It is pretty simple, see if all my school is about to pay off:
The equation for a spring (specifically a linear coil spring) is: F = -kx
This is commonly referred to Hooke's Law.
F is the force (in Newtons, or Pounds: N or Lbs)
K is the spring constant, in Newtons per meter, or Lbs per foot: N/m or Lb/in.
Typically we offroaders refer to a springs K property as the PPI rating of a specific springs. For example, my Skyjacker springs are 415PPI (pounds-per-inch), or more accurately, 415 Lbs/in.
Basically what this means is that in order to compress the spring 1", it takes 415lbs of force to do so (neglecting the 1:1.5 lift ratio of the TTB beams b/c of where the coil sits on the beam in relation to the pivot point and the distance to where the weight is applied at the wheels).
The beauty of the above equation is this. If you know any two of the 3 variables, you can find the missing one. And most of the time you will know the K value, so that means you only need one other property, either the force or the X.
X is the distance the spring has been compressed or stretched, in Meters or Inches. Expressed as Delta X, or the change in spring length between free length and compressed length.
The negative value out front of the K is there to indicate that the spring always WANTS to return to its free length. Compression is positive X values, and extension is negative (for trucks you won't need to know the second statement since we are only interested in finding the compressed length).
So, as an example, using my springs:
I know several things.
The free length of my spring: 21"
The compressed length as it sits in the vehicle at static ride height: 17"
The PPI or K value: 415Lbs/in.
This means I can easily solve and find out how much my truck weighs at each wheel (times two to get total nose weight at front tires).
Original equation:
F = -KX --------> Assume compression is the positive direction, and drop the neg. value out front eliminate places to mess up.
Sub in my known numbers:
F = (415lbs/in)(21in-17in)
F = (415lbs/in)(4in)
F = (1660lbs) <------Notice how the "in" units canceled out (in/in = 1)
That's the force at one coil (if I was SAS)...but since this is TTB we are interested in, it gets even better:
The TTB suspension has (roughly) a 1:1.5 lift ratio, meaning that it takes a stiffer coil on TTB than on a SAS rig to accomplish the same thing. Meaning, if I was SAS and wanted a coil that acted, and road exactly like my current coils do, I would need a linear coil with a rating of around ~280lbs/in...this would guarantee that the ride I perceive sitting in the driver's seat would be exactly the same with both types of suspension.
415lbs/in / 1.5 = 276.67lbs/in just like I said above.
To get the true weight at the wheels (for TTB) you would need to take the 1660lbs calculated from above and divide it by 1.5:
1660lbs / 1.5 = 1106.67 lbs at each wheel. Meaning between both tires, my front end on my truck weighs ~2213 lbs (1106.67 x 2).
Keep this in mind.
Now if I wanted to calculate how much my truck would compress a coil of 190 lbs/in (K value of EB deaver superflex coils) I would do the following for each coil:
True weight at one coil (leverage effect of beams): 1106.67 lbs
1106.67lbs = 190lbs/inX
Now solve for X. Divide both sides by 190lbs/in to isolate the wanted variable to its lonesome.
1106.67lbs / 190lbs/in = X
5.82458in = X <------------Notice the "lbs" units canceled out (lbs/lbs = 1) leaving you with inches, the correct unit for the change in spring length.
So If I were to run the above mention coils, each one is going to sag ~5.8in with the weight of my pickup on them.
For SAS rigs, you neglect the 1:1.5 ratio step. But for TTB rigs you will need to multiply or divide to get certain values.
As you can see, you can solve for whatever you need, as long as you have 2 pieces of the 3 for the puzzle...
How'd I do? Did anybody follow any of that :icon_confused:
