that is really incorrect. you only lose about 30% when shocks are at 45*.
try sin(45) degrees, lol.

 

It took you two months to come back and respond to a post I made in June of last year?

So the shocks at an extreme angle are still crappy?

 

Actually, 50% damping loss is about right at 45 degrees. Yes, sin(45) is 0.707 but the shock "velocity" is also reduced by the same ratio, so assuming a linear damping rate, 0.7x the velocity and 0.7x the vertical force is ~0.5x the effective damping.

 

And hilarity ensues. Just wanted to make sure you tried this too.

 

Ummmmmm.... the angle of the dangle is equally proportionate to the heat of the meat. I'd say 50% as well.

I have my shocks raked back. The shocks are fully charged with nitrogen (200 PSI) with stock valving and it feels fine to me. They are also mounted inboard on the axle so that they can allow for articulation.

The downside is body roll. I'm working on that one now.

 

yeah i was looking at an old thread so what.
youre all still wrong.
sin 45 is .707

pg 36. of Chassis Engineering, by Herb Adams backs up my answer.

 

sorry but unfortunately, the vertical force is the same.

angling your shocks doesnt change the sprung mass, lol

 

No arguing sin(45) is 0.707 and a shock laid over at 45 degrees has a vertical damping factor of that same amount. But there are two factors to how much damping a shock provides. One is basic statics, the vertical and horizontal components of the shock damping forces due to it's angle of mounting. The other is dynmanics, or the change in length per unit time (i.e. the velocity), which is also influenced by the angle the shock is mounted at.

Shocks are velocity sensitive, that is force = velocity times the damping coefficient. You also need to look at the fact that the velocity of the shock is also reduced by the same 0.707 factor as well. If a shock is vertical, the shock "sees" 100% of any up-down velocity. In the extreme, if a shock is horizontal, it "sees" none of the up-down velocity and provides no damping. At 45 degrees, it "sees" 70.7% of the up-down velocity and that means the damping force is 0.707 of a vertical shock (assuming a non-progressive shock). Then you need to take the vertical component of that lessened damping force, which is 0.707 as well. 0.707 * 0.707 is 0.50 or 50%.

Now if you had a coil spring mounted at 45 degrees, then yes, it's upward force would be reduced to 0.707 since a spring is not velocity sensitive. Or if you had one of the old friction dampers (pre-hydraulic shock technology) where the damping force is not velocity sensitive, then that too would show the 0.707 factor as well.

 

Correct, the unsprung mass is essentially the same. I never mentioned unsprung mass in my explanation. And yes, the vertical component of a given force changes with the angle it is applied and if you push straight up w/ 100 lbs. force, then you get 100 lbs. upward component. Push at a 45 degree angle w/ 100 lbs. and you get a 70.7 lb. vertical component (and a 70.7 lb. horizontal component).

But with shocks, the force depends on the velocity and the velocity changes with angle. You have angle and velocity that the shock sees, and assuming the same vertical motion/velocity, the angle shock sees less velocity than the vertical shock and unless you have some inverse-progressive shock (one that damps harder for slow speeds and softer for high speeds) then a shock moving slower provides less damping force than a shock moving faster.

To prove it, take an old shock and slowly push/pull on it and note the force it takes. Now try and move it twice as fast then twice as fast again. You'll soon find out how fast the force needed to move the shock increases.

Actually most shocks are progressive, in that the damping rate increases with higher velocity, that way you have a softer ride when tooling around town, then a firmer ride on the highway.

 

im an ME so im fully aware whats going on here.

ill just agree to disagree, and you should write the book company and tell them their author is incorrect.

its more like this:

(c)(xdot), where xdot=velocity

so its .707*[(c)(xdot)]

thus 70.7% effective.

i think you might be thinking that the damping coefficient C changes with angle, when it doesnt.
C is a constant for a damper, meaning that only velocity can change the force exerted.
you are right that the velocity_of_damper is velocity*sin(theta)
but since C doesnt change, the force exerted is then C*Vsin(theta)


youre trying to count the angle in there twice.

 

you said it yourself:
"At 45 degrees, it "sees" 70.7% of the up-down velocity and that means the damping force is 0.707 of a vertical shock"

that is the vertical component. you then try to count the angle in there again for some reason which is totally incorrect.

 

Yep, angle figures twice. You are assuming velocity in total stays the same. xdot may be the same, but if you take the velocity the shock sees, that is not xdot (unless the shock is vertical). Shocks compress along their long axis, if that axis is not aligned with the vertical axis, the change in length per unit time of the shock's length is not equal to xdot as you call it.

If you want references, Vector Mechanics for Engineers: Statics and Vector Mechanics for Engineers: Dynamics by Beer and Johnston, published by McGraw Hill.

And to be clear, I AM NOT saying C changes, but it is velocity that is V*sin(theta) as you note. So the force of the shock changes with the slower velocity. Then that smaller force is acting at an angle, once again V*sin(theta) so you have to multiply that factor twice. If the shock is vertical, then sin(theta) = 1.0 and you have the full force acting directly up and down. But at 0 degrees (shock horizontal) sin(theta) = 0.0 so you have no compression in she shock and thus no damping force and even if you did the shock could not push up or down.

 

im sorry, youre still wrong. and i have both of those books.
have you taken engineering classes? further past dynamics?
you wouldve learned the xdot im talking about is velocity, its dx/dt in a class like system dynamics and then further in vibrations.

velocity in the Y direction that the shock sees is xdot.
the change in length per time, again dx/dt is V*sin(theta)

im not assuming that velocity is the same.
velocity any way you look at it is velocity of the body*sin (theta). that way the sin function is 1 when vertical.

 

here maybe since you dont believe me, youll believe this:

first google search entry.
search: shock absorber effectiveness at angle

http://www.4x4review.com/feature/shock-genius.asp

 

Um .... dfoxengr ... you're making an ass out of yourself. Some might say "you've been owned"....

As a mechanical engineer and an impartial reader of this thread, Roger's argument makes a lot more sense than yours. In addition, he's been "doing' engineering a long time, near as I can tell from your sig, you're still in school.

Don't worry - we all think we know everything at some time in our lives...

 

Yes, taken more classes than those. Yes, xdot is dx/dt but you need to put that in the frame of reference of the shock, it's X-axis is along it's length. But that axis is not aligned vertically with respect to the frame of reference of the suspension (assuming we are talking about one angled 45 degrees). So you must account for the change in angle both in terms of the dx/dt of the shock length as well as the static analysis of the force applied at an angle. At an angle the shock compresses slower than one that is vertical. Slow compression speed = less damping force. Then if you push on something at an angle, only part of the force applied acts vertically.

 

That analysis is ignoring the real world behavior of the shock absorber. If you have a purely resistive (friction only) shock, then what they say applies. But shocks (I am assuing a commonly used hydraulic shock absorber that is in common use on vehicles these days) is not a purely resistive device. It has a velocity sensitive nature to it's resistance to motion. Move slow and get a little damping, move faster and you get more damping. Put the shock at an angle and it moves slower for a given vertical motion, slower motion = less damping force.

A real shock probably has friction effects (seals and what not), velocity effects (viscous fluid resistance) and probably position effects (spring constant due to internal gas pressure) not to mention any progressive damping features that increase damping constant in relation to velocity. So the only way to knwo for sure on a given shock is to put it up on a chassis dyno and measure the forces in the shock and in the suspension.

I have run my rear shocks at various angles and I can tell you the fall off in damping is way much more than due to the single sin(theta) effects. ANd this is with the same shocks and the same springs, only change is the angle of the shocks.

 

1. The article itself says the numbers are "rules of thumb"

2. I don't trust anything about forces that uses the word "dampening" (to make wet) instead of "damping" (reduce the amplitude of oscillation)

 

Yeah, and I am a certified idiot, but this is hilarious.

Quote the bible for all I care.

 

I am not an engineer.

I am, however, two months away from getting a degree in physics. I'm only mentioning this because it will assure you that i have had through at least partial differential equations, so i can handle algebra.

Since it is the only thing with which i have experience, we will purely assess your math.

you have written .707*[(c)(xdot)]

You are defining xdot as velocity, which is standard, and c as your damping coefficient.

You then claim it to be 70% effective.

In order for this to be true, .707*[c*xdot]=.707

Thus, c*xdot is 1, by your arguement. However, this means that C is no longer a constant coeffiecient but actually the inverse of velocity. Since velocity will vary widely, c must be some function we are unaware of. So, there must be an equation for C that you have excluded. Perhaps i am confused by shorthand, but i think this isn't likely.

Please post the function that is represented by C, and we will continue to break this down.