DEVELOPING MORE TORQUE

[9PLOTTS]

HI.

IM AN ELECTRICAL GUY WITH A MECHANICAL QUESTION.

IM BUILDING A EXCERSISER FOR DOGS
ITS BEING DESIGNED AFTER THE OLD HORSE WALKERS.
THE PROBLEM IS NOT ENOUGH TORQUE AT THE END OF A 12 FOOT ARM THAT ROTATES IN A CIRCLE.
IM USING A .75 HP MOTOR @1725 RPM {AC}
60/1 GEAR REDUCER
AND NEED AN RPM OF BETWEEN 8 TO 14 MAX
SO FAR I HAVE USED A 9 INCH AND A 2INCH PULLEY TO GET TO AN ACCEPTABLE RPM. HOWEVER I DONT HAVE ENOUGH TORQUE TO KEEP THE DOGS MOVING. EACH DOG IS ABOUT 60 LBS. NOW I DONT NEED ENOUGH TORQUE TO PULL THEM ALL AROUND THE CIRCLE BUT I FEEL THAT 30 TO 50 FT. LBS. WOULD DO THE TRICK. NOW I THINK THAT I CAN DEVELOPE MORE TORQUE BY USING A LARGER PULLEY ON THE MOTOR BUT I ALSO NEED TO KEEP THE FINAL DRIVE PULLEY AT AROUND 9 OR 10 INCHES. RIGHT NOW I HAVE IT SET UP IN THIS SEQUENSE .75 HP MOTOR W/ 2.00" PULLEY TO A 9.00" PULLEY THAT DRIVES THE INPUT SHAFT ON A 60/1 GEAR REDUCER THAT HAS THE OUTPUT SHAFT COUPLED TO THE FINAL ROTATING SHAFT.
SHOULD I USE A JACK SHAFT WITH MORE PULLEYS ?
SHOULD I USE 2 PULLEYS FOR FINAL DRIVE INSTED OF A RIGID COUPLING?
IM ASKING FOR ANY ADVICE TO HELP ME SOLVE THIS PROBLEM.

THANKS
9PLOTTS

 

[sacem1]

9PLOTTS:

Right now you should be getting around 6.38 rpm in your device if the pulley diameters are in the primitive diameter, not outside diameter, if they are outside diameter then you should be getting around 5 rpm. Probably your speed is even less because you should be having slippage in the small pulley (motor side).

If the dog is not moved around but your motor overload protector (which you should have installed, I suppose) does not trip (stop the motor because of an overload) then definitly you have a transmission power problem from the motor to the speed reducer, if it does trip then you have an undersize motor (and probably also an undersize speed reducer) have in mind that by reducing the input speed of the reducer with the pulleys you are multipliying the input torque by the same rate as the speed reduction in the pulleys and you should get a reducer that is rated for at least 4 HP @ 1725 rpm input.

You stated that you could work with a final speed of up to 14 rpm and if your motor power is enough, no overload tripping, you could try increasing your motor pulley in diameter up to around 4 inches and you should be getting around 12/13 rpm final output speed, if at that speed your motor does not trip or exceed rated amps then you are set, if it does change to a bigger motor lets say 1.5 HP with a speed reducer rated for 3-4 HP @ 1725 rpm input speed, that coupled with the 4" motor pulley and 9" input pulley in the speed reducer should do the trick.

Its difficult to pre-establish your final torque requierements because we can not assume values for the resistance offered by the different dogs that might go even to one that wants to go the opposite way and clings with its claws to a surface that we do not know its characteristics either, so field experiment should do the trick.

Good luck.

SACEM1

 

[dvd]

30 to 50 lb-ft of torque at the end of a 12 ft. arm only produces a pull of 2.5 to 4.17 lb. That doesn't seem like enough to keep a 60 lb dog interested in moving. How many dogs are on this at a time?

You haven't described how you have supported the arm. It may be that you have too much friction involved in the support of what I assume is a cantilevered arm. Have you used a force gauge to determine how much of the motor torque is required to rotate the arm? A better description of the apparatus would be helpful.

 

[ietech]

Very interesting from an engineering point of view, please keep us posted on how this works out. I'm getting pretty old and someday I might need something to help me walk my dog.

For now I will enjoy her companionship and take her for a 30 minute run, although I'm not sure how many more years I can keep running.

ietech

 

[sacem1]

Sorry dvd but I disagree with you, a simple stage worm gear speed reducer with a ratio of 60:1 at an input of 777 rpm with a 1750 rpm motor with a 4" pulley driving a 9" pulley at the reducer input shaft should produce an average of 2000/2300 in-lb torque depending on reducer design (3.25"/3.50" between shafts center distance) which divided by 12 (inches to a ft) gives you around 180 ft-lb torque on the shaft and a pulling force at the end of the 12' arm about 15 lb net pull, thats why, before, I suggested a jump to a 1.5 HP motor which will give around double that pull to 30 lb which should be a good starting point.

That's of course asuming that you only walk one dog at a time and as you stated, that the shaft does not have a high friction, maybe a counterweight at the other side of the arm could do the trick easily.

Cheers,

SACEM1

 

[dvd]

The 30 to 50 lb-ft of torque was what the originator of this thread stated. I was just remarking on the author's proposed torque requirement.

 

[sreid]

With the motor and gear reduction stated (with no losses) I get about 50 pounds force at the end of the arm. You could put a big fish scale on the end of the arm and see what force you are actually getting.

 

[desertfox]

Hi 9PLOTTS

I agree with sreid, I calculate your force at the end of the 12Ft radius arm to be 51.075lb (force) assuming no losses through the gear reducer and pulleys and this would equate to about 612.9 lbft (torque).

regards desertfox

 

[9PLOTTS]

thanks for all the info guys!

sacem1 after i made the post i was working on the project and smoked the gear reducer that i was using [you were right it was undersized but i got it really cheap 10.00 so i thought id give it a try. you have to start somewhere.]
Do you guys think that a 4/5 hp rated reducer is large enough. I am going to try to stay at 3/4 to 1 hp max for an ac motor primarly due to supplying power to the unit with some sort of extension cord. I may have to try to run the moter @ 220 volts to get a better ratio [more torque?]between voltage and speed. at one time i saw a chart that gave the breakdown , but i cant seem to locate it now that i need it. also i was wondering about the location of the gear reducer [just so i understand]. Where is the best location for the reducer [for maximum torque} coupled to the motor or should i reduce speed with pulleys first?

as for the arms they are supported on a 20"round 1/4" plate that is on a 1" shaft,the shaft is supported by 2 pillow block bearings the arms are then guyed to a point 36" above the base amd guyed to one another there is very very little friction.

I just wondering if there would be any other better way
to increase the torque. some type of other transmission device ?

 

[sreid]

It appears to me that the 3/4 HP motor is big enought, you might be able to go smaller. Her's the problems I see;

New gear boxes are not cheap.

Most high reduction gear boxes are limited by the output shaft torque (in your case 600 lb-ft, can you say "Hemi").

My suggestion-build your own "gear box" using belts and pulleys. Four stages of a 4 to one reduction would be a 256 to one reduction. I would use V belts for the first two stages and toothed belts for the last two. These are availabe from Stock Drive Products. You could also get an 8 pole motor and start with a lower starting speed (625 RPM). Check at the Baldor Web Site.

There are also differential planetary gear box arrangements that give large reductions in two stages (sort of like a differential hoist-chain fall). You might be able to emulate these with belts. Life's too short to not think about fun things like this.

 

[sacem1]

Sreid: you are disregarding frictional losses with the pulley arrangement, on the first stage you can work with one type "A" belt, probably on the second you will have to have 2 type "A" belts or one type "B" (that would force you to have a minimun 3" diameter on the small pulley) on the third step you will have at least 3 to 4 type "A" belts or 2 type "B" belts and on the fourth you would need at least 4 type "B" belts, if you add the cost of the 8 pulleys most of the with several belts on, the cost of the set-up to hold the pillow blocks, the cost of the pillow blocks, the tensioning devices and the cost of the belts themselves, for sure a worm gear reducer will be more cheaper, cleaner and compact, all that said I can tell you that 3/4 HP will not move the whole array of belts/pulleys/pillow blocks so as a Mecano/Erector device it would be nice to assemble, as a practical working solution, forget it.

For 9plotts: ALLWAYS put pulleys on the first stage of reduction, then you put roller chains and finally you put reduction gears. The reason is that the order is dictated by the natural efficiency of each type of transmission, belts work best at high peripheral speed, chains have more slippage resistance and effective power transmission at lower speeds, and finally gears have the best efficiency and duration as they go slower so we locate them at the end of the power transmission train.

As for voltage requierements, the power of a motor is not dependent on the voltage you connect it to (assuming its a multiple voltage motor) it will still be 3/4 HP at 220 or 110 Volts, sure your Amp load will be double at 110 than what it is at 220 but any 1 1/2 HP - 110 Volt motor can be fed with a 12 gage electrical extension cord wich can be bought at any hardware store, so better check first which voltage you have availiable at your location.

Hope this clears things a little bit.

SACEM1

 

[dvd]

You may also want to consider using a soft-start to bring this up to speed. If you just turn the motor on at full speed the inertia of the arm, depending on it's weight and geometry, could require a very high starting torque. Look at a VFD, fluid coupling or friction coupling.

 

[sreid]

SACEM1 is quite correct, a worm reduction gear box makes more sense. In computer peripheral design, gear boxes are ususally not perfered but for this application, the output torque requirement makes belts impractical. And it would undoubtly be more expensive. However, I don't think that efficiency alone would make the belt drive impractical.

DVD. Inertia is reflected by the square of the gear box reduction and at 270 to 1 squared, there would need to be a rather large flywheel at the output shaft for the motor to even know it is there.

 

[dvd]

Wow this is a tough crowd.

The author states "arms" in his description of the apparatus. I don't know if this is two or twelve. At a diameter of 24 ft the inertia of this could be significant. Starting torque could actually be quite high. I wouldn't discount high starting torque until I knew more about the total mass and geometry of the machine.

 

[sreid]

DVD

I didn't mean to be a tough crowd. And you are correct that statements without numbers are not science. And to be fair to 9PLOTTs we should see if starting inertia is a problem. So here's an analysis for starting torque.

Assume two dogs to start with. Two seconds to 6.4 RPM. A 24 ft. pole rotating through the center. The pole is 2 in. O.D., 1/8 in. wall and made of steel.

6.4 RPM = 0.11 Rev/sec = 0.67 Rad/sec = Omega

Alpha = Omega/t =0.67/2 = 0.335 rad/sec^2

The pole weighs 64 lb. = 2 slugs (mass)

The inertia, J, is

J = mL^2/12 = 2 x 24^2/12 = 96 lb-ft sec^2

Torque = J x Alpha = 96 x 0.335 = 32 lb-ft

This is about 5% of the available full load torque (with out losses). Even for eight dogs (4 arms) the starting inertia should not be a problem. And clearly there are ways to make the arms much lighter (aluminum trusses?).

 

[dvd]

Thanks, Sreid. The author's mention of a "smoked" gear reducer led me to question the starting torque, since the motor wasn't exceeding the gear reducer rating unless the pull-up torque was the cause. I also ran the numbers, but used a heavier tube (as is my nature). Additionally, the author states that a maximum desired speed is 14 RPM. My worst case scenario would have the motor at full speed in half of your estimate and with a much higher torque requirement. Thus my apprehension and recommendation for soft starting. Plus I think the dogs would appreciate it.

I think that M. 9PLOTTS should pony up some more information for further comment.

 

[9PLOTTS]

i will pony up asap
just dont have time right at this time
but i sure like all the input!

9plotts

 

[sreid]

I agree that the pooches would probably appreciate the soft start but on the other hand they would probably learn the drill pretty fast without too much cruelty.

For the "variable speed", instead of some variable speed drive or belt pulley changes, how about simply making the radius variable using sliding tubes and "Pip Pins."

 

[9PLOTTS]

HI GUYS

I FINALLY HAVE A LITTLE TIME TO GET BACK TO THE DISCUSSION.
LETS SEE : 1. THE ARMS THAT I HAVE NOW ARE 1"X 1/8" SQUARE
TUBING{STEEL}.

2. SREID IS RIGHT THE DOGS WILL CATCH ON FAIRLY QUICKLY. THE TORQUE THAT IS NEEDED WOULD BE TO KEEP A DOG GOING ONCE THEY ARE STARTED.I DONT FEEL THAT I NEED A SOFT START AT LEAST AT THIS STAGE OF THE DESIGN.

3. I THINK THAT A 3/4 HP MOTOR WILL WORK AND A 1HP AT THE MAX. THE GEAR BOX THAT I SMOKED WAS REALLY TO SMALL. IT ONLY HAD .25HP INPUT RATING SO I WAS NOT SUPPRISED WHEN IT DIDNT HOLD UP.

WHAT I AM TRING TO DO NOW IS LOCATE A REASONABLY PRICED GEAR REDUCER RATED 4 TO 5 HP [PER SACEM1'S SUGESTION].
ANY HELP ON THIS WOULD BE GREAT. "USED" IS OK
AND ANY SUGGESTIONS ON WHAT TO LOOK FOR OR TO AVIOD WOULD BE HELPFUL.
4. WHAT DOES A DOUBLE OR TRIPPLE REDUCTION GEAR BOX DO FOR TORQUE ? DOES IT MULTIPLY THE TORQUE AVAILABLE AT THE OUTPUT SHAFT?

5. I WAS THINKING ABOUT SOME SORT OF SPRING ASSIST BUT HAVENT FIGURED OUT HOW TO HOOK ONE UP YET.
IS THERE ANY IDEAS ON THIS?
IS THERE ANY TYPE OF PRODUCT ON THE MARKET THAT WOULD WORK IN THIS SITUATION?

IF YOU WERE NOT SURE WHAT I MEAN BY SPRING ASSIST I WAS THINKING ABOUT SOMETHING THAT WOULD ALLOW THE MOTOR TO KEEP TURNING UNTIL THE SPRING WOULD DEVELOPE ENOUGH FORCE TO GET THE DOGS MOVING AGAIN. JUST A THOUGHT ANYWAY

THANKS AGAIN FOR THE INPUT.

9 PLOTTS

 

[sacem1]

9 PLOTTS:

A doble or triple reduction gear box is just how the manufacturer gets the reduction he needs in plain words the torque will be multiplied by the same factor as the reduction in speed that is a 25: 1 reduction in speed will multiply your torque by a factor of 25 (this is disregarding efficiency in the transmission which should be about 98% per stage of reduction) for your practical purposses just consider the reduction as the factor of torque multiplier.

For those who want to be more exact a doble reduction will have lets say a 98%*98%=96.04% efficiency and a triple with the same final ratio will have 98%*98%*98%=94.12% efficiency so for the same ratio a double will be more efficient than a triple reduction but usually for high reductions it is cheaper to have a triple reduction instead of a double, usually it is recommended not to go over 5:1 in each reduction stage so doubles stay in the 25:1 range and triples in the 125:1 range.

You know its fun to cross information with all you guys (and gals)

Cheers

SACEM1