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Discussion Starter #1
I am trying to work out how much my shock absorber stroke will reduce when it is mounted at x number of degrees from vertical. (this is so I can buy the correct dimension shock absorbers when they are relocated).

If you imagine that the stroke is 100% when the shock absorber is mounted vertically, if the shock absorber then is leaned back by, for example, 30 degrees (from the back of the A-arm to the new top mounting on the car's chassis), how much less of a stroke will there be? I expect there is a formula for working this out, but I don't know it :( .

Taking it to extremes to try and guess, if it was mounted at 90 degrees from vertical, there would be no stroke, so I thought that at 45 degrees the stroke might be halved, but having just experimented with two pens and a piece of cardboard, I appear to be completely wrong :( .

Someone save me with a formula, or I'm going to have to make a miniature model to experiment with :) .
 

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Face Puncher
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trigonometry, my friend

actually, the stroke of the shock will remain unchanged, but the effective stroke will be what is affected..

but, alas, Im no trig expert... but a Yahoo search should prove helpful.
 

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Discussion Starter #3
I must be psychic! :cheesy: I thought to myself 'I predict that 216RIDER will answer this one!' :biggrin: And see what happened! :biggrin:

Maths isn't my strong point either. I can see that at 90 degrees from vertical, the end of the shock will be moving up and down the full travel of the stroke, but not (or hardly) pushing the rod into the chamber. It's the bit between 0% and 100% stroke that is confusing me at the moment :angry: .

My relocated front shock absorbers will lean back approximately 30 degrees from vertical, so I expect that it will not be a major problem. However, I like to think things through and thought up this problem :biggrin: .
 

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Face Puncher
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so, go in steps..

if vertical is 6 inches and horozontal is 0, then it stands to reason that every 10° of angle will subtract 0.66 inches of effective stroke..

so, 30 degrees from vertical will give you an effective stroke length of 4"

you and your psychic self..
 

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Discussion Starter #5
:biggrin: :biggrin: :biggrin:

I did think that, and was very pleased with myself. That was until I tried to put it into action with two pens and a piece of cardboard on my desk surface.

Admittedly this wasn't an ideal experiment, but the results proved that I was either:

a) wrong, or b) in need of a different piece of cardboard.

:uh:
 

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EXAMPLE ONLY...................

If your shocks are mounted next to your Lower ball joint (@90 Deg )
Then if your wheel moves up 1" your ball joint moves up 1" then your shock will move up 1"..

Now if you mount shock ( @90 Deg) half way between lower ball joint and lower arm pivot point.....
When wheel moves 1" up..Ball joint moves up 1" Shock will move 1/2" ( shock is now only 1/2 as efficent )
 

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Come on. Have you guys forgotten about the Pythagorean Theorem?

For a right triangle... A-square + B-square = C-square.
The square of a hypotenuse is equal to the sum of the squares of both sides.

To figure out the max stroke of the new shock (if the current one actually IS vertical...

1) Square the max length of your current shock
2) Square the distance that you move the new mount
3) Add those two
4) The square root of that sum will be the max length of the new shock.



Last edited by BOUNCIN89MERC2LOW97LHS at Nov 3 2003, 12:16 PM
 

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In basic terms shocks can be calulated by Ratio::

If your shocks in std for work @ 1" wheel travel to 1" shock travel.

you will require twice shock [email protected]
1" Wheel travel to 1/2"travel..

on orignal question..

shock @ 0 Deg ( upright ) = 100% Effective
shock @ 10 Deg Incline = 98% Effective
shock @30 Deg Incline = 86% Effective
shock @45 Deg Incline = 70% Effective.

Results curtasy of " Chassis Engineering" HERB ADAMS

And he knows what he is talking about !!!!!!!!!!!!!!
 

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Discussion Starter #10
Thanks for the latest replies! I knew this question would attract some technicians! :cool:

That's an interesting result. Like you jjj, I thought 45 degrees would halve the stroke. But the example shows 86%.

Thanks everyone!

:cool:
 

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Find your househole calculator and do the following.

Find your normal stroke lenght first of all, lets say 3 inches. Then find the calculator's "cos" command and enter the angle you will use. then multiply that by your stroke length.

EX
45º degrees
3 inches of stroke

enter cos45 * 3
 

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Discussion Starter #12
Thanks drboogie! :cheesy:

I haven't got a scientific calculator! If anyone has got one, could they run the example for a 30 degree angle (which mine is!)?

Thanks :)
 

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Face Puncher
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Originally posted by BOUNCIN89MERC2LOW97LHS@Nov 3 2003, 11:14 AM
Come on. Have you guys forgotten about the Pythagorean Theorem?

For a right triangle... A-square + B-square = C-square.
The square of a hypotenuse is equal to the sum of the squares of both sides.

To figure out the max stroke of the new shock (if the current one actually IS vertical...

1) Square the max length of your current shock
2) Square the distance that you move the new mount
3) Add those two
4) The square root of that sum will be the max length of the new shock.
isnt that the same as the 3-4-5 rule?
 

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Originally posted by 216RIDER+Nov 3 2003, 05:38 PM--></span><table border='0' align='center' width='95%' cellpadding='3' cellspacing='1'><tr><td>QUOTE (216RIDER @ Nov 3 2003, 05:38 PM)</td></tr><tr><td id='QUOTE'><!--QuoteBegin--BOUNCIN89MERC2LOW97LHS@Nov 3 2003, 11:14 AM
Come on.  Have you guys forgotten about the Pythagorean Theorem?

For a right triangle... A-square + B-square = C-square.
The square of a hypotenuse is equal to the sum of the squares of both sides.

To figure out the max stroke of the new shock (if the current one actually IS vertical...

1) Square the max length of your current shock
2) Square the distance that you move the new mount
3) Add those two
4) The square root of that sum will be the max length of the new shock.
isnt that the same as the 3-4-5 rule?[/b][/quote]
Yes, it is. 3 and 4 being the sides of a right triangle and 5 being the hypotenuse...

3sq + 4sq = 5sq
9 + 16 = 25



Last edited by BOUNCIN89MERC2LOW97LHS at Nov 3 2003, 10:38 PM
 
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