Quentin Posted June 7, 2013 Report Share Posted June 7, 2013 I am a bit confused about quoted spring rates. The standard rear springs on a TR6 in the Moss catalogue (ref: 216275PR) are 350 ft/lbs....per inch I assume. However, in an effort to measure my existing spring rate (which I think are too soft), my calculated spring rate for the above springs (new and out of the box) is 177 ft/lbs/inch. I am very confident about the calculation. I used the following calculator: http://www.efunda.com/designstandards/springs/calc_comp_designer.cfm#calc This implies that the quoted rates are for a pair. Is that correct? It does nearly make sense I think as the free length of the above referenced spring is 11.6" and the compressed length is 9" (Moss catalogue again). This 2.6" compression requires a weight of 460lbs for each rear corner. That said, I believe it should closer to 539lbs assuming a 49% rear weight distribution on a total weight of 2200lbs? I only really wanted to compare new vs existing to help inform me on whether to uprate slightly (too many bottomings out on my sports exhaust). Whilst I can do that anyway without knowing the answer to my question I am just interested to know really and suspect this is an easy question for many. Thankyou Quentin Quote Link to post Share on other sites
AlanT Posted June 7, 2013 Report Share Posted June 7, 2013 First thing is its just lbs/in. Its the weight load on each spring divided by the vertical movement of the centre of the spring when the weight goes on to it. Say the all up weight of the car was 2000lbs. Say it splits so 60% is on the front wheels and 40% on the back. So each rear spring carries 400lbs. Say the spring moved up by 2in compared to being jacked up. Then this is a rate of 200lbs/in. Round numbers used to make the arithmetic easy. Fill in the real values as required. Quote Link to post Share on other sites
Peter Cobbold Posted June 7, 2013 Report Share Posted June 7, 2013 Quentin, Spring rates are in lbs/inch, sometimes written lbf/inch, the 'f' indicating force. To calculate your own spring rate the wire diameter has to be measured precisely as the rate rises with the fourth power of the diameter. So if we were comparing diameters of,say, 2cm and 2.1 cm then the wire diameter factors would be respectively 16 and 19.44. So that small wire diameter difference of 5% makes a 21% difference in the calculated rate. Maybe that 177 figure is a bit higher?? - did you use a micrometer? The rate is given per spring not per pair. Peter Quote Link to post Share on other sites
Quentin Posted June 7, 2013 Author Report Share Posted June 7, 2013 Hmm ok. That suggests that much of what I have been thinking is bo@#*ks and Suggests my measurements may not have been accurate enough. It was not a micrometer but it was a caliper thingy. I measured an 11mm wire thickness and used an outer maximum diameter of 99mm. 7 active coils and free length of 290mm. Quote Link to post Share on other sites
Peter Cobbold Posted June 7, 2013 Report Share Posted June 7, 2013 Hmm ok. That suggests that much of what I have been thinking is bo@#*ks and Suggests my measurements may not have been accurate enough. It was not a micrometer but it was a caliper thingy. I measured an 11mm wire thickness and used an outer maximum diameter of 99mm. 7 active coils and free length of 290mm. The coil diameter is measured to the centre of the wire. So your coil diameter should be 290 - (5.5 x2 ) = 279. The equation uses the cube of than number. So using that new figure would raise your rate from 177 by 290^3 / 279^3. So 177 rises to 198 lbs/in. Getting closer.... Quote Link to post Share on other sites
Quentin Posted June 7, 2013 Author Report Share Posted June 7, 2013 I didn't use the mean diameter because the formula calls for the outer diameter. I have been playing with the formula and even if I was 0.2mm wrong on the wire thickness I can't get near 350 lbs/inch. must be doing something wrong but can't work out what. Quote Link to post Share on other sites
Peter Cobbold Posted June 7, 2013 Report Share Posted June 7, 2013 The formula is: K= ( W^4 x G) / 8 x N x D^3 where K is the rate lbs per inch W wire diameter inches G 12,000,000 for spring steel N number of active coils ( ie no free coils plus 1/2 coil) D diameter of coil measured to centre of wire, inches Quote Link to post Share on other sites
Quentin Posted June 7, 2013 Author Report Share Posted June 7, 2013 Ok. I will play with that thanks. I understood that for ground flat ends you subtract 2 coils. Quote Link to post Share on other sites
Peter Cobbold Posted June 7, 2013 Report Share Posted June 7, 2013 Ok. I will play with that thanks. I understood that for ground flat ends you subtract 2 coils. Fred Puhn's book, source of that formula, says " The number of active coils is the number of free coils plus one half coil. The free coils are the ones free to move, usually all but the two end coils". No explanation given... Peter Quote Link to post Share on other sites
john.r.davies Posted June 7, 2013 Report Share Posted June 7, 2013 Formula - schmormula! Nothing like real world measurement! Stand the spring on the floor and measure its free length. Now load the spring with something heavy, whose weight you can measure - like you. Stand on it. A stout plank will let you and friend stand on it, compressing it further into the real world. But include the weight of the plank. Now have a second friend measure the compressed length of the spring. Calculate the weight required to compress it by 1". Spring rate in lb/in = Your weight/compression distance (=free - compressed height) Voila! John Quote Link to post Share on other sites
Quentin Posted June 7, 2013 Author Report Share Posted June 7, 2013 I reckon you're right John. Not least as I expect the old springs may be a bit shagged. In which case the formula is irrelevant though I did anticipate that the calculation would be be more straightforward. I do take the point however that unless my measurements are spot on its a waste of time. Quote Link to post Share on other sites
TR NIALL Posted June 8, 2013 Report Share Posted June 8, 2013 (edited) Quentin,bring the Springs to someone who duz Rally/Race prep,they should have a device for measuring/testing spring rates. Have a looksie on DemonTweeks Website to get an idea of what you need for Measuring Spring Rates accurately. Edited June 8, 2013 by TR NIALL Quote Link to post Share on other sites
Peter Cobbold Posted June 8, 2013 Report Share Posted June 8, 2013 Zebedee springs to mind Quote Link to post Share on other sites
john.r.davies Posted June 8, 2013 Report Share Posted June 8, 2013 Spring has sprung, the sprang is shagged The rear suspension's parts are dragged Along the ground But what is this? With newer springs, the car has riz! Nhoj Quote Link to post Share on other sites
Peter Cobbold Posted June 8, 2013 Report Share Posted June 8, 2013 I reckon you're right John. Not least as I expect the old springs may be a bit shagged. In which case the formula is irrelevant though I did anticipate that the calculation would be be more straightforward. I do take the point however that unless my measurements are spot on its a waste of time. The springs may have taken a 'set' but the rate wont change, the steel's elasticity stays the same. Quote Link to post Share on other sites
Quentin Posted June 8, 2013 Author Report Share Posted June 8, 2013 Zebedee it is then. An old spring strapped to one foot and a new one to the other then see which way I boing. Quote Link to post Share on other sites
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