Coil Spring Free Length
#3
is there a spring calculator that will help determine the free spring length needed to adjust ride height. iexample: i want a one inch change in ride height. this isnt going to equate to a spring with a one inch longer free length. spring rate will determine a comression factor. sorry about no caps to start sentence, mobile phone makes it a pain.
#4
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you can use Shaikh's spreadsheet.
But there is roughly 560lbs on each front corner and 530 lbs on the rear corners.
a 700 lb spring will compress only about .8" inches in the front. A 450 lb spring will compress around 1.2". Less if the car is acutally lightened.
I tried 7" 550s in the front and i could barely lower it enough...so i'm assuming 700s need to be a little shorter to have some up/down wiggle room to adjust.
But there is roughly 560lbs on each front corner and 530 lbs on the rear corners.
a 700 lb spring will compress only about .8" inches in the front. A 450 lb spring will compress around 1.2". Less if the car is acutally lightened.
I tried 7" 550s in the front and i could barely lower it enough...so i'm assuming 700s need to be a little shorter to have some up/down wiggle room to adjust.
#5
you can use Shaikh's spreadsheet.
But there is roughly 560lbs on each front corner and 530 lbs on the rear corners.
a 700 lb spring will compress only about .8" inches in the front. A 450 lb spring will compress around 1.2". Less if the car is acutally lightened.
I tried 7" 550s in the front and i could barely lower it enough...so i'm assuming 700s need to be a little shorter to have some up/down wiggle room to adjust.
But there is roughly 560lbs on each front corner and 530 lbs on the rear corners.
a 700 lb spring will compress only about .8" inches in the front. A 450 lb spring will compress around 1.2". Less if the car is acutally lightened.
I tried 7" 550s in the front and i could barely lower it enough...so i'm assuming 700s need to be a little shorter to have some up/down wiggle room to adjust.
(560 lbs x 1.6) / 700 lbs/in = 1.28".
Defection at the wheel (from unloaded) for that 700 lb spring will be:
1.28" x 1.6 = 2.05", more or less.
#8
The quick way is to measure the total length of the lower arm, then divide the length of the shock to the inner bolts of the lower arm by it. You then have to figure out the degree the spring is laying, and add that into the equation....... Or just use the figures Shaikh has already blessed us with.
#11
The mechanical motion ratio is the ratio of the shock mount location to the overall length of the control arm, so whether you use it as a number greater or less than one, depending on which number you choose for the numerator/denominator, either number is the reciprocal of the other. The reciprocal of 0.66 is 1.52; the reciprocal of 0.735 is 1.36.
Shaikh reports both mechanical and bounce motion ratios; I wonder if this is where the difference is, not that I know the difference between the two. I haven't found the ratios you've mentioned; could you refer me to a cell location?
Cool cat fight, btw.
Last edited by Thucydides; 06-02-2010 at 10:33 AM.
#13
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So hypothetically if I wanted to lower my ride height by 1/2" all around, I would need to adjust my coilover perches downward by...
.5" / 1.34 = .37" front
.5" / 1.26 = .40" rear
That is a question. I'm not really going to lower it another 1/2" just yet but I want to make sure I'm understanding the concept correctly.
.5" / 1.34 = .37" front
.5" / 1.26 = .40" rear
That is a question. I'm not really going to lower it another 1/2" just yet but I want to make sure I'm understanding the concept correctly.
#14
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Between the tubular sway bar stuff in yellow and springs & sway data.
Spring MR (bounce) | Leverage ratios (mechanical motion ratio)
Front 0.72 | 0.66
Rear 0.88 | 0.735
So hypothetically if I wanted to lower my ride height by 1/2" all around, I would need to adjust my coilover perches downward by...
.5" / 1.34 = .37" front
.5" / 1.26 = .40" rear
That is a question. I'm not really going to lower it another 1/2" just yet but I want to make sure I'm understanding the concept correctly.
.5" / 1.34 = .37" front
.5" / 1.26 = .40" rear
That is a question. I'm not really going to lower it another 1/2" just yet but I want to make sure I'm understanding the concept correctly.
#16
So hypothetically if I wanted to lower my ride height by 1/2" all around, I would need to adjust my coilover perches downward by...
.5" / 1.34 = .37" front
.5" / 1.26 = .40" rear
That is a question. I'm not really going to lower it another 1/2" just yet but I want to make sure I'm understanding the concept correctly.
.5" / 1.34 = .37" front
.5" / 1.26 = .40" rear
That is a question. I'm not really going to lower it another 1/2" just yet but I want to make sure I'm understanding the concept correctly.
#17
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Okay, glad I understand the theory. However, in practice I'll probably just drop all four coilovers approximately the same amount, then take the car to get corner balanced again and the individual corners' ride heights will end up wherever the scales tell them to be.
#20
When setting a car up on the scales, you might have one corner .25-.50 higher or lower then the other side. Unhooking the end links, set the one side up till the bar is parallel with the pavement (or close, this part has little science behind it), then adjust the other end link to where the bolt slips in and out without any force, which gives you zero preload.
I assume you know what preload is?
http://www.longacreracing.com/articles/art.asp?ARTID=30
This is a good read and it explains why I'm able to get away with my 400F/225R setup as a DD/AutoX car
I assume you know what preload is?
http://www.longacreracing.com/articles/art.asp?ARTID=30
This is a good read and it explains why I'm able to get away with my 400F/225R setup as a DD/AutoX car