Wheel Balancing Double Check

[Kriegen]

I need a double check of my math to make sure I am getting the right number.

As you will see in the picture point A is a fixed point wheel axle which line AC is free to rotate about point A. Point B and C are also wheel axle points. When line AC is horizontal Wheel C collides with the ground. I want wheel B to collide with the ground at the same instant, but want wheel B to press down with enough force to almost lift wheel C off the ground. So to do this we have attached wheel B to an arm with a torsion spring pre-loaded at 69deg at point D. If point C is pressing down with 33 lbf then wheel B should press down with
(33lbf*36.625in)/22.5in =53.72 lbf
to balance out. So the reulting force that the arm of the spring should be pressing down with is
53.72lbf/cos (18.5deg) = 56.66 lbf
This gives that the spring rate should be
[56.66 lbf * (10.25in/cos (18.5deg))] / 69deg = 8.88 lbf*in/deg

This should just be a math and statics double check

 

[israelkk]

Is this a student assignment?

 

[desertfox]

hi Foofire

You need to tell us what exatcly the application is.
Israelkk I don't think this is a homework question.

regards

desertfox

 

[Kriegen]

No this is not a homework question.
If you can see the picture, then you are looking at the side view of a wheeled apparatus. The far left point being the rear axle, the far right point being the front axle. The middle point (B) is the middle axle. All three axles need to glide along the ground together, but for people to be able to handle this better the front needs to be balanced with the middle. Line BD is the pivot arm for the B axle, and point D is the axis for the torsion spring. I need to double check the forces and the needed spring rate for everything to balance out.

 

[zekeman]

OK except remove all terms in cosine; the torque at the end is 53.72*10.250
The middle wheel does not have a horizontal component,except for negligible friction.

 

[zekeman]

OOps, another minor point.
Since your total angle of motion appears to be the complement of 18.5 deg, the total motion of the arm from the zero torque position is 71.5 deg , which should be the divisor in the spring rate equation you developed; this assumes that the zero torque position of the arm is vertical, and at 69 degree preload, it rotates an additional 2.5 degrees after making initial contact with the ground?

 

[Kriegen]

Well the resultant downward force of the arm is 53.72lbf, but since the arm is at an angle of 18.5deg, then the arm has to produce 56.66lbf at that length. Then the spring rate is calculated with that force.
As for the angle, when the all three wheels are in contact with the ground then the angle of the leg of the torsion spring is 69deg from zero position.
So all things considered here, would the sprung rate be 8.88 (lbf*in)/deg ??

 

[zekeman]

No. Look at my post again.
Also, if it is preloaded at 69 degrees as you stated earlier, and you have 3 point contact at 69 degrees, you have zero force on the mid axle. Think about it.

 

[desertfox]

Hi Foofire

Firstly I agree with the force at point B but not with your spring force.
If the arm of the spring is not acting directly at point B then the spring needs to exert a lot more force than 56.66lbf. You don't appear to give a length of the spring arm.
Also if point B is pressing down on a wheel and is horizontal with point A then imagine the wheel was solid it will not lift point C, so as I see it the only slight lift you will get is dependant on the spring rate of the wheel at point B but that would mean points A and B could not be horizontal.

regards

desertfox

 

[Kriegen]

Well after reviewing some comments, I realize that my original information is incorrect. The angle should be 71.5 for pre-load from zero. I assumed the info to be correct and didn't check it.
The arm length is given, not directly, but if there is a length of one side and an angle then you know the rest of the triangle. So the arm length should be [10.25in /cos (18.5deg)] = 10.81 in .
Getting this far to me is easy, but I have never done torsion springs and believe the spring rate (k) to be equal to (Force desired * Length of arm)/ degrees of load. Now considering that the load angle is actually 71.5 that would bring the spring rate to (56.66lbf * 10.81in)/ 71.5 deg = 8.5664 lbf * in / deg ??

 

[desertfox]

Hi Foofire

I meant the length of the torsion spring arm and not the length of the arm between points B and D.
Also if and when the tyre at point B moves downward what ensures that point A doesn't lift slightly as you have shown no external load on a but have 33lbf acting down at point C.
I have attached a rough sketch of what I think is happening
if I have missed something then please explain.