TOPIC:

As long as it's not transporting live subjects, a good estimate beats a bunch of math any time.
I think vector calculation, needed for number 4 in your post, is actually one of the more straightforward disciplines.

I wrote:

I remember the academic book on the subject (dynamics) being very thick - and nearly everybody flunking it.

They had a really stringent way of correcting the papers as well.
You had to hand in all the calculations but put the final answers on a separate sheet.
If anything was wrong with that, zero points (no matter if 99% was correct and you made a small final error).
With a correct answer, they'd look if anything was off with the calculation so you could get less or no points anyway.



Reasoning was obvious though - what if the calculations were meant for a commercial plane...

Made an Inkscape drawing of the basics.
If the shock is under an angle with the vertical position (drawn in red), the force it can apply (the diagonal arrow) will be divided into a vertical and horizontal component :



If my basic math is still correct, this is the correlation between them :

vertical component = diagonal*cosinus(shock angle)

www.rapidtables...ulator.htm

So if your shock rate is 10 and the angle 45 degrees, the vertical rate it can deliver will be :

10*0.7 = 7

What a beauty

andyaus wrote:

What a beauty ...

I think that needs preserving for posterity ... the roof isn't half bad & is quite an authentic shape (when this type of caravan was built, the shape & size of the arch was based on the father of the family, so he could stand in the middle with arms outstretched & his fingers & head not quite touching the roof and sides), but I've got to say the rest looks like it's been badly made from scrap

If it was (very) local I'd be tempted to buy it & fix it to a tree as a bird nesting box

Edou's got the right idea for working out the axial force exerted on the shock.
cosA=adj/hyp, so hyp=adj/cosA.... so for 1kg mass (9.81N weight) of car resting on that shock your spring's seeing about 14N force (I think...)
Young's modulus & the equation that goes with it might be useful here too (Sorry, my memory's broke & I don't remember it... modulus= force x extension?).

Springs can be bought by diameter, length, wire diameter & force, but you'll need to be shopping at an engineering supplier's who have some idea about the subject (Won't find any like that around here, & RS don't really get it either)

If it were me, I'd be delving into my big box of springs & find something suitable by trial & error - it would be far quicker & cheaper.
You could always look at springs which fit inside eachother, or even hide smaller springs INSIDE the dampers (A'la highlift).
OR, stop being a pussy & make a leafer rear end! (Just add leaves if it's too soggy). Metal strap-banding is ideal material for the job, cuts fairly easily with aviation snips, easy to drill, FREE if you look in the right places...

eddrick wrote:

so for 1kg mass (9.81N weight) of car resting on that shock your spring's seeing about 14N force (I think...)

I think you are correct. I've drawn the shock response here because it is easier to visualise.
But since action = (-)reaction, the vectors could be drawn in opposite direction to express the force on the shock.
Seems counterintuitive but indeed the force on the shock will be greater than the vertical component when it is under an angle.

I was thinking about it last night for exactly that reason - seems wrong that a semi-sideways component would have a stronger force than its vertical origin, but if you look at it in reverse... Consider when your suspension is too hard what do you do? - lay the shocks down at a shallower angle, ie same force can more easily squash the shocks the further layed-down they are, SO the shocks would have to put more effort into stopping that happening

I had my doubt because the thing with vectors is that they have to 'add up' and that doesn't quite seem to be the case with the forces/arrows on the previous image if you put them in the opposite direction. That should probably look like this :





Gonna have to reduce where this discrepancy comes from...

I wrote:

Put them together for comparison...

I think they might both be correct but just different situations.
The one on top would merely be about weight and the second about forces redirected onto the chassis.
An action + reaction = 0 should be in there somewhere but it's really too much of a flashback to go looking for it.

The bottom diagram is right, but it depends on how you read it - if you made your angled arrow red, you could say it's the resultant force of the horizontal & vertical black forces combined (Or reverse the red arrow to show it cancelling out the horiz/vert components).
BUT, what we're looking at is Newton's 2nd law where if something is in equilibrium then all forces acting on it are equal but opposite. ie your 14N acting at 45 degrees upwards has to cancel out the 1kg (9.81N) acting vertically downwards for the car to remain stationary up/down-ways.
The shock tower is resisting being bent by the horizontal component, in other words it's exerting the same sideways force on the shock as the shock is exerting on it, or it would be accelerating in one direction or the other