scootz
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| posted on 22/5/12 at 11:13 AM |
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Spring Rates...
I'm trying to work out what lbs/inch rate I'm roughly looking at for the springs on the reverse-trike. I guess there will be many
variables, but can anyone offer a guestimate based on a finished weight of 300kg (distributed evenly across the three wheels).
Cheers! 
It's Evolution Baby!
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Dangle_kt
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| posted on 22/5/12 at 12:40 PM |
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is there a linkage on the swingarm?
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v8kid
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| posted on 22/5/12 at 01:03 PM |
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Ok assuming you have 100kg on the rear wheel
If you have a 1g bump without hitting the bump stop thats 100kg force on the wheel. (for 2g 200)
and if the wheel moves say 4" before hitting the bump stop thats 100/4 =25kg/in if you will excuse the mixed units.
from there you need to know the lever ratio which is the distance between the spring mounting point and the swinging arm bearing divided by the
distance between the axle and the aforsaid SA bearing .
Say it is 16" divided by 24" or 0.66666
divide the 25kg/in by the ratio and you will get 37.5kg/in
But that is assuming the spring is vertical so if it is at 45 degrees say you have to divide it by the sine of the angle 0.707 in this case
37.5/0.707 = 53 kg/lb
or in sensible units 116lb/in
Now 4" travel for 1g deflection is fairly soft usually sports cars are around the 1.5 mark which makes it 2.66 times stiffer (300lbin) and i
think you are optomistic with the 100kg.
Also if your spring is closer to the vertical say 60 degrees then the 0.707 correction factor becomes 0.866.
Any help?
Cheers!
You'd be surprised how quickly the sales people at B&Q try and assist you after ignoring you for the past 15 minutes when you try and start a
chainsaw
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hughpinder
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| posted on 22/5/12 at 01:25 PM |
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I think you need to make a couple of decisions!
This is all approximate:
100kg per wheel = 220lbs
If the spring mounts a fraction of the way along the arm divide this weight by that fraction to account for the leverage,e.g if it mounts 80% of the
way along the arm divide by 0.8, if it mounts to the top of the upright like the haynes rear end divide by 1.0
I'll assume 80% (divide by 0.8) = 275 lb seen by spring mount point
Now allow for the spring angle, by dividing this by SIN(angle to horizontal). If its bolt upright thats a 1.0, if its at 60 degrees its 0.86, 45
degrees = 0.707, at 30degrees = 0.5.
I'll assume 45 degrees = 275/0.707 approx 400 lb seen by spring to provide this vertical force
Then work out how much spring travel you want to use up to take this load - divide by the expected full stroke length of the spring/shocker and then
divide by the fraction you want to use - say its a 4 inch stroke which is quite common) and you want to use 1/3 of the spring to take this (so 2/3 of
the suspension travel is bounce, 1/3 for droop).
For my assumptions then
(((220/0.8)/0.707)/4/(1/3)) approximately 295lb/inch spring rate
for a spring mounted on the top of the upright, at 90 degrees (upright) with a 5 inch stroke and 1/2 of that used:
(((220/1.0)/1.0)/5/(1/2) approx 90lb/inch
So a wild guess would be between 90 and 300 lb/inch!
I hope this actually helps (and I haven't made a mistake or missed anything important)
Regards
Hugh
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scootz
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| posted on 22/5/12 at 01:59 PM |
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That helps me understand the process much better guys...
Thanks V much!
It's Evolution Baby!
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v8kid
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| posted on 22/5/12 at 02:17 PM |
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Err summat I forgot to mention 'cos I did not want to get all complicated.
What is really important is the suspension frequency. Don't worry what it is it could be doughnuts for all it matters the important thing it is
transferrable between very different cars.
So if we have a suspension frequency of 2 hz front and 2.2hz rear that would suit a mid engine car around 50/50 weight distribution regardless of
the actual weight. It is not weight dependant.
Auto engineers use this a lot so there is loads of info on what frequency may be sucessfull for your specific needs.
Now the easy peasy bit is the suspension frequency is directly related to the distance the wheel will deflect with a 1g load and you use a graph or
lookup tables to determine it. ( jack the car up and the distance the chassis moves when the weight is just off is the 1g distance)
You should be able to google the graphs.
So pinch somebody elses data (uni papers usually give it - look sae papers) apply it to your own circumstances to get a wheel rate then translate this
into a spring rate.
Should be in very much the right ball park.
Then you will need a roll bar for a 3 wheeler
Simples
Cheers!
You'd be surprised how quickly the sales people at B&Q try and assist you after ignoring you for the past 15 minutes when you try and start a
chainsaw
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scootz
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| posted on 22/5/12 at 02:34 PM |
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Cheers!
I'll be running a 3/4" ARB inside one of the beam tubes.
It's Evolution Baby!
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mikeb
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| posted on 22/5/12 at 04:29 PM |
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Glad someone mentioned calculating using the frequency.
Higher the Hz rate the stiffer the suspension.
1-1.3Hz is soft for off road type stuff, 2-3Hz stiff for track focussed stuff. freq = 1/2pi x sqrt (stiffness/mass), as people have said you need to
turn these wheel rates i.e. stiffness at the wheel to spring rates by applying whatever suspension ratio/angles you have. This is from distant memory
so may have forgotten the formulas and can't be bothered checking
Other option is to work out how much body roll you want and come up with a roll gradient deg/g then work this back to suspension stiffness including
your anti roll bars.
have fun
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