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Spring compression force? High school physics...

T

tractorsam

Well-known Member
Hello, I'm busily pondering useful modifications to an old tractor and was wondering if anyone can recall what the formula for how much force it takes to compress a spring is? If I'm thinking correctly, if you compress a spring say 1" and it takes 50lbs of force then compressing it 2" would require something rather more than that 50lbs? Many thanks, Sam
 
If I remember right it is something along the lines of every time you cut the distance in half (1" to 0.5") you double the amount of force (50lbs. to 100 lbs)

I will dig up my physics binder later and double check.

Tom
 
Spring force in Newtons is F=kx where k is some spring constant in newtons/meter and x is the displacement (how much it is compressed) in meters. If you change k to lbs/inch and x to inches, then the force in lbs would be calculated with the same formula.

If it takes 50lbs of force to compress 1 inch, then if the spring is an ideal spring, it will take another 50 lbs to compress another inch since the spring constant is 50 lbs/inch.

50=k(1), therefore k=50 lbs/inch
F=50(2 inches).... F=100 lbs

Hope that makes sense.
 


Sam ,

Springs are not linear in the amount of
force required. Much depends on the wire dia
and number of turns and most important is the
heat treatment ,the temp at which it is drawn
makes a difference in the spring rate.

If you are thinking about adding tension to
a set of valve springs in an old tractor,don't
do it as it will increase the lobe and lifter
wear rate.

george
 
Thanks to all who've replied. It appears that I was thinking something along the right lines. For my purposes, the exact figures are relatively unimportant, more the approximate theory.

The reason for the question is that myself and a friend were discussing adding a 3pt hitch to older Farmall tractors (H/SM). Due to me being me I'm intending to have a bash at doing it myself, not because it'll be easier or cheaper but because I want to. Anyhow I was rolling around (just for an added challenge) the idea of draft control and would it be possible and how would you do it. I got to thinking that if one had a (slightly) compressible top link then you could use that motion to actuate a hydraulic valve to initiate draft control. But then how does one adjust the draft control (to control depth, while plowing, say) and so the question of the nature of springs came up. Needless to say I've got a cold and am not actually doing anything, and thus spending hours thinking of ways to bring a 65 year old tractor to a level of technology that was obsolete at least 35 years ago! Sam
 
I would just put what they call a "Hydraulic Accumulator" on the pressure side of the cylinder holding it in the ground. This is a gas filled canister (with a diaphram to keep the nitrogen gas and hydraulic oil sperate). It takes up shock produced in a hydraulic system. They work great and come in various sizes. They put them on lift cylinders of big fork lifts. It absorbs shock when the machine goes over bumpy areas. It turns you hydraulic cylinder into a gas shock kind of is the easiest way I can put it. They also put them on giant sprayers with 90ft booms. This saves the frame work of the booms and main frame from cracking and bending. There is a lot of shock in a boom that's hanging out 45ft on each side of the frame when your traveling 10-15 miles an hour in a no till field that's washed out in spots.
 
Hate to throw in my 2 cents without 100% proof.

But, I believe spring force is this equation...

F = k(Lorg-Lcomp)

K is the spring constant is derived from a big formula that not many people care about (wire diameter, material type, # of coils etc)

Lorginal-Lcompressed is just the displacement on the spring measured in inches, mm, etc.

You can look at www springmasters com or www efunda com for tons and tons of info on spring design and formulas.

Rick
Untitled URL Link
 
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