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Old 15-03-2010
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Northy Northy is offline
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Default Measuring springs

Anyone had any reasonable success measuring/comparing springs at home with just general equipment lying around?

I have a couple of ideas how to do it but would like to hear other peoples ideas.

I'd like to basically compare one spring against another of a different diameter.

Cheers,

G
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Old 15-03-2010
Alfonzo Alfonzo is offline
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Some sort of thingy which applies load and allows you to measure deflection. I think I once saw such a device specifically for doing this job, possibly from one of the suppliers of set-up stations, fancy tools etc. Imagine a column that sits between and in-line with the springs. At 90 deg to this is an arm that sits across the top of the 2 springs. The column is threaded, and as you tighten it up it pulls the arm which compresses the two springs. The arm is free to pivot, revealing which spring is resisting the compression most.

I might have made the above up..

Nahh, it's real.
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Old 15-03-2010
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Bed springs? Have you been rutting to hard and broken them?
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Old 15-03-2010
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I'ts brutal, it's ugly, and it's coarse... but I sort my springs out into a relative order by bunging one under a side of my laptop and measuring how far it compresses...

There are spring reference charts around, and you can compare a mystery spring's behaviour to that of known one's to get an idea of its rating.

There must be more accurate ways of going about it, but I did this the other night to figure out the apporx weights of some random springs I picked up in a parts lot.

Cheers,
Ty

Quote:
Originally Posted by Northy View Post
Anyone had any reasonable success measuring/comparing springs at home with just general equipment lying around?

I have a couple of ideas how to do it but would like to hear other peoples ideas.

I'd like to basically compare one spring against another of a different diameter.

Cheers,

G
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Old 15-03-2010
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Battle_axe Battle_axe is offline
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you cold compress the spring against scales by 1 inch and mesure the force in lbs and that would be the rate i think? however smaller diamiters get harder faster than larger ones
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Old 15-03-2010
Alfonzo Alfonzo is offline
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If you can lay your hands on a some measuring verniers, I guess just putting weights on the springs and measuring the deflection. So just need to find a way of supporting the spring and weight properly.
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Old 15-03-2010
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A quick and crude way to compare mystery springs is to thread both springs onto a stick (pencil, screwdriver, whatever) and compress them against each other. The one that compresses fully first is the softer one.

Only works with springs of similar length though.
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Old 15-03-2010
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Seriously, though - you could get a pretty good reading using a ruler and any object of a suitable known mass.

If you don't have digital balance at home, try taking an object to your local post office and weighing it. Even something like a rock from your garden or a red brick would be fine. Come to think of it, a red brick (or a piece of one) would be perfect...

1)measure the spring unloaded.
2)balance the known weight on top (a little guidance with your free hand to keep it balanced won't make any difference to the reading)
3)measure the spring again.
4)do the maths...

Cheers,
Ty
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Old 15-03-2010
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Quote:
Originally Posted by Battle_axe View Post
... however smaller diamiters get harder faster than larger ones...

Quote of the day!!!
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Old 15-03-2010
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I use my bench drill in the shed and a cheap digital scale from argos.
Compress the spring 25mm and read the weight off the scale.

Gives you grammes load per 25mm,then label them in boxes.
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Old 15-03-2010
Alfonzo Alfonzo is offline
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Might be easier to find an object that will sit on top of a spring and let you tie a piece of string which you can pass through the center of the spring and tie a weight to, therefore hanging down and compressing the spring. Measure the length vs the original uncompressed length.

You need a piece of string in there somewhere!
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Old 15-03-2010
Alfonzo Alfonzo is offline
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Quote:
Originally Posted by sparrow.2 View Post
Quote of the day!!!

I was thinking the same thing! But far too polite to point it out of course

Last edited by Alfonzo; 15-03-2010 at 07:58 PM. Reason: spellll
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Old 15-03-2010
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Quote:
Originally Posted by sparrow.2 View Post
Quote of the day!!!
ahh damm you saw it :P i didnt know what to put as stiffer wasnt any better
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Old 15-03-2010
Alfonzo Alfonzo is offline
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Quote:
Originally Posted by Dingbat View Post
I use my bench drill in the shed and a cheap digital scale from argos.
Compress the spring 25mm and read the weight off the scale.

Gives you grammes load per 25mm,then label them in boxes.

I like.
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Old 15-03-2010
pro4nut pro4nut is offline
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going a little more complicated,

drill a 4mm hole in a piece of wood
put a spring on the piece of wood with the hole dead centre
put a shock shaft with its bottom still on it on the top of the spring
attach some fishing line to the end where the piston would normally be i reckon leave on circlip on for this attach the other end to a fishing sprung scale basically under the piece of wood
pull down on the scale to a fixed load, measure the compression
or pull down to compress spring to a fixed compression and read off the mass from the scales
does this make sense to anyone but me??
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Old 18-03-2010
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Isnt a spring just a coiled up roll bar? Its basicaly twisting a rod of steel over its length when you compress it.

So if you measure the wire thickness and the length of the wire (number of coils x Circumference = π × diameter) you should be able to make a chart??

Lets pretend they all use the same gauge wire, the longer the length the more twist you can get out of it ie the softer the spring??

Hmm think that'll work
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  #17  
Old 18-03-2010
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a old shock no oil or seals in, nothing to increase Resistance,clamp in vice from bottom eyelet, the use known weights attached to top eyelet, job done
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Old 18-03-2010
TSewell TSewell is offline
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Most everyone here is on the right track, but just because:

For a linear (not progressive, easily recognized by change of pitch of wind) spring, the equation for spring rate is

F= K (X2-X1)

Where:

F= Spring Force (equals zero at static length)

K= Spring Rate

X2= Compressed height

X1= Uncompressed ( or compressed less!) height

What this tells us is that the best way to measure spring rate is to apply a known force (known weight) to the spring and measure its' deflection (how much it compresses).
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