Thread: load = ride comfort = what spring rate?

Originally Posted by Bush65

IMHO the ride you felt with the heavier load can be replicated by changing the spring rate to achieve the same natural frequency.

Natural frequency is proportional to square root (spring rate / weight)

From this you can determine the required spring rate knowing the rate of your current springs and the weight on the springs when you were loaded and normal.

i.e. sqrt(K1/W1) = sqrt(K2/W2) where:
K1 is your current spring rate
K2 is the rate that you want for better ride with normal load
W1 is the weight on the spring when you were loaded
W2 is your normal load

The natural frequency (normally measured in Hz which is cycles per second) for rear springs should be a little less than that for the front springs. This is because the front wheels hit the bump before the rears but you want the motion caused by each to finish at close to the same time - obviously different speed and wheel base affect this.

Now your comfortable ride was achived with rear spring rate of 250 lb/in and weight on rear wheels is 940 kg plus 150 kg of steel (assume the 250 kg is carried entirely by the rear springs) then:

Neglecting unsprung weight (rear axle and wheels, etc)
sqrt(250/940+150) = sqrt(216/940) therefore use 216 lb/in spring rate for comfort with no load.

Now assume unsprung weight is 200 kg
sqrt(250/(940-200+150)) = sqrt(208/(940-200)) therefore use 208 lb/in spring rate for comfort with no load.

Now assume unsprung weight is 250 kg
sqrt(250/(940-250+150)) = sqrt(205/(940-250)) therefore use 205 lb/in spring rate for comfort with no load.

Recommended natural frequencies are:
For rock crawling: 0.75 Hz front, 0.93 Hz rear
For off road tracks up to 40 or 50 kph: 1.1 Hz front, 1.375 Hz rear
For general on and off road driving: 1.35 Hz front, 1.688 Hz rear

Now for spring rate K in lb/in and weight on springs W in lb, natural frequency f in Hz is:
f = 3.1269 sqrt(K/W)

As a very rough guess assume unsprung weight is 250 kg front and rear.

For weight on front wheels of 1300 kg, then load on each front spring is:
(1300 kg – 250 kg) / 2 = 525 kg (1157 lb)

For weight on rear wheels of 940 kg, then load on each rear spring is:
(940 kg – 250 kg) / 2 = 345 kg (761 lb)

For front spring rate 210 lb/in, rear spring rate 250 lb/in and comfort rear spring rate 200 lb/in

Front natural frequency = 3.1269 sqrt(210 / 1157) = 1.332 Hz

Rear natural frequency = 3.1269 sqrt(250 / 761) = 1.792 Hz (with 250 lb/in spring rate)

Rear natural frequency = 3.1269 sqrt(200 / 761) = 1.603 Hz (with 180 lb/in spring rate)

Required free-length for a new spring with different spring rate is:
desired static ride height + static deflection
where static deflection = load (lb) / spring rate (lb/in) i.e. 761 lb / 200 lb/in = 3.8 inch

(edit) sorry I stuffed up by confusing the numbers for your load of steel I used 250 kg should have been 150 kg. So spring rate will be 205 lb/in if the unsprung rate is assumed to be 250 kg. (end edit)

In my previous posts I focused on natural frequency. IMHO it is the most appropriate measure for selecting spring rate or comparing spring rates for different road surface and load conditions.

Natural Frequency is the term used in engineering and science for the un-damped frequency of vibration of a mass supported by a spring, etc. For vehicles, suspension frequency is commonly used in place of natural frequency.

If you bounce a corner of a car without shockies up and down then release it, you could measure the frequency (number of cycles per second (Hz) or minute), or the period (time for one complete cycle) for the spring.

The table below summarises recommended suspension frequencies for a few different applications.

The spring rate required for a particular load and frequency is:
K = L x (SF / 3.1269)^2
K is spring rate in lb/in
L is load on spring in lb
SF is suspension frequency in Hz (cycles per sec)
3.1269 is a constant to take care of the units given

This can be re-arranged as:
SF = 3.1269 x sqrt(K / L)

The period is: T (sec) = 1 / SF (Hz)

I have been told that you dont want the spring frequency to be the same as the tyre frequency....maybe its not an issue due to the fact they would be so different in our vehicles.

the million dollar question though, is my butt tuned to the same frequency as yours John

you have again provided great material for me to play with and make a decision on. Then its a matter of suck it and see (what will the compromise be???)

Originally Posted by Bush65

In my previous posts I focused on natural frequency. IMHO it is the most appropriate measure for selecting spring rate or comparing spring rates for different road surface and load conditions.

Natural Frequency is the term used in engineering and science for the un-damped frequency of vibration of a mass supported by a spring, etc. For vehicles, suspension frequency is commonly used in place of natural frequency.


[snip]

Natural frequency is a starting point but with something as dynamic as a car, and as there are so many variables involved nothing beats just trying different spring rates to see what feels the best, and often what feels nice in ride is too compromised in rate for actual use in load carrying/pitch/squat/roll/road holding.

This is why car companies spend huge $ and sometimes years on test drivers pounding around test tracks and differing road surfaces/environments around the world, coupled with LVDT's on the suspension wired back to a data logger to come up with the best compromise, and why race teams go to race tracks with a pile of springs with varying rates, and sometimes can change them by several hundred lb's to get the ride rate suitable.

And yes, as mentioned you don't want the springs natural frequency to match the tyre's natural frequency.
It's usually hard to do but can happen, speaking from experience
With underdamped dampers you will bounce clean off the road.

Trying to find out the tyres natural frequency from a manufacturer is all but impossible and in practical terms something we wouldn't normally have to worry about.