Torque created around offset wheel

[fleebell]

This is probably obvious to the folks good with math but it's been 40 years since school for me and I was never a math wizard.
I basically need a formula to figure out the picture below.
3000 lbs pushing down on an incline of 18.1 degrees - I need to know how much force would be moving the triangle to the right assuming no friction between the bottom and whatever it would be setting on
or
3000 lbs of force toward (and pinned) to the center rolling around an 8" fixed wheel offset by 2 inches in the center. - how much sideways force would be generated? Either one has to roll 12.56 inches and move 4" toward the center if that makes any difference.

The pic on the right is actually what I'm working with but I figured the other would give me an average

Is there any simple formula I can plug the numbers into to get a torque or ft lb push value?

What I'm trying to figure out is the output torque on a simple hot/cold water motor I want to build. It will run off the psi differential between a cold water tank and a hot water tank sides using either propane or freon to pressurize some auto air springs mounted in a wheel frame. The tank is split with simple plastic hinges that allow the bottles in the water to switch from side to side as the wheel rotates around the center axle. The air springs are made from nitrile rubber which is pretty much immune to either type of gas I could use. The hot water will come from a solar panel and the cold from our well if possible. I want to be able to water my wife's garden using a trickle system by driving a small piston pump with it and possibly spin a small wind turbine type home made alternator if it will create enough torque once geared up to about 500 rpm. The output will be taken from a #50 or #60 chain gear on the top cover

The picture below is a top down view. The axle will be fixed to a very heavy metal frame supporting the circular water tank and motor wheel itself at both top and bottom of the axle. The top and bottom covers not shown in the pic will have the bearings for the wheel. It rotates around the fixed axle and offset wheel in the center. 3000lbs force will be the maximum allowed to each air spring. I'm need to find out the torque for one and I can figure out the rest. Yea, this thing is going to be a bit heavy as that outside frame is probably going to be 7/16" thick 8" wide steel strips bolted together with the covers at least 1/4" thick...There will be a lot of force on that axle/offset wheel once it gets going and it's got to be a strong frame...It will probably only turn about 2-3 rpm max due to the heat transfer time from the water to the little paintball CO2 bottles with the gas in them that will be setting in the water tank.

 

[Mech_Engineer]

In this picture (which you claim is what you're aiming for) the net torque on the larger wheel would be zero:

This is because the vertical height difference between the line of action for the force and the centerline of the axle is zero.

I see what you're aiming for with the design, it's a sort of a radial engine with air springs. I'm thinking it might be overly complicated though, if you have pressurized steam there is likely an easier way to extract work energy from it?

 

[fleebell]

I think you misunderstood my drawing there. That larger wheel is fixed on an axle that doesn't turn. The small wheel that will be attached to the air spring which is attached to the outer 8 sided wheel frame is what will be rotating about the fixed wheel in the center as shown in the big picture of the whole thing. I'm not using steam. The pressure will be supplied by either propane or freon inside the small bottles setting in the water feeding into the air springs and will go up and down with the water temp in the doughnut shaped tank as the outer wheel turns. This thing is horizontal too, not vertical.

 

[Mech_Engineer]

So if I understand correctly, your goal is to utilize vaporization and condensation of compressed propane in the center "springs" to rotate the whole assembly? Have you considered doing something a bit simpler like a Stirling engine?

 

[fleebell]

Oh yea. There is an old design out there that worked I could copy if I wanted to but I want to run this on solar hot water. That would take a big stirling engine to get any real power out of it. I've got a better design for a simple Minto wheel too but that would cost me about $1100 to build like I want and that's a bit beyond what my current budget will handle unless I wanted to take at least six months to build it.

I built a little version of this a couple of years ago and it worked OK but I was using some small surplus air cylinders instead of the air springs and It leaked so badly from the shaft seals on the cylinders I never got more than 3 rotations before too much gas had leaked out for it to operate. Since setting in a cloud of propane didn't really appeal to me I stopped working on it and stripped it back down for parts. I'm just building a bigger heavy duty version now.

I know how to figure a out a normal crank. This design has basically got 4 inches of 'piston' travel back and forth so even with the fixed center wheel being 8" in diameter it's still effectively a 4" crank. I'm just not sure whether 'effective' and actual would use the same math in this case That's why I asked here.

 

[Dr.D]

Looks like a stationary eccentric circle cam with pivoted roller followers. This is kinematically equivalent equivalent to having the cam rotate and the outer frame remain stationary.

There are well developed analyses for systems such as this, but they are not simple.

In your top view of post #1, it appears that the two horizontal cylinders are (momentarily) contributing nothing at all to the torque (they are at TDC and BDC), and whatever torque there is arise from the pressure differences in the two vertical cylinders.

I have a lot of doubts about the workability of this design.

 

[fleebell]

Dr.D
All I am asking for is the basic math to figure out if I push down with 3000 lbs on a wheel rolling down a side of an 18 degree angle how much force would be trying to move the angle sideways if there was no friction and the wheel can only go up and down. It's been 40 years since school and I simply don't remember the formula for it and can't seem to find it on here when searching.

If anyone can answer that simple question I would really appreciate it.

 

[Dr.D]

Neglecting the weight of the wheel, and assuming that there is no acceleration (quasi-static) motion, the system is in vertical equilibrium. For vertical equilibrium,

Sum Fvert = N*cos(phi) - W = 0
N = W/cos(phi)

where
W = 3000 lb
N = normal force
phi = 18 deg

The sidewise component of the normal force is then

Fside = N*sin(phi) = W*sin(phi)/cos(phi) = W*tan(phi)

where Fside is the required sidewise force.

 

[fleebell]

Thank you!

The wheel is not vertical but horizontal but that should still get me pretty much the answer I need.