Please report the findings and how he did the testing. I know Associated says the taper goes down and Team C says up. Never got around to testing which is better on the same car.

Im mumbling NO as I hang my head low and kick rocks! Already had the buggy built by the time I came apon this thread. I plan to this weekend since I'll be emptying the shocks out any way. I do have a full set of springs, so I plan to find a balance using all the rates. Hopefully!

I assume you are talking about springs? From the spring selections available I'm using relatively light LOSI and Associated springs or heavy TEKNO springs in the rear. Check the TEKNO SCT410 thread. A very helpful fellow has put together a list of front and rear springs from many different sources and put them in order by spring rating. I printed this and take it everywhere. Since you said your chassis isn't slapping the ground much you are probably running much heavier rear springs than me (or have a flatter track). I don't have LOSI cars so my spring selection won't help you but I think you said you have three different LOSI springs, I count seven LOSI rear spring rates ranging from 2.50 to 4.30 (6 front spring rates). If you don't know the spring ratings of your springs check out the list I referenced. The heaviest rear spring I use in any of my cars is 3.55 in my 1/8th scale buggy.

Don't forget that where you mount the shocks (top and bottom but more so the bottom) affects the spring balance. You can use that to fine tune the balance if you don't have the exact spring you need. Have fun & let us know what you come up with and how it worked on the track.

Let's see if we can work this out together. I'm going to post some illustrations and we'll talk about them.

The first is a simple illustration showing two different shock mounting angles. Can we all agree that the shock that is vertical has more leverage on the arm than the shock that is leaned over? If we can agree to this, then we should be able to deduce that the shock that is leaning over would have to have a stiffer spring to maintain the wheel rate to compensate for the mounting angle. Agreed?

Are you referring to his post from 22 months ago on page 44, post #659?
Is there a better page on this thread to look at?

I'm running the 3.0 buggy, they have a tapered spring. Losi only has 3 different rated springs for front and back. I meant the weight of the buggy, I'm wondering if im using too heavy rated spring. Since my buggy is weighing in on the light side. Im going to get my spring balance right this weekend, then try different weight oils. If I understand Freds method correctly, too light oil I'll be bouncing on the springs still. To heavy, my buggy will do what it is now. Bouncing the wheels off the track. If there is nnogrey area in between, then my springs are too stiff. Am I correct?

I made a quick layout drawing of this. Add another arm shown at 1 degree movement, another at 9 and another at 10. Show an upper shock mounting position at a small angle, maybe 10 degrees. Draw a line from the upper shock mount to the 4 lower mounts (0,1,9, and 10 degrees). You can measure the distance between each of the lower mounts to the upper mount. You can then find how much the shock compresses for the first 1 degree stage (0 to 1 degree arm movement) and the final 1 degree stage (9 to 10 degree movement). I found that the final stage compressed the shock .0022" more than the first stage.

Then do another drawing leaning the shock over an additional 15 degrees. I found the final stage compressed the shock .0032" more than the first stage.

I believe This would mean that the more leaned over shock gets, the more progressive it gets.

I would post the drawing but I do not know how.

What those numbers tell you is that you increase pack as they lay over.

I think the amount of shock shaft movement for a given amount of arm movement also correlates to leverage. With the shock mounted vertical (or perpendicular to the arm) for the greatest leverage, the shaft movement will be more than when leaned over.

That's a good point.

Also, wouldn't a greater shock angle means the piston isn't going to be moving as fast, or as much, as a more vertical shock, all else being equal? Less piston speed should also cause the shock to pack less, and later in the compression stroke.

If the shock length increases, piston speed also increases and hence pack will as well. This is why when people run the same pistons front and rear, the rear of the car will have more pack than the front. This is the reason why I say "run more pack" in front than rear. It's a technically wrong term but most people understand it. What I mean by that is to run less piston hole area in front than at the rear. Then balance wheel rate back out with shock oils.

Let's not get pack and spring leverage confused. A vertical spring will have more leverage than one that is laying over. Lay one over 90 degrees and it'll have no leverage at all. You need to think in terms of things being equal. One thing I keep saying over and over again is that changing one thing almost never results in one thing being affected. If we want to keep the same wheel rate then we must change spring rates with shock angle. That's the whole point of moving the shocks mounting points around until we find balance.

There are two things going on in a shock and they need to be thought of separately. One is spring rate and the other is pack. Shock travel does determine leverage but we can just change springs to compensate so that is really fairly irrelevant. In the same fashion, since pack is also different based on stroke length, we can again adjust pack with pistons so that too is really nothing to concern ourselves with when it comes to geometry. My diagrams are only concerned with the effect on spring rate based on shock orientation since that's all that really matters. More vertical is more leverage. More leaned over is less leverage. Less leverage requires a stronger spring to maintain the same wheel rate.

Preferably start at the beginning of this thread and just start reading.

I made a similar rig about 6 months ago that worked pretty well. The idea was based on a "pull down rig" used on full scale race cars to measure the wheel rate. Proper pull down rigs us a hydraulic ram under a scale pad to compress the suspension with the chassis fixed, and measure the load and displacement of the wheel to generate a rate curve. I used block of wood to clamp my buggy to a bench, then placed a kitchen scale under a tire to measure the load. I insereted washers of a known thickness between the scale and the tire to compress the suspension then recorded the new scale reading with each new washer added (minus the weight of the washers). When I plotted the change in load at the scale vs change in position of the wheel, I had my rate curve. The rates were progressive of course, and sorry Fred but I never tried laying the shock over at the top to measure the change in progresiveness. I was using the data to validate my setup spreadsheet. Maybe another day. I will warn that if you are going to try this make sure the suspension is as free as possible with no shock oil in the shocks. Any friction in the system will make it tough to get good scale readings. Good luck!

Hideeho
I figured it out!
I went looking for info on levers. I didn't find what I was looking for, but did learn the a suspension arm switches between being a 2nd & 3rd class lever depending on which way it is moving. All the lever info refered vertical forces, not what we are looking for. What we want is forces at an angle or a vectored force. Once I found the correct page for that all my old trig classes started to come back to me. The 1st thing I remembered is the points on a lever (suspension arm) do not define the forces involved, only the range of motion. 2nd, the actual position of the arm in relation to the ground or the chassis is irrelevent, only its relation to the forces involved is important & they can be moved around to make it more convenient to visualize as a snapshot of an instant of its motion. 3rd any vectored force is comprised of a verticle & a horzontal force. 4th the spring on the shock is not a constant (it changes depending on it's length), the force being applied to the wheel is the constant.

Putting all that together & the snapshot you end up with is a 2nd class lever with the fulcrum being the chassis hinge pin & the effort being the wheel hinge pin & the load being the shock end. to define the vector of the shock angle the line describing the force acting on the wheel can be moved to the top of the arm at the shock end. The angle between the vertical force & the shock can be used to find the actual force applied to the shock. I found a handy trig calculator to help with this.

http://www.carbidedepot.com/formulas-trigright.asp

On this calculator, for our purposes, you 1st have think about the triangle in the picture being inverted so the right angle is to the top left instead of bottom right. This will put the hypotenuse (c) at an angle going up to the right (this is the vector of the shock). The side b will be the verticle force acting on the lever. Side a is irrelevent for our discussion. Angle A is the angle the shock is off vetical. To find what we are looking for only 2 things need to be defined & put into the calculator. 1st is the force being applied to the lever (pick any number) & enter that in the side b blank. Next is the angle the shock is off vertical which is entered in the angle A blank. For comparison, enter a different, larger angle A number (the angle will increase after the lever moves up some). The answer returned as side c is the force acting on the shock.

As an example enter the side b as 100 (n, oz, gr, lbs, tons, what ever...) & start with angle A as 10 (nearly verticle shock). This makes the force acting on the shock (side c) 101.54, slightly over 1:1. Now do it again with b=100 & A=80 (the shock is almost completely laid down). The force acting on the shock is 575.87. To find what we are looking for this will have to be done 4 times, 1-arm down, shock up; 2-arm up, shock up; 3-arm down, shock down; 4-arm up, shock down. The difference in side c from run 1 to run 2 is the increase in force needed or what we are reffering to as progressivness of the stand up shock. The difference from run 3 to run 4 is the progressiveness of the laydown shock. The larger number has more progressiveness.

run 1 b=100 A=20 c=106.42
run 2 b=100 A=25 c=110.34
run 3 b=100 A=45 c=141.42
run 4 b=100 A=50 c=155.57

run 1 to run 2=3.92
run 3 to run 4=14.15

My angle A numbers are made up & would be MUCH more accurate if it were done with numbers from a setup like fred did (hint, hint, hint ). That said, this does indicate the forces involved do get progressive as the shock is laid over.

If you have read all of this you have way too much time on your hands & tons of patience. Now go poke holes in this & tell me (politely please) what I did wrong .