750.
However instead of a one inch shaft I will have a 3 inch pipe to wrap the cable on so that will have 1/3 as much or 6250 lbs.
Are these assumptions correct? Really appreciate any help
|Feb4-12, 09:45 AM||#2|
Your assumption that changing shaft diameter affects torque is incorrect.
Keep showing units with answers to all calculations.
In the chain drive, calculate chain tension (force) using the equation:
force = torque / distance
In this case, the distance will be the radius of the sprocket or:
F = T / r
After calculating chain tension, convert back to torque in the axis of the 10" sprocket.
T = r X F
You will need to convert from torque to tension and back twice (once for each speed reduction stage).
In the end, the torque at the axis of the 2nd 10" sprocket will be equal to that in the 3" pipe. I would convert the torque in the pipe to force in the cable attached to the pipe (use the equations above).
I hope that helps with your initial torque calculations.
The sizes that you suggest for your components may not be appropriate for the torque and tension values that you calculate.
The size of the cable on the 3" pipe will affect the tension available.
3000 ft*lb at 650 RPM is equivalent to 370 horsepower; seems like alot (I'm not a tractor expert).
After you calculate the tension available, you should ask yourself if it is practical.
|Feb4-12, 01:33 PM||#3|
Thanks for your answer. I guess this is where I am getting confused - you say that changing shaft diameter will not affect torque -
This is what I dont understand - on the last pulley if I had a 1" shaft sticking out you are saying that this 1" shaft will pull with the same force as a 10" shaft -??
sO SAY THE 1" SHAFT CAN PULL 7500# ARE YOU SAYING A 10" SHAFT COULD PULL 7500#?
SAY THE RPM OF THE SHAFT IS 1OO
THE 1" SHAFT WOULD THEN PULL IN A 7500 # LOG 314" IN ONE MINUTE
WHILE A TEN INCH SHAFT PULLING 7500# WOULD PULL THE LOG IN 3140 INCHES IN ONE MINUTE
IT SEEM TO ME THAT THE FORCE WOULD BE MUCH LESS WITH THE TEN INCH SHAFT THEN THE ONE INCH SHAFT ?? aM i USING THE TERM TORQUE WRONG ??
ps THE 300 FT LB IS JUST A hypothetical
|Feb4-12, 11:11 PM||#4|
Torque calculations for logging winch
|Feb5-12, 07:30 PM||#5|
Thanks for the response . Sorry about the letters. I have been reading about torque and I think I am beginning to understand it. Let me ask you this - is torque only measure at the shaft ?
If that is the case then I can see how torque at the shaft never changes. If I understand it correctly torque is a force - however one must take into account the length of the lever arm - a longer lever arm moves more but with less force at the outside of the circle when compared to a shorter lever arm it moves less but with more force - each cause equal torque AT THE SHAFT.
Having said that - what do you call the force at the OUTSIDE of a gear on a shaft ? IS that just force in pounds??
OK so now if we come back to my original post - that last 10" sprocket
I understand what you are saying that the size of the shaft deosnt matter when measuring torque because torque is measured at the center of the shaft?? Is that correct? HOwever the pulling force at the outside of the shaft does vary by diameter of the shaft or gear that is attached ?
|Feb6-12, 08:01 PM||#6|
It looks like you already have lots of good answers, but let me see if I can simplify it for you.
You have two 5:1 gear reductions for a total of 25:1.
So you will have 25 times as much torque coming out as you put in, ignoring frictional losses.
Your first small sprocket sees the same torque as the motor.
Your first large sprocket sees five times the torque.
The second small sprocket sees the same five times.
The second large sprocket sees 25 times the torque, as does your winch drum. That would be 75,000 ft-lbf.
You have a 1.5 inch radius on your drum. Let’s assume that you have half inch wire rope on the drum, so the effective radius is 1.75 inches. So your maximum expected line pull is:
75,000 / (1.75/12) = 514,000 lbf.
Just broke your half inch rope. It was rather large for a three inch drum anyway.
|Feb6-12, 08:04 PM||#7|
I've got no indea how much force it takes to drag a 7500 pound log. Depends on what it gets hung up on. So I did not go there. But your drum will turn at the speed of the motor divided by 25.
|Feb6-12, 09:27 PM||#8|
Hi Thanks for the response - I do see how you are stepping up the gears but let me ask you this. You say the first small sprocket sees the same torque as the motor
that has to be at the shaft not at the outside of the gear - I guess this is where I ma getting confused - torque must be measured only at the shaft? - it cant be at the outside of the gear because if it were you could just put a huge gear on there and get speed and the same pulling force at the same time and we know this must not be true - So when you say the first small gear has the same torque as the shaft you must be talking at the shaft? IS this the definiton of torque?
as far as the pulling force at the first small gear it would half to be 1/2 as much as the shaft but twice as fast ?
So I think the key in my understanding torque is that it is by definition the force that occurs at the shaft taking into account both the lever arm and the force applied - talking about torque at the outside of gear doenst make any sense - you can only talk about force there not torque ?
|Feb6-12, 11:28 PM||#9|
One thing to consider is that torque doesn't just appear out of nowhere. It must be applied to the shaft of the motor before it can be used to move a load. Inside any rotating type motor, torque is developed as a distance multipied by a force.
|Feb7-12, 05:53 AM||#10|
Torque is not a force. Every sprocket, motor, or drum will see the same torque as the shaft it is attached to.
|Feb7-12, 06:42 AM||#11|
I just want to say be careful, as noted above, 3000 ft-lb at low rpm will generate quite a bit of torque. That 2" pulley on your initial axle will generate around 9 tons of force on the cable. Be sure you use a cable/strap that is rated to that.
|Feb7-12, 12:13 PM||#12|
Here's how I think about torque, maybe this explanation will make it easier to understand:
The first thing to note are the units: its a force at a distance, like ft-lbs. Thus a shaft that has a torque of 3000 ft-lbs means if you stuck a lever arm on it 1 ft long, it would lift/pull 3000 lbs from that one-foot-out point. If you go 2 ft from the center, it lifts or pulls 1500 lbs. (Because 2ft * 1500lbs still equals 3000 ft-lb) At a half foot out, it would pull 6000 lbs. And so on. Note it doesn't matter if the shaft is 1" or 6" or whatever. The torque measurement is saying "*if* you go one foot out, this shaft would pull such-and-such amount of force".
Same thing if we're in different units, like N-m (newtons are force, meters are distance).
Anyway, that's how I think of it. I'm not 100% sure I'm correct with this, but if not I hope someone will correct me!
|Feb7-12, 10:10 PM||#13|
Yea I finally think I see whats going on
OK so if you look at my orginal drawing if i have the motor that has a 1 inch shaft and has 1000 ft lbs of torque that means it can pull at 1000 # at the shaft.
Ok put a 2" gear on that shaft - the torque is the same but since the lever arm is twice as long its pulling force is half as much
SO that 2" gear is now only pulling at 500# but does it twice as fast
OK so here is where i was getting confused.
NOw we hook that 2 " gear to a 10" gear by a chain. I thought that since that gear is bigger - the pulling force must change -- but it doesnt ... because that 10" gear is moving the same amount as the 2" gear its attached to. ANother words the 2" gear goes around 5 times to move the ten inch gear around once. SO the two inch gear has covered 5 * 2PI while the 10" gear moves 10pi - so they each move the same distance so they each have the same pulling force !
NOw if on the shaft that the ten inch gear is attached to you hook another 2" gear at the other end - THAT 2" gear has 5 x the force because the distance it moves is 1/5th of the ten inch gear !
Now it all makes sense!!
Thanks guys I love to learn and understand something new - cool !!
|Feb8-12, 06:00 AM||#14|
I think you are still confusing torque. Apples & oranges. Also we are looking at it backwards. The load generates all forces and torques. The motor matches what the load requires up to its max capacity.
|Feb8-12, 06:02 AM||#15|
I meant confusing torque & force.
|Feb8-12, 08:27 AM||#16|
If I understand it correctly torque consists of a measurement of force over a distance thus you have the pushing force on the lever. Simply we always look as the pushing force being 90 degrees that would be T=F * r where F = pushing force and r is the lenght of the lever. However that pushing force could be at an angle and that is why the complete formula for toque takes that into account such that torque = FR Sin (of angle of push)
For this discussion we will assume that the force pushes at 90 degrees
Thus the motor has a torque of 1000" acting over 1 " - if we put a gear on that that measure two inches the torque is the same - but since the radius is twice as big the force is half as much or 500# over 2 inches
when you attach the 2" gear to a ten inch gear via a chain the distance they move is the same therefore the force has to be the same - this is equivalent to having two gears mesh - the torque is the same for each and the distance they each move respective to each other is the same therefore the force must be the same
|Feb9-12, 09:23 PM||#17|
Hi MSN56 -
This isn't correct:
OK so if you look at my orginal drawing if i have the motor that has a 1 inch shaft and has 1000 ft lbs of torque that means it can pull at 1000 # at the shaft.If a motor is rated at 1000 ft-lbs of torque, and has a 1" shaft, it means that *if* you put a pulley with a 1ft radius (2ft diameter), it would pull 1000 lbs on a rope wrapped around that pulley.
But your later comments give me the impression that you do understand, and that earlier comment was just a typo. (Then again, I may not be getting it quite correctly either!)
So back to your original problem: when using pulleys and gears, the ratio changes the torque, but not the power. Neglecting efficiency losses, the power in remains the same no matter how much you change the gear ratios. Thus to maintain the same power, as you change the ratio, if the torque goes up, the speed goes down, and visa-versa.
In your original post, look at the overall ratio change. In the first stage, you go from a 2" pulley to a 10" pulley. That's 1:5. The torque goes from 3000 ft-lbs to 15000 ft-lbs. Note the RPM's drop by a factor of five to 130 rpm
Then you do it again, 2" to 10" which is 1:5. Torque now goes up 5x more, to 75000 ft-lbs, while rpm is down to 26. (Total power still hasn't changed, except for friction losses).
At this point the torque is unchanged whether you end with a 1" shaft or a 3" pipe. It's still 75000 ft-lbs, meaning if you attach a 1ft lever arm on that shaft or pipe it would pull 75000 lbs. (see why the shaft diameter doesn't matter for *torque*?). What the shaft/pipe diameter *does* affect is the force - not torque - you can pull at the circumference of the shaft, such as using a cable wrapped around it.
So, 3" is 1/4th of a foot, thus the pulling force is 4x the 75000 = 300,000 lbs force. By the way, that's a helluva lotta force -- I think you'll see practical limits come into play here, like crushing your bearings or snapping the cable.
Hope this helps, (and hope I got it right)!
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