Drag Spring Rates

SUPRSLO · Post by **SUPRSLO** » Thu Mar 23, 2006 8:14 am

Guys, it seems there is alot of confusion regarding spring rates, well atleast on my part anyway. I constantly hear talks of potential energy in the spring. OK, now lets consider the higher rate vs. lower rate spring on the front of the same car with the same front weight, cg, yadda yadda yadda.

You have the first spring which is a 275 pound spring. According to most this spring does not have as much potential energy as the 300 pound spring, but is it that easy? The 275 pound spring will compress more in an at rest position then a 300 pound spring which means the front of the car will be pushed up for a longer time when launching with the 275 pound spring then a 300 pound spring which is not compressed as much right? The question then becomes though does the 300 pound spring have enough potential energy to actually aid in pushing the front end up more or less then the 275 pound spring. To answer this question wouldn't you have to consider the average force the spring exerts during launch? Maybe in terms of force/distance or something?

Also, this will have an effect on the landing of the front after the wheely as well in terms of unloading the rear tires. It seems to me that the 300 pound spring would unload the tires less as it does not want to compress as easily and let the front of the car fall down and unload the rear tires.

Any suggestions, answers, or comments are appreciated

BillyShope · Post by **BillyShope** » Mon Apr 03, 2006 10:42 pm

The potential energy, in a compressed coil spring, equals kx^2/2, where "k" is the spring constant, "x" is the deflection, and "^2" indicates that "x" is to be squared.

For a given load, the product "kx" will be the same for the two springs you mention. But, since the deflection of the lower rate spring will be greater, the potential energy of the 275 pound per inch spring will be greater than that of the 300 pound per inch spring.

SUPRSLO · Post by **SUPRSLO** » Tue Apr 04, 2006 7:45 am

BillyShope wrote:The potential energy, in a compressed coil spring, equals kx^2/2, where "k" is the spring constant, "x" is the deflection, and "^2" indicates that "x" is to be squared.

For a given load, the product "kx" will be the same for the two springs you mention. But, since the deflection of the lower rate spring will be greater, the potential energy of the 275 pound per inch spring will be greater than that of the 300 pound per inch spring.

Excellent! Thanks for the detailed reply!

One quick question though. You say that the product of kx will be the same for both springs but then go on to say the deflection of the lower rate spring will be greater (which in turn would make the product kx greater). I'm confused here.

BillyShope · Post by **BillyShope** » Tue Apr 04, 2006 11:45 am

Remember: You're going to multiply by "x" TWICE. The product of "k" and "x" equals the load. So, since the deflection of the lower rate spring is greater, the potential energy of the lower rate spring is going to be greater.

Just realized that you might be confused by my definition of "k." This is not a material constant. It is the spring rate. So, you'd substitute 275 and 300 for the two different cases.

SUPRSLO · Post by **SUPRSLO** » Tue Apr 04, 2006 12:21 pm

BillyShope wrote:Remember: You're going to multiply by "x" TWICE. The product of "k" and "x" equals the load. So, since the deflection of the lower rate spring is greater, the potential energy of the lower rate spring is going to be greater.

Just realized that you might be confused by my definition of "k." This is not a material constant. It is the spring rate. So, you'd substitute 275 and 300 for the two different cases.

Ahh, gottcha. Thanks again!