I have been trying to design an inexpensive and simple lift to lift vehicles about 3 1/2 ft high to quite easily get under them, to work on them, or to rotate tires etc. I was thinking of a electric winch and cable system to create the lift, or a threaded rod or shaft system to create the lift. I was wondering if anyone on here knows how many ft lbs of torque it would take to lift a ton on a threaded shaft or rod, if it was oiled and greased well with quite deep and course threads, or know where I can get the information. Our van jack used a threaded rod to create the lift. This would be much the same as knowing how much squeezing pressure a bolt and nut apply to two pieces of metal when tightened to a particular torque or tightness. The type of threads and lubrication would make a big difference. Thanks.

>>I have been trying to design an inexpensive and simple lift to lift vehicles about 3 1/2 ft high to quite easily get under them, to work on them, or to rotate tires etc.<<

Ask your local H minister he probably knows all about simple fits that lift things up so he can get in there and check the tires.

> I was thinking of a electric winch and cable system to create the lift<

They have plenty of winches always adjusting the tension.

>>or a threaded rod or shaft system<<

I've heard they know a lot about shafts.

>> to create the lift. I was wondering if anyone on here knows how many ft lbs of torque it would take to lift a ton on a threaded shaft or rod,<<

It takes about as much as the Holdie preacher has got with his jack boot.

>>if it was oiled and greased well with quite deep and course threads, or know where I can get the information.<<

It's all in the BD&P.

>>Our van jack used a threaded rod to create the lift. This would be much the same as knowing how much squeezing pressure a bolt and nut apply to two pieces of metal when tightened to a particular torque or tightness.<<

Hey that's back room talk..

The type of threads and lubrication would make a big difference. Thanks.

It depends on the minister.



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Amos, I haven't done these for twenty years, but I will try to get you started here. The work performed in cranking the screw will equal the work done in lifting the vehicle (assuming perfect efficiency with zero friction on the threads).

Work1 = Work2 where 1 denotes the work applied to the screw handle, and 2 denotes the work in lifting the vehicle.

Work is defined as a force applied over a distance. So...substituting (force) x (distance) for work gives us:

(force1) x (distance1) = (force2) x (distance2)

force1 is the force applied at the point of contact of the screw handle.
distance1 is the radial distance this point of contact traverses.
force2 is the weight of the vehicle.
distance2 is the distance it raises for every distance1 that the screw is cranked.

If you need more help, let me know, or Stan can help too, or if I have missed something, please point that out to me also.

For maximum efficiency approaching zero friction, please make sure your threads are wetter than the BD&P government-regulation kiss

>>For maximum efficiency approaching zero friction, please make sure your threads are wetter than the BD&P government-regulation kiss<<

,,,lol


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I have had contact with 2 different people who have died with support system that failed. If you have a cement or smooth surface a hydraulic lift on wheels could be purchased for a couple of hundred dollars and if the vehicle is supported very well it can be quite save, I doubt if one could manufacture a screw system that would be nearly that efficient. If you know someone who is studying Power or Steam Engineering the formulas for what you want should be in the lessons.
Howie7

Amos, why invent the wheel when they sell safe quality engineered osha approved ones them so cheap?


click here

May God richly Bless your day!

Amos,

To find torque in foot-pounds to turn a screw, divide the weight you are lifting by 75.4, and then divide again by the threads per inch on your screw.

So if you are lifting 2000 lbs, and you have 8 threads per inch on your screw, the torque would be 2000/75.4/8 = 3.31 foot-pounds.

This assumes no friction. I'm just guessing you might have a coefficient of friction of maybe .3 to .5, which means you would add another 30 to 50 percent to that figure. Maybe around 4.3 to 5 foot-pounds.

To calculate horse power, multiply your torque times your desired RPM's, and then divide by 5252. So if you wanted to lift one ton 42 inches in 1 minute, you would need 336 RPM. 5 * 336 / 5252 = 0.32 HP.

Thanks for your replies

Seems like when I start wondering how something could be made, I can't quit thinking of different and better ways of doing it. Trying to get the most out of the least. It's a kind of exploring that intrigues me.

I was thinking of something that would be portable and could be moved out of the way, would not need high cealings, yet could easily get a vehical up to a nice height. Yet I know saftey is a very big issue in something like this.

It is amazing how cheap some things are that have been made for many years and they are all set up to do it. Sometimes something better and simpler can be more expensive to make simply because of not being experienced and set up efficently to make it.

<<< Sometimes something better and simpler can be more expensive to make simply because of not being experienced and set up efficently to make it. >>>

Molding your own tires or Pentium IV would be good examples of that.

Thanks so much Stan. I had to figure out why your formulas worked. And they do make sense. You divided by 75.4 since that is the amount of inches around a 2 foot wheel. You also divided by 5252 instead of 33000 because a ft lb of torque in one rotation moves 6.2832 ft the distance around a 2 foot diameter wheel. 6.2832 x 5252 =33000 the amount of ft lbs in one hp.

I personally was thinking their would be so much efficiency lost in friction in the threads that it would hardly be practice to calculate it out. Yet even if one half the torque is lost in friction, I think you could still lift 2000 lbs with 6.6 lbs of force on a two foot diameter wheel or 40 lbs of force on a 4 inch wheel which would not be bad. The top end of the rod supporting the weight could rest on a vertical pressure bearing, minimizing that friction. Yet I still slightly fear more force than expected would be lost in friction within the nut and thread even if well lubricated. Further this might involve less then 8 threads per inch and thus also demand more torque. Interesting. I used to have sort of a metal fabrication shop and making something like this would have been quite easy.

Thanks again

I know there is a tendency sometimes for people to write Amos off as way out, but I think if some of you could see some of the things he has created you would stand amazed. WE have a 40 yr old man with cerebral palsy here, that never fed himself until he was in his 20's and Amos put his mind to work and created a mechanism that allows him to have that independance that was formerly denied him.
I know. I know sometimes it would be nice to get him a little more down to earth though.
He does create some amazing things though.

I remember one of Amos' early inventions, a solar device that tracked with the sun and helped heat the water for the heating system in his father's house. I may still have some pictures of that invention. If I remember correctly he created that device when he was about 15.

Amos, you calculated that correct. I realized after I posted my earlier figures that I didn’t calculate the friction correct. I wasn’t going to correct it because I didn’t think anyone would be looking at it, but since you have proved plenty of competence here in picking through my work, I had better correct it.

The friction will be a percentage of the force on the threads, not the force required to lift the weight. The friction component will be nearly the same, regardless of the threads per inch. I just checked my handbook and the friction coefficient for lubricated steel on steel is 0.16, meaning the friction force will be 2000 * 0.16 = 320 lbs. For a 1” diameter screw, the distance to push that force one revolution is 1" * pi = 3.14”. So you have done work of 320 * 3.14/12 = 84 foot lbs. This compares to the 2000 * 1/8” = 250 foot lbs you did for the lifting.

So the frictional component turns out to be about another 34% of the torque to lift the weight. So add 34% to 3.31 for a total of about 4.43 foot lbs of torque. Pretty close to my original estimate, but for different reasons.

It sounds like you are naturally predisposed to engineering. I never was much of a tinkerer or inventor before I went to college. When I was a kid I dreamed about perpetual motion machines and tried to build a model once, but never tackled anything practical.

In college, I didn’t really have anything special that grabbed me at first, but just kind of ended up in mechanical engineering based on a lot of circumstances and a proven competency in math. I think it was after looking at the physical world through a math and physics perspective that I really started to enjoy applying these new tools to design and invention.

I’ll bet you would enjoy taking some courses in calculus and physics, and probably would be very good at it!

Stan and Amos, I am amazed at how formulas work. calledoutPTL