dfoxengr

right still an EE though.

Erich_870

I hope some day you finish your education with a little more respect for other engineering disciplines and other people's points of view.

The whole point of this thread was to come to a conclusion through working together and sharing ideas, was it not? All you have to do is share some more data from all of the references you've cited so far. You've listed numerous books and said "I've read them, you should believe me."

Let us see what you've found instead of taking it so personal.

Speak to your professors next week, and post back what they say. The more input the better. I guess I'm looking at this situation as if we are all on the same design team and we're trying to come to the final conclusion. Keeping an open dialog and open mind is very important, and a team member needs be able to communicate thier design point without turning it into intellectual attacks.

I have to agree with Roger on this one and reply that you are still missing the Dynamics at work with the shock. You're spot on with the Statics, but the web page Roger cited explains that the geometry of the shock will also effect its overall damping. Its cycle is shorter AND slower so the effective damping is LESS.

Here's the real question. Where do you start when choosing a shock? How best do you calculate the rate of damping that you want? You can tune your geometry once you've picked a shock, but what if you're starting from scratch?

Erich

dfoxengr

im not saying believe me. i opened the books as i was posting that to double check. and im not going to try to explain anymore because none of you have the background to understand apparently.

i know he's smart, i was simply saying i wouldnt get into an electrical arguement because i know i dont know as much as EE's do.
im taking vibrations right now, and i say that qualifies me a ton more to understand this stuff than someone who has never taken that class or a systems class, and who has been out of school for a long time.
memories fade, i know i dont remember everything i learn.

starting from scratch, you look at a few things like damped natural freq. desired, suspension geometry, and shock dynos.

"I have to agree with Roger on this one and reply that you are still missing the Dynamics at work with the shock. You're spot on with the Statics, but the web page Roger cited explains that the geometry of the shock will also effect its overall damping. Its cycle is shorter AND slower so the effective damping is LESS."

again, read my posts and youll discover that theres only one equation that determines damper force. AGAIN it is C*V=F. thats the only "dynamic" thing. the velocity. so of course im accounting it. velocity along an inclined line is the velocity of the point of interest*sin(theta). so like ive been saying. i account that, lol.

statically a shock doesnt apply hardly any force. thats what a spring does.

dfoxengr

"The automotive type maintains a damping force which varies in direct proportion to the velocity of the piston, "

from roger's link.

exactly what ive been saying. linear.
F= C*V

so force doubles when velocity doubles, since C is constant.
just like ive been saying the whole time.


this will visualize some of the general eqns. ive been telling you.

http://en.wikipedia.org/wiki/Damping

4Crawler

iTrader: (3)

Exactly what I have been saying as well, F = C*V and if V changes because of the angle of the shock, then F changes as well, so an angled shock produces less damping force inside itself than the same shock that is mounted vertically.

tc

I think the big difference here is that dfoxengr says C = constant, and Roger says C is a function of velocity (xdot), so dfox comes up with xdot, Roger comes up with xdot^2.

Roger's argument makes more sense.

dfox: a little hint before you have to learn the hard way like I did. Listen more, talk less. Arguments like you're making here come off as a petulant little kid in business and WILL limit your advancement opportunities or get you fired. Start working now on how to tell people they're wrong without telling them they're wrong... :2cents:

4Crawler

iTrader: (3)

No, not saying that at all. I assumed (and stated so) that C is a constant. What I say changes is the velocity of comprssion of the shock when it is at an angle. At 45 degrees, given the same up-down velicity in the suspension, the shock will only compress at about 71% the speed that the suspension does. I posted a link earlier with shows this graphically and numerically, check it out, it shows exactly what is happening and no higher math is involved.

So what I am saying is that with C constant, if V changes so does F. And it is this lessened F that acts at the angle of the shock and the vertical component of that lessened force is what damps out the suspension.

Robinhood150

Does that mean somebody who has designed and built a high performance suspension and is currently a mechanical engineer and runs a vibrations laboratory for a major aerospace company is tons^2 more qualified than somebody who has not completed a vibrations class? BTW, there's not much vibrations needed to answer this thread topic. It's all statics and dynamics.

If you're arguing this to design a real world suspension for a private vehicle, then all of it is irrelevant. If you're arguing this for purely academic reasons, then go for it. Everyone here knows I love a good academic debate and have had plenty.

After reading this thread, I have to side with Roger (and yes I had an open mind about this). I'll continue to think about this because I know it will bother me. Post up what your profs say.

You might also want to pose your question to www.eng-tips.com.

dfoxengr

roger and i are completely on the same page, except that he says the force resultant is still applied at the 45 angle thus you must for a second time take the sin function of that force and that is the end-all force.

i say that the angle is already accounted for by the first sin function in F=C*Vsin(45)

we both agree upon the eqn just posted, its just after that he says one thing and i say another. no huge deal. dont think that i think he's not smart, because of course he is. but dont think im not as well, since you dont know me.

and please dont think things that are out of the scope of your(whomever) knowledge. if you dont have any background in this i would say, dont form an opinion.
and no, it is not static, because like i said, shocks dont do anything statically. its more of a system dynamics problem, which is not your basic dynamics class. basic dynamics does not cover dampers at all in my experience. youre introduced to that at the next level.

Robinhood150

All of these dynamic forces can be used as a static force at any instant in time. The only thing that is vibratory in this problem is the F=CV. At the instant in time T the only thing you care about is F. At that instant in time, it is a statics problem that can be solved with a FBD.

4Crawler

iTrader: (3)

Show me an automotive type shock that can apply force in a direction other than it's long axis and then what you saw might apply. All the automotive shocks I have generally have freely pivoting bushings at both ends so that they can only generate force along their long axis. If that long axis is at a 45 degree angle to the axis of motion of the suspension, then that force in the long axis of the shock must be split into vertical and horizontal components (relative to the suspension).

Note, I am simplifying the system at this point into a static system, replacing the shock absorber with it's equivalent force that is applied at an angle to the supension components. I am not saying that the shock applies any force when it is at rest, this is not the case unless you have a high pressure gas shock. Just makes this easier to analyze by first determining the forces involved in the frame of reference of the shock, then transform that frame of reference of the shock into the frame of reference of the suspension. And, at least on the suspension on my 4Runner, if I push horizontally on the frame where the shocks attach, it does no useful work. If I push up and down on the frame, that is the way it wants to move. So I find that only the vertical component of the shock's damping force does me any good.

dfoxengr

yeah i guess everyone else is lying to people.

quick question roger: if we were discussing a linear spring, how would your arguement for that go?

4Crawler

iTrader: (3)

Would be a similar type of effect. The spring itself would generate a force in line with it's long axis, force of a linear spring is a function of it's spring constant and it's displacement. Since the spring, at an angle, is compressed less for a given vertical displacement of the suspension, the force in the spring would be reduced by this reduction in compression. Then, since the spring's force is applied at an angle, only the vertical component of that force works to support the load on the suspension. In order to support a similar weight, the suspension would need to compress farther in order to compensate.

surf4runner

if you angle the shocks on the the wont fly

dfoxengr

can you then explain why both my book, and the site i showed say 70%, and there hasnt been shown any that say 50%?

i sent erich_870 a picture of the figure in my book, which he can post if he'd like to.

4Crawler

iTrader: (3)

It depends on how the problem is stated and what they are calculating.

If you are looking at a force applied at an angle, say 45 degrees, then 70.7% of that force it directed vertically. If that is what they are calculating in the picture, then that is correct.

If they are calculating how far a spring compresses at an angle or how fast a shock compresses at an angle, again, say 45 degrees, then yes it compresses 70.7% as far/fast as a spring/shock that is vertical. If this is what they are calculating, then that is correct as well.

Or maybe they are saying that the spring or shock at an angle is still being compressed as far/fast as it would be vertically, and thus would be providing the same force as it would be vertically, and that only the change in angle and the reduction in force transferred vertically is being accounted for.

If they are looking at the whole system, as I am, then you need to combine the two factors, since you have both a lesser force being applied at an angle. My definition of effectiveness is that for a given vertical deflection of the suspension, what will a given spring/shock do for me in terms of resisting that deflection. So my constants are that the vertical deflection distance/speed is the same in all cases and that the sping/shock constants stay the same. All depends on what they are calculating and depicting in the picture.

And the reason for keeping the shock or spring specs and the suspension travel the same is to follow the scientific method. That is change only one variable and observe the effects of that one variable change on the system as a whole. So, in my case the one variable subject to change is the angle of the shock or spring relative to the direction of motion of the suspension, but the same up-down motion (speed or distance), same shock or spring constant.

Erich_870

Erich

4Crawler

iTrader: (3)

Yes, as I thought, that figure is labelled "Vertical Load Factor", so all they are figuring is how much of the force in the shock is vertical and yes, that is totally correct in that sense. At 45 degrees, 70.7% of the force in the shock is directed vertically. Totally agree.

mastacox

Man, this thread's all over the place. Here's my take:

You have a two shock setups:

1) Vertical damper, damping coefficient of C

2) Damper angled at 45 degrees, also damping coefficient of C

(Just so you know, in this post, sub d denotes damper, sub a denotes vertical axle movement/force. So Fa is the vertical force on the axle, Vd is the velocity of the damper, etc.)

So, if you have an induced vertical (y-direction) velocity on the axle, Va, which is equal to the velocity the damper sees so Vd=Va, then the force the vertical shock will produce is Fd= C*V, easy enough. Also, since the shock is straight up and down, the force it produces is in the (-y) direction, so Fy=Fd. Also easy.

Now, you have a damper angled at 45 degrees. The speed of the axle is the same, but the OBSERVED AXIAL speed at the damper is less, Vd=Va*sin(45). So the force the damper produces is Fd=C*(Va*sin(45)), 70.7% less than the force produced by the vertical damper.

Now, the tricky part is that the force produced by the damper is AT 45 DEGREES. The damper only creates axial forces, so we must once again decompose it into it's components, one perpandicular to the ground, and one parallel to the ground. The vertical (perpandicular) component of the damper's force is the only one we're interested in, since it is what affects the axle's movement. So, Fa=Fd*sin(45), the vertical force on the axle is 70.7% of the total force put out by the damper, the rest of the force is absorbed by the axle itself.

Therefore, the total vertical force imparted on the axle by the damper is affected two-fold by the angle which it is at. So for the two cases:

1) Vertical Shock:

Vd = Va
Fa = Fd

Therefore: Fa = C*Va

2) 45 degree shock:

Vd = Va * sin(45)
Fa = Fd * sin(45)

Therefore: Fd = C*Vd ->
Fd = C * (Va * sin(45)) ->
Fa = (C * (Va * sin(45))) * sin(45) ->
Fa = C * Va * sin(45) * sin(45) ->
Fa = C * Va * sqrt(2)/2 * sqrt(2)/2 ->
Fa = (C * Va) * 1/2

So for the angled damper, Fa = (C*Va)/2, or 50% as effective as the vertical damper.

Triumph! Flame away, but I'm pretty sure I'm right (Mechanical engineer here by the way)!

4Crawler

iTrader: (3)

Bingo, we have a winner!
Very well stated.