Accountant needs help with spring

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In summary, The conversation is about a 3dof platform project for a flight simulator. The prototype uses 16ga 1" square steel tubing and 3 small 12vdc 50rpm gear motors. The motors can induce pitch and roll movements but need a spring to lift the upper platform. The specified compression spring needs to be able to lift the 70lb platform 5" above and below the 8" deflection point. The spring is assumed to be approximately linear with a spring constant of 1.532 N/mm. This means that an additional force of 194.6 N (43.6 lbf) is required to lift the platform 127 mm and the same force is required to lower it
  • #1
binto100
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Greetings all.

I'm working on a 3dof platform project as the basis for a flight simulator and am pretty much out of my element.

Here's is a picture of the typical platform: http://www.flickr.com/photos/46403450@N02/4420935812/

I have constructed the following lightweight prototype using 16ga 1” square steel tubing. Here is a quick vid:

There are two center column tubes in my prototype: The tube attached to the weld yoke from the underside of the upper platform fits inside the slightly larger tube attached to the lower platform, this allows for up and down movement.

I have three small 12vdc 50rpm gear motors that that are powerful enough to induce pitch and roll movements in the upper platform (which sits on a u-joint) but they are not strong enough to lift the upper platform; therefore I need to introduce a spring to the system.

I have initially specified the following:

Compression spring:
Inside dia: 3.00” min
Free length: 24”
Load 70 lbs +/- 10% at 16” (8” deflection)
Solid height max: 7”
Closed ground flat end’s
Material: music wire.

I want the upper platform to be able to heave 5" above and 5" below the 16" deflection point of the spring at 70lbs (10 in of travel). I’ve constructed mounts for the gear motors, reduced the rpm to 25 using pulleys and a 2L belt, and attached a parallel lever to the shaft of the large pulley with holes for all thread at 1”, 2”, 3”, 4”, and 5”. When a perpendicular shaft of all thread is attached to the 5” hole then the up and down heave will be 10” per complete rotation.

My question is how much force, based on the above sprint specification, is required to lift the 70lb platform up 5” from the 8” deflection point and how much to lower the 70lb platform down 5” from the 8” deflection point? My very limited understanding of springs is that they act linear on the middle 60% of travel (?) so is it fair to say that the force required per inch of travel will be the same?

I would appreciate any feedback and please remember the simpler the better.

Thanks.

Bruce
 
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  • #2
binto100: The spring can be assumed approximately linear, having spring constant k = 1.532 N/mm. Therefore, the additional force required to lift your platform upward 127 mm will be 194.6 N. The additional force required to lower your platform downward 127 mm will be 194.6 N.
 
  • #3
Thank you very much for your reply.

Using convert.exe can I say that 194.6n = 43.6 lbf and if I have three motors all working equally then they must be able to supply 14.54 lbf ea to lift the platform? In other words there must be 14.54 lbs of force produced at some point on the perpendicular arm that is attached to the motor shaft?

Bruce
 
  • #4
That is correct (or actually, 43.75 and 14.58).
 

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