Sizing a Shock Absorber

[PCarson85]

I'm trying to size a shock absorber for a seat. I found a helpful website but when on step 2 shown below, I don't see what is meant by the drive force. Is this just the weight x 9.81?

Isn't the weight already covered in step 1? Also, if there is a g force. Isn't the weight in step 1 and 2, multiplied by the g force?

 

[hydtools]

I don't see it as all that helpful.
FD would be the force of the shock absorber moving through distance S. The work done by the shock absorber equals the kinetic energy of the moving weight.

Ted

 

[GregLocock]

That's rather a strange way to size a shock, I don't think it applies to road type inputs.

Cheers

Greg Locock


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[zeusfaber]

The first sentence in step 2 suggests this method is for some sort of application where you're applying power through the shock absorber (presumably to drive the object at the other end once the absorber has closed up as far as it's going to).

Looking at the shape of the equations, I think the first step covers only kinetic energy (all that talk about weight is because they couldn't be bothered to talk about mass). It's not written at all clearly, but I think the formula is trying to be

E[sub]K[/sub] = (W.V[sup]2[/sup])/722

In your application, step 2 could conveniently cover the potential energy released by seat and rider dropping the distance of the piston stroke (so F[sub]D[/sub] would be the weight of the seat and rider in lb).

You really need to watch your units: You can't talk about force being Weight x 9.81 (g is not dimensionless) and use a website that talks about weights in lb without coming severely unstuck.

A.

 

[MintJulep]

Weight x acceleration is meaningless.

Mass x g is a kitchen chairs. Probably doesn't need a damper.

 

[GregLocock]

So, OP, describe the application

Cheers

Greg Locock


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[saplanti]

The following link may help you with the calculation:

http://www.magral.com.br/public/imagens/dinamics/modelos/Enidine_folha calculo_HD-HI.pdf

 

[rb1957]

I hate when people say "Weight ... mass in lb" ... weight and mass are different.

KE = Wt*V^2/722 ... not so sure about the factor ... definitely nothing to do with 9.81 m/s^2 it'll start with 32.174ft/sec^2 changing to 386in/sec^2 and then there's the 1/2 ... KE = 1/2*(m*v^2) which makes it 772 ... so I think another gremlin.

so what does that make the units of KE ... Lbf/(in/sec^2)*(in/sec)^2 = lbf*in ... not a pure unit (should be lbf*ft) but it'll do.
what this means is S is in inches (as noted).

It would be "better" if weight in lbf, V in ft/sec, D in ft (these units are consistent) ...
and you can figure out the changes in the coefficients.

another day in paradise, or is paradise one day closer ?

 

[3DDave]

I find that forcing calculations to use slugs slows people down from equating mass to weight. As much as people hate them, it really exposes where the gravitational constant needs to be used. Even the metric system messes up by not requiring scales to be calibrated in newtons. In common use people don't put their weight in newtons or refer to their mass in kilograms; they mix those up.

Drag slugs into the equation and people's brains suddenly have to turn a corner and actually spend some time thinking about the units. Sadly, though, there's still the pound-mass hanging around to spoil things.

 

[rb1957]

I don't need slugs (or nails) ... I just mean the words ... don't say "mass" when you mean "weight".

Using mass is more correct, but recognising you're using weight and so when you need mass in an equation well that's weight/g. At least you're understanding where the "g" is coming from when it's "randomly" added to the equation.

Using metric weight, kgf instead of N, is ok in the real world, but in our world I think it can lead to trouble.

another day in paradise, or is paradise one day closer ?