brianm Posted January 3, 2003 Share Posted January 3, 2003 I want to check the rating on a number of spare/used coilovers I have stuffed in a box in the garage Now I've got access to a compression tester, which although not a bespoke spring tester, it will cover the range required. What I dont know is how to arrive at the rated figure. For instance do you compress the free length by say 25%, or say 25mm to arrive at the rated poundage? Anyone know? Brian. Quote Link to comment Share on other sites More sharing options...
Blatman Posted January 3, 2003 Share Posted January 3, 2003 There is a guide/article here................. Quote Link to comment Share on other sites More sharing options...
Keith Scarfe Posted January 3, 2003 Share Posted January 3, 2003 I have measured springs using a home diy method with reasonable accuracy. Obviously nothing like a proper digital spring rater jobbie but it gives you an approximate reading (to about +- 10 LbIn): What you need is your work bench (mine is made of steel) bolted to the floor, a set of bathroom scales, a trolley jack and a measuring stick (tape, ruler,calipers). With the trolley jack on floor, put the scales on top then the spring on the scales then pack the gap from the top of the spring up to the base of the work bench (or what ever you are pressing against (wall etc) using wood etc. Measure to spring length with 0 reading on the scales (adjust to 0). Jack up the trolley jack until the scales read for example max deflection and measure the new (shorter) length. Then do the sums to work out the pound inch value or kg cm as you wish. If your bench isn't bolted to the floor I guess it will work horizontally e.g. between a door opening. You just need to be able to compress the spring by a few inches in a controlled way (trolley jack). If you have access to a press at work or something even better. It doesn't matter how much you compress the spring you simply multiply the distance change (Open length - squashed length) by the force measured. e.g. OL = 12", CL = 10" force = 100lb, therefore spring rate = 2*100 = 200LbIn. HTH. Keith. Quote Link to comment Share on other sites More sharing options...
brianm Posted January 3, 2003 Author Share Posted January 3, 2003 e.g. OL = 12", CL = 10" force = 100lb, therefore spring rate = 2*100 = 200LbIn. HTH. Keith. -------------------------------------------------------------------------------- Thanks blatman, Keith, but in the example above wouldnt the spring rate, (by the method stated in the link), be 50lb as the measurement is stated to be over exactly one inch which would be load divided by distance in inches? ie 100/2 =50 lb in? Brian. Quote Link to comment Share on other sites More sharing options...
chazpowerslide Posted January 3, 2003 Share Posted January 3, 2003 Please be very very carefull, Ive seen a number of serious workshop accidents involving road-springs, slipped spring compressors ect. A fully compressed road-spring stores a massive ammount of energy, that can be lethal if released in an un-controlled way. Chaz Quote Link to comment Share on other sites More sharing options...
brianm Posted January 3, 2003 Author Share Posted January 3, 2003 Chaz, you're right to highlight the risk. There is a massive amount of stored energy in a fully compressed road spring, hence one of the reasons for my original query, I didnt want to overdue it. But in my case dont worry too much, I've run an engineering workshop for the last 15 years and have a background of over 32 years in the field I,ve managed to retain all my digits , sticky out bits etc, although probably minus a few brain cells by now. (Why cant they make them more alcohol tolerant?) Anyhow the tester we use was designed and built by me a few years ago. It is a hydraulic unit, load sensor etc and is fully enclosed. Brian. Quote Link to comment Share on other sites More sharing options...
markg Posted January 3, 2003 Share Posted January 3, 2003 i just retreat to a nice malt! and have a think! Quote Link to comment Share on other sites More sharing options...
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