I havea no experience with the BMR bar. I used the ATR bar mercilessly and it performed very well. I don't drag race the car at all.
This maybe too much tech but this helped me attach numbers to both bars and then you can add your spring/shock combo to it............
The stiffness "I" is a function of the diameter^4.
My automotive engineering school books tell me to multiply by 32 rather than 2 (since for a round bar: I= pi/32 x D).
Further, it calculates the twist angle (in radians) of the bar. For us, it's probable more interesting to know the height difference of the end length (in order to make it useable for calculating wheel rate or something), for which you need to multiply it by the arm length. That means that if you were to express it as lb/inch, it would look like this:
Force/distance = (pi x diam^4 x material modulus)/(32 x arm^2 x length), with distance being the difference in height of the outer ends.
Actually, this equation assumes stiff arms, but in practice, most are made of the same spring steel as the rest and consequently twist as well.
Here's another example:
A 293 lb load at 10” from the bar center gives a 2930 in-lbs torsion in the bar and produces a particular twist in a bar of given alloy and length that results in a 1” deflection at the point of application (endlink) of the 293 lb load. You don’t need to know how much this twist is to compare the effects of varying the arm length.
With 12” arms on that same bar, you only need 2930/12 = 244 lbs to achieve the same twist.
However, at a radial distance of 12” from the bar, that same twist gives you 1.20” of deflection.
To bring the deflection at the 12” distance along the arm back down to 1” so you can express that new rate in terms of lb/in you need to divide that 244 lbs by the bar end deflection of 1.20”.
Which gives you about 203.5 lb/in.
Now, you can go get coffee, cause you just had a great nap!