Worm Gear Calculations.

Hello,

There are a couple of things that I don't see in your calculation. first thing is effiency, the highest effiency rating is 60% on a worm gear. Second is shock load.

Hi Thomos:

You r right. Efficiency for my case ia around 60% & I think there is no shock load.

But considering the 60% efficency still I can achieve required output torque (600 lbf*in)with this gear ratio.

But I am worried about gear forces and tooth strength.So do u have any formulae handy for this or is there anything wrong in my calcualtion.

Thanks

I think you're making this way too hard. If you are lifting 200lbf on a 6" pulley, and that pulley is connected on the same shaft as a 3.3" worm gear, the force on the tooth (or teeth) of the worm gear is simply 200*6/3.3=364 lbf plus a bit for friction, lets say a total of 380lbf. You did say there was essentially no shock load, so I'm ignoring acceleration.

The output torque of the worm gear is then 380lbf * (3.3"/2) = 627 in-lbf.

A frictionless 20:1 ratio would require 627/20 = 31.4 in-lbf of input torque. With an 8.5" lever, the frictionless force required at the handle is then 31.4/8.5 = 3.7 lbf.

If you are having to apply 30 lbf to the handle, then your efficiency is 3.7/30, or 12.3%, not 60%. Lubricate the thing!

Dick

Hi:

As u have understood my problem, I want to say more about the situation here.

30 lbf applied at the handle is the maximum force a human can apply, but depending on the person it will vary , I dont know may 10, 20,25 or 30lbf(Only thing the input torque should be more than the output consdering losses).

So efficiency of 12.3% may not be right.

Your calculation say just 3.7 lbf on the handle is enough to lift 200lbf. U are right in this but we don't know how much a person can apply(min should be 3.7 lbf).

But according ur calculation load on the worm gear tooth could be around 380lbf.That is u are doing reverse calculation, but as u know force on the handle changes(max 30lbf) hence input torque on the worm changes, which in-turn rotates worm gear & this torque on worm gear is much more than 627 in-lbf(ur calculation)& hence the input force on worm gear.

So I think this is situation where may be just for example to lift 200lbf we are appling 1000lbf(just a random number) on the worm gear.

so dont u think load on the worm gear tooth will be what i calculated if u dont go from reverse side?

By the way what is the relation behind this(200*6/3.3=364 lbf ) equation.

thanks

First, the 3.7 lbf was the force required in the absence of friction. If the efficiency of the worm gear is 60%, then the actual force required with friction would be 3.7/0.6 = 6.2 lbf.

The 200*6/3.3=364 lbf equation is simply a solution of torque in = torque out. Torque is force * lever arm. Since the rope and the worm both push essentially tangential on their respective 'wheels', the lever arms are simply the radii of the wheels.

So F1*L1=F2*L2, or F1*R1=F2*R2 ( F=Force, L=Lever, R=Radius)

Since the radius is half the diameter, F1*(D1/2)=F2*(D2/2). (D=Diameter)

Multiplying both sides of this equation by 2 eliminates the 2's, leaving F1*D1=F2*D2

Dividing both sides of the equation by D1, we get F1=F2*D2/D1.

Substituting the correct values, F1=200 lbf * 6 in /3.3 in =364lb. This is the total force applied to the teeth of the worm and gear. In any well-made worm gear set, there will always be at least a couple of teeth in at least partial contact, so the maximum force applied to any one tooth will be significantly lower than 364 lbf.

Now, just because a person can exert 30 lbf on a (locked) handle, does not mean he/she actually will, especially when the handle is free to move. In this case, when the person exerts just over 6 lbf on the handle (assuming the force is perpendicular to the lever), the handle will begin to move (away from the hand), turning the worm and beginning to lift the 200 lbf weight. The person must then adjust the direction of his push or pull to continue approximately perpendicular to the crank (otherwise the lever arm will soon be zero (in 1/4 revolution of the handle), and no amount of force will cause rotation except breaking the handle), making it difficult to apply additional force. If the person is able to continuously apply more than the force required to turn the handle, and keep that force perpendicular to the lever, the additional energy will be used to accelerate the system, and the person must turn the handle faster and faster to keep applying that force. The inability of the person to move fast enough will automatically self-regulate the force to a value just above that required to overcome friction and lift the weight.

Your worm gear is quite adequate! Relax and build the thing!

Hi

Thanks for ur reply.

Just one more thought.Lets say the handle get jamed when rotating & then will there be high load suddenly dumped into the gear.

Thanks

If the handle gets jammed by a foreign object between the worm and gear, the jamming force will be applied to a different tooth (or teeth) and in a different direction than the lifting force, so the teeth doing the lifting will not be affected.

If the handle gets jammed by something getting in the way of the handle, very little, if any, of the jamming force will be transferred to the gear.

You're still a go on building it!

Dick

Hi,you may use a software for your calculation.You can download the software from http://www.hexagon.de/zar3_e.htm .I think it may help you.