Force in the springs after a jump 2

[RegisAlstom]

Sorry for the newbie question again.

How to calculate the force that is actuating in the springs after a jump?

Lets supose that I have a car with 2700lbs (Unsprung) + 800lbs sprung. so the total weight of the car is 3500lbs.

Lets supose that I have 650 lbs per corner unsprung in the front, and 700lbs per corner in the rear. The sprung weight is 200lbs per corner.

Now lets supose that this car is travelling at 65mph, and it jumps 40" height.

The question is what is the force applied to the spring when the car hits the ground?

I was suposing to be F=mgh, but somebody said that it is completely diferent calculation.

Can you give me a light about that?

My intention is to have a clue about the spring rate I need to deal with during a rally, and how this spring rate will affect the ride. In other words, how stiff should be my spring and the bumpstop.

Thanks

Regis

 

[patprimmer]

You are presuming that it lands evenly on all 4 wheels. I think you need to allw for the fact that one end will take a lot more load than the other at times, and that there will be a lesser, but still real eveven force from side to side, ie one wheel might take 60% or so of the load at times.

Regards
pat

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.

 

[patprimmer]

I should have said, I am not a suspension guy, and the above answer is from intuition, and NOT based on data

Regards
pat

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.

 

[RegisAlstom]

I agree with that, but I am intending to assume that it will land evenly on all 4 wheels just to have a starting point

Is the F=mgh valid for that?

Thanks

Regis

 

[IRstuff]

No, mgh is the potential energy gained in the jump.

You'll need the time constant of the suspension+tire spring system to get the average force.

You'll need to get the step-function response of the suspension+tires to get the instantaneous force during the transient.

TTFN

 

[RegisAlstom]

Sory for my lack of understanding, but what is the time constant for the suspension + tire spring system?

Lets supose the tire+spring constant is 300 lb/in.

Can you explain better what is the step-function response of the suspension + tires that will give the instantaneous force during the transient?

Sorry again for my lack of knowledge, and thanks for the information

Regis

 

[stingc]

The characteristic angular frequency of a spring-mass system is sqrt(k/m), where k is the spring constant and m is the (sprung) mass. The quick and dirty answer to your question is therefore:

Favg=(momentum change)/(time change)=dp/dt

dt=2pi*sqrt(m/k)=0.3 s

dp=M*v=M*sqrt(2gh)=7000 kg m/s

Favg=20,000 N=6000 lb

(all springs added together)

This answer is *extremely* rough...

 

[bhart]

Through conservation of energy I came up w/ this, which you will see is not very realistic at first glance, but the truth is the problem is much more complicated than a simple equation will describe.

T_1+V_1=T_2+V_2 where T_1 = kinetic energy at max height and T_2 = kinetic energy @ min height (fully compressed). Since the vertical velocity at both states 1 and 2 are assumed to be zero then the T_1 = T_2 = 0.5*m*V^2, so the kinetic energy terms drop out of the energy equation. That leaves V_1 = V_2 or total potential energy at the highest point the vehicle reaches equals the total potential energy at the lowest point the vehicle reaches.
T_1 = W*h_max where W=mg=weight, and h_max is the height of the bottom of the chassis when the car is at the top of its arc.
T_2 = W*h_min + 0.5*k_total*delta_x^2, where h_min is the height of the bottom of the chassis when the suspension has compressed the amount required to stop the chassis from moving toward the ground, k_total is the total ride rate (including suspension springs and tires), and delta_x^2 is the amount that the vehicle travels from when the tires just touch the ground to when the vehicle stops moving downward (the amount that the k_total spring compresses).
Solving for k_total,
k_total = W*(h_max-h_min)/(0.5*delta_x^2)
Now you have to know something else about h_min in order to solve this equation, so consider h_min to be the difference between the free length of the total ride spring minus delta_x or h_min = L_0 - delta_x (I said consider it to be this way, but this is actually how it is).
so k_total = W*(h_max-L_0+delta_x)/(0.5*delta_x^2).
Delta_x essentially represents the full suspension travel if you assume that the suspension springs begin to compress as soon as the tires re-contact the ground, so you or I can pick the travel that we know our suspension is designed for, and solve for the spring rate that would be required to use all that travel.
A few more assumptions need to be made before you can use this equation for k_total, mainly that all four spring rates are equal and unsprung weight (wheels, tires, hubs, brakes, etc) is negligable compared to sprung weight. You have given what seems to be an unusually high unsprung/sprung weight ratio, but if you are sure that is how it is (monster trucks), then stop reading now b/c this equation is junk (which it might be anyway b/c of all of the assumptions required).
So w/ your numbers, and assuming that the entire suspension travel of 15 inches is used, you get
k_total = 2700lb * (40in-15in+15in)/(0.5*(15in)^2)
k_total = 960 lb/in.. since this is the total ride spring, you divide it by 4 to get the ride spring @ each corner, k_ride_@_corner = 960/4 = 240 lb/in. k_ride_@_corner = the equivalent spring between the tire and the wheel rate, if you assume a tire rate of around 1000 lb/in (total assumption), then the wheel rate of your suspension should be around
k_wheel_rate = 1/(1/240 - 1/1000) = 316 lb/in. Now divide by your suspension motion ratio squared to get the spring rate that is required.
There is the solution to your problem if you ignore the shocks, bump rubbers, asymmetric CG location, asymmetric spring rates, multi-stage springs, and of course assume unsprung weight is negligable (sounds like a bad assumption in your case). At least its a start.

 

[bhart]

Sorry but the equations that I wrote above as:
T_1 = W*h_max
T_2 = W*h_min + 0.5*k_total*delta_x^2
should be
V_1 = W*h_max
V_2 = W*h_min + 0.5*k_total*delta_x^2
I had already stated that T_1 and T_2 (kinetic energy) drop out of the energy equaiton, leaving only the potential energy terms, V_1 and V_2, which must be equal

 

[desertfox]

Hi RegisAlstom

Well I thought I would throw my hat into the ring and probably get my head blown off but here goes:-

Assume the car leaves the ground 40"=1.016m

Assume car hits the ground like a solid block with no bounce

therefore :- Impulse Force =m*u

where m= mass of car = 1590.9kg (3500lbs)
u= velocity of car on impact

using energy analysis = gh=0.5*U^2

u= 4.4647m/s

impulse force = 1590.9kg*4.4647m/s=7102.95N/s

This a very crude analysis and does not address other things people have posted but with the information you have given its about the best I can do.


regards desertfox

 

[IRstuff]

The math is correct up to the last unit conversion, it should be N*s, since N=kg*m/s^2 and m*v is only kg*m/s

Additionally, the terminology is a bit suspect, since m*v is momentum and impulse force must be integrated over time to get momentum change, hence the resultant unit conversion of N*s.

As usual, this the point at which this type of analysis dies, since there is no way to determine the duration of the impulse without knowledge of the spring time constant.

TTFN

 

[desertfox]

Thanks IRstuff

Your right it should have been Ns not N/s thankyou for pointing it out.

Regarding the terminology I was working from a reference
in a book "Mechanics for advanced level" by Bostock & Chandler which states:- Where a large force acts for a very short time so that neither force nor time for which it acts
can easily be evaluated seperately, eg a bat hitting a ball etc:- the change in momentum caused by the impulse however can be used to evaluate the impulse.It then goes on to give lots of examples like a stone hitting a ground etc without bounce. So thats the reason the momentum equation is present.

thanks again
regards desertfox

 

[IRstuff]

Agreed... but then again, it's not different than using the potential energy expressed as N*m since energy is force integrated over distance.

The terminology is different than that used in the rocket industry; units of N*s is the momentum change and the actual force applied over a short time is the impulse force.

My reading of your citation is that your last expression should have been delta_momentum, which is equal to impulse_force*delta_time. Unless, the text is defining impulse force as force*delta_time, which smacks of misleading terminology.

TTFN

 

[GregLocock]

None the less, in the context of the original question, the maximum force is fairly easy to calculate if you know the velocity as each axle strikes the ground, and the rate of the springs between the axle and the body.


Cheers

Greg Locock

 

[desertfox]

Hi IRstuff

My understanding of impulsive force F is:-

impulsive force F= delta momentum
--------------
time taken


impulse of the force = F*delta time = delta momentum

so in my above post I should have said Impulse of the force
equals change in momentum.
Thanks IRstuff, that hopefully clears that up now.

regards desertfox

 

[IRstuff]

works for me.

TTFN