turbo time
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| posted on 9/12/04 at 08:06 AM |
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Is Leverage linear or exponential?
I didn't happen to know off the top of my head if the calculation for leverage is a linear or exponential function. I was thinking ahead to my
inboard suspension bellcranks, which happen to have 800 lb/in springs (for the rear, and 550 front). Obviously I won't want that unless
I'm going for Kart-like, tooth rattling ride. I was planning on having 4 holes drilled on the inside part of the bellcrank, so that I could
unbolt the coilover unit and move the entire unit (so that it would be completely vertical in each position) to whatever hole equated to the spring
rate I wanted, whether it be a track day, auto-x, or country drive. But I don't want to make the unit based on outer to inner length ratio based
on linear calcs, then find out it's an exponential function  .
-Thanks
[Edited on 9/12/04 by turbo time]
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marktigere1
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| posted on 9/12/04 at 11:58 AM |
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If a solid rod was bolted in the middle and a force exerted on one end, the resultant force on the other end would be the same ie. linear. Moments I
think are linear.
A spring I believe is exponential.
Add the two together and my brain hurts
Mark
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tom_loughlin
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| posted on 9/12/04 at 12:07 PM |
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as far as i know its linear, as it uses the principal of moments.
moment =force. distance.
a spring however is exponential, as it uses a constant in the expression F = ke
thats my 2p's worth.
tom
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locogeoff
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| posted on 9/12/04 at 12:17 PM |
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Springs are linear (Hook's law) i.e. double the force double the extension, unless progressively wound, my gut reaction is that things are more
or less linear unless you have large amounts of suspension travel whereby things will become a bit sinusoidal, though I think we're getting into
failure scenarios there.
[Edited on 9/12/04 by locogeoff]
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marktigere1
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| posted on 9/12/04 at 12:43 PM |
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F=ke is linear. If you plot a graph of F against e you get a straight line of gradient k.
I take back what I said about springs being exponential as my typing finger engaged before brain.
Cheers
Mark
If a bolt is stuck force it.
If it breaks, it needed replacing anyway!!!
(My Dad 1991)
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Mikey G
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| posted on 9/12/04 at 12:51 PM |
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You guys are on about springs. TT is on about keeping the spring a constant and alter the positioning to give a different spring rate.
So in answer to your question..... i'm not sure how to explain it!
Basicaly on your bellcrank arrangement, every time you half tha distance to the pivot point it halfs the force upon it. or the other way around say
you have a bar and socket, if the bar was twice the lenght you would need half the force to turn the bolt to the same poundage.
I think thats how it goes..
Mike
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DaveFJ
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| posted on 9/12/04 at 01:37 PM |
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quote:
Originally posted by Mikey G
You guys are on about springs. TT is on about keeping the spring a constant and alter the positioning to give a different spring rate.
So in answer to your question..... i'm not sure how to explain it!
Basicaly on your bellcrank arrangement, every time you half tha distance to the pivot point it halfs the force upon it. or the other way around say
you have a bar and socket, if the bar was twice the lenght you would need half the force to turn the bolt to the same poundage.
I think thats how it goes..
Mike
however it also reduces th actual travel at the spring which must effect things also....?
Dave
"In Support of Help the Heroes" - Always
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marktigere1
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| posted on 9/12/04 at 01:52 PM |
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Mike
Are you not talking about moments?
In which case the answer is linear aint it?
Cheers
Mark
If a bolt is stuck force it.
If it breaks, it needed replacing anyway!!!
(My Dad 1991)
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skinny
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| posted on 9/12/04 at 01:58 PM |
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both springs and levers are linear. nice and simple.
if you don't fail, you aren't trying hard enough.
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liam.mccaffrey
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| posted on 9/12/04 at 02:14 PM |
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it all depends whether the applied force and the output force are allowed to act parallel to each other, if so then the relationship is linear. if
not the relationship wont be linear but it won't be exponential
i reckon anyway
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chrisf
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| posted on 9/12/04 at 02:17 PM |
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Have a look at this thread. The formulas given should help you along.
If you look in my photo archive (or my website) you can see that I tried the same thing. I managed to get it to work in the rear, but not the
front.
Out of curiosity, how did you come accross 800 lbs springs?? Do you have the dampers to match?
--Chris
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turbo time
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| posted on 10/12/04 at 04:28 AM |
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quote:
Basicaly on your bellcrank arrangement, every time you half tha distance to the pivot point it halfs the force upon it. or the other way around
say you have a bar and socket, if the bar was twice the lenght you would need half the force to turn the bolt to the same poundage.
Thats the info I was needing!
The 800 lb rears are from a Honda bike, and have about 2-3" of travel, so it's a good thing that I need to move them in anyways....then I
both reduce the spring rate to something usable, and increase the suspension travel to something more car-like/ usable on in the real world of our
poorly maintained roads.
[Edited on 10/12/04 by turbo time]
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Mikey G
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| posted on 10/12/04 at 10:11 AM |
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Working on that theory with your 800lb spring:-
Equal length bellcrank and wish bone would mean a spring rate of 800lb and travel of 3" at the wishbone.
Half the bellcrank length to the wish bone would mean a 400lb spring rate and double the travel at the wishbone end.
Again half that or a quarter of the original bellcrank would make a 200lb spring rate and 4 times (12" ) the travel at the wishbone end.
It would be interesting to see a plot of the above as at first glance it is linear but thinking about it more it would actually be
exponential?!?!?!?
Mike
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chrisf
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| posted on 10/12/04 at 11:10 AM |
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Not really
The push rod angle can greatly decrease the effectiveness of the whole thing. I also think the damper travel is 1.5-2", so he should be OK.
Still, 800lbs sounds really high.
--Chris
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Peteff
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| posted on 10/12/04 at 11:32 AM |
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It's giving me a headache.
If your leverage is 1:1 and your damper is upright the force needed will be 800lb to compress the spring. Honda bike springs with a taper are
progressive so if you have these you will need to take this into account as well. Wouldn't halving the bellcrank length on the wishbone side
double the effort needed to compress the spring Mike?
yours, Pete
I went into the RSPCA office the other day. It was so small you could hardly swing a cat in there.
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Mikey G
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| posted on 10/12/04 at 01:20 PM |
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quote:
Originally posted by chrisf
The push rod angle can greatly decrease the effectiveness of the whole thing.
--Chris
I was working on the principle of all things straight to explain leverage forces.
quote:
Originaly posted by Peteff
Wouldn't halving the bellcrank length on the wishbone side double the effort needed to compress the spring Mike?
Yes, sorry i wasnt a bit more clearer in my post relating to bellcrank and wishbone, when i refered to the bellcrank i was supposed to refer to that
in being the side attached to the damper. The wishbone side then being a constant fixed point and the damper side of the bellcrank could be made
adjustable in that multiple fixing points can be made to fine tune the bellcrank ratio.
Mike
[Edited on 10/12/04 by Mikey G]
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