Quote:
Originally Posted by 135
This is an excellent thread - one of the best technical threads I've read. You're meticulous in your approach, you have documented it in a very thorough manner and it's well articulated such that a lay person can (mostly) understand it. Also, the diagrams are extremely useful to assist in visualising what your explaining. I had to "appreciate" quite a few posts to give you credit where credit is due. Well done!
My questions are mostly regarding your calculations - I'd like to understand these in more detail so I can apply the same to my own case. This topic interests me quite a lot so I've spent a lot of time formulating my thoughts and addressing a number of points that you raised, the result being my post is v-e-r-y long!
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With all those "appreciates" I guess you have compelled me to reply! But I do appreciate that you are paying attention and keeping me on my toes. I am going to pick at this a bit at a time though, so that my reply is easier for me to keep track of.
Quote:
Originally Posted by 135
When discussing the rear springs, you refer to the difference between kg/mm and N/mm and how it can impact the conversion to lbs/in. I'm not sure what units the Ohlins spring rates are measured in but the Swift Spring rates are, as per your example for the Z65-228-120, specified as both 12.0 kg/mm and 672 lbs/in according to the Swift Springs website. 12 kg/mm converts to 117.68 N/mm, which then converts to 671.97 lbs/in, but, in your example, you've used 120 N/mm and 684 lbs/in. Since you mentioned the differences between the units of measure, I thought you may have just rounded the N/mm measurement so, when I saw that you referenced 684 lbs/in in the associated spreadsheet calculation and graph, I was a little confused. Am I missing some minor or obvious detail? |
I had that Swift data in front of me too. Strictly speaking they refer to "kgf" not "kg". A "kgf" (kilogram force) is not a normal metric unit, but in the same spirit as a "lb" (pound) is a unit of mass and a "lbf" (pound force, shorthand "pound" to confuse things) is a unit of force and is the force exerted by one lb when acted on by 1 g, a kgf is presumably the force exerted by one kg when acted on by 1 g.
In that scenario, the conversion from kgf/mm to lbf/in is kgf/mm x 2.2046 (lb/kg) x 25.4 (mm/in).
If you do the math 12 kgf/mm = 12 x 2.2046 x 25.4 = 672 lbf/in
That agrees with the Swift website data. The only problem is that it does not agree with my measured Swift spring rate data.
The correct metric unit of force is the Newton, which is the force acting on 1 kg when accelerated at 1 g.
1 Newton = 1 kg x 9.81 m/s^2 = 9.81 kg.m/s^2
Assuming "12 kgf/mm" really means "120 N/mm" we convert to lbf/in:
120 N/mm * 2.2046 lb/kg * 25.4 mm/in / (9.81 N/kg) = 685 lbf/in
That does agree closely with what I measured whenever I have measured Swift springs.
The "6.0 kg/mm" springs (stamped "60") should be 342 lbf/in if they are really 60 N/mm. I have measured between 344 lbf/in and 346 lbf/in. (3 springs).
The "12.0 kg/mm" springs (stamped "120") should be 685 lbf/in if they are really 120 N/mm. I have measured 684 lbf/in (1 spring).
By "measured", I mean I compressed the springs with a hydraulic cylinder and measured the resulting load with an electronic load cell. I measured the deflection with a caliper. I took readings every 10 mm (approx) and continued to a total compression of at least 100 mm. I plotted the results and used Excel to plot a linear regression line on the points from 20 mm to 100 mm. I rounded the slope of that line to 3 significant digits and that is the measured rate I am reporting.
It is too late to say "in short", but, in short the Swift catalog appears to be wrong (ie. produced by marketing not by engineering). I have to conclude from my testing that the Swift springs are manufactured in increments of 10 N/mm, not increments of 1 kgf/mm.
Incidentally, the Ohlins springs in the kit are 60 N/mm front = 342 lbf/in (I measured 344 lbf/in), and 70 N/mm rear = 400 lbf/in (I measured 399 lbf/in).
Quote:
Originally Posted by 135
Also, while it wouldn't be too difficult to replicate, would you be able to attach the spreadsheets that you use for your calculations?
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I have described the method so I will let you replicate it if you wish. I figure by the time you have done that you will understand it backwards and forwards. To be meaningful you will need some test data. Try this data for a Z65-178-060:
Force Compression
(lbf) (in)
0 0.000
147 0.374
250 0.681
355 1.004
451 1.295
552 1.606
652 1.906
752 2.197
851 2.476
950 2.756
1050 3.035
1152 3.327
1255 3.606
1349 3.862
1461 4.169