## Spring Rate Theory

### Motion Ratio and Lever Arms

The springs do not act directly at the wheel, they act through the lower control arm. Since the wheel is essentially pressing on a lever tied to the spring, the wheel gets an advantage. The spring is also not generally mounted exactly at the vertical. So we have to correct for that. The force on the spring is the spring rate multiplied by the amount the lower spring perch rises vertically multiplied by the cosine of the spring angle from vertical. $F_{Spring}=k_{Spring}y_{Spring}\cos \alpha$

The wheel force is the effective spring rate at the wheel multiplied by the vertical movement of the wheel. $F_{Wheel}=k_{Wheel}y_{Wheel}$

Assuming small vertical motions, geometric similarity gives us: $\frac{y_{Wheel}}{y_{Spring}}=\frac{d_2}{d_1}$

Balancing the moments on the control arm: $F_{Wheel} d_2=F_{Spring} d_1$

Solving for $F_{Spring}$ and substituting into the first equation: $F_{Wheel}\frac{d_2}{d_1}=k_{Spring}y_{Spring}\cos \alpha$

Using equation 3 $F_{Wheel}\frac{d_2}{d_1}=k_{Spring}y_{Wheel}\frac{d_1}{d_2}\cos \alpha$

Then equation 2 $k_{Wheel}=k_{Spring}\left(\frac{d_1}{d_2}\right)^2\cos \alpha$

gives us the wheel rate in terms of the suspension geometry and the spring rate. $\left(\frac{d_1}{d_2}\right)^2$ is called the “motion ratio”.

But we really can’t easily measure wheel rate directly. But we can measure the sprung weight and the amount of droop we have before the spring takes its share of the sprung weight. So, $k_{Wheel}=\frac{F_{Wheel}}{y_{Wheel}}$

where $F_{Wheel}$ is the Sprung Weight and $y_{Wheel}$ is the droop. Note that this assumes that at full droop the spring is fully extended, and has no preload.

### Suspension Frequency Calculation

Another common measure of a suspension is its natural frequency. If a suspensions natural frequency is too high, the ride is perceived as being harsh, if is too low motion sickness can ensue. Typical road cars aim for suspension frequencies between 1 and 2 Hz. These frequencies are difficult to achieve with cars with limited suspension travel.

The suspension is similar to a damped harmonic oscillator. With no damping the natural frequency is $\omega=\sqrt{\frac{k_{Wheel}}{m_{Car}}}=\sqrt{\frac{m_{Car}g/y_{Wheel}}{m_{Car}}}=\sqrt{\frac{g}{y_{Wheel}}}$

where $k_{Wheel}$ is in Newtons per meter and $m$ is in kg. $\omega$ is in radians per second. This equation shows an interesting effect. Remember the the wheel rate, $k_{Wheel}$ , is a function of the weight (mass) of the car and the suspension travel. The mass of the car in the denominator cancels out the mass in the numerator showing that suspension frequency is solely a function of suspension travel. This is the same as saying that a pendulum’s natural frequency is only dependent on its length, and not on its mass.

### For further details see:

Competition Car Suspension Design Construction and Tuning, Allan Staniforth, 1999
Good coverage of suspension tuning, not as much design information as I would like, but this deserves a place on the bookshelf of anyone who wants to build their own car.

### 9 Responses to Spring Rate Theory

1. Braxton Seraphin says:

How would mass related to the spring constant… that is the mass of the spring?

2. Braxton Seraphin says:

Is that spring supporting an axial or a bending load.

3. enderw88 says:

Good questions, I didn’t explicitly state my assumptions.

First, the model ignores the unsprung mass by assuming is it substantially smaller than the sprung mass. If you take it into account, the calculations get substantially more complicated but the correction is at most second order. The mass of the spring itself is another order of magnitude lower than the unsprung mass.

If you get to where the unsprung mass is more than a few percent of the sprung mass then the wheel frequency calculation here will be slightly off. That is serious race car territory.

I don’t quite understand your question regarding what type of load the spring is supporting. There will be a bending moment about the lower spring joint, but the load through the spring is purely axial. The load IN the spring is torsional…

4. Vincent says:

I did not understand why you used y_s*cos(a)
I thought that y_s is a oposite cathetus on rectagle…

5. enderw88 says:

Y_s is vertical, cos alpha rotates finds the component in the direction of the spring.

6. Vincent says:

hmm… sorry,I cant see that!!!!

Another question: where I can find material about that (books, sites)?

Thx

7. enderw88 says:

Draw it out on paper, the lower point of the coilover moves vertically (actually on an arc, but use small angle approximation). If the lower point moves directly up by Y_s, how does the length of the spring change?

Check the list of car design and tuning books I have (https://enderw88.wordpress.com/automotive-theory/car-design-and-tuning-books/). I have never seen another detailed derivation like this, that’s why I posted it.

8. Paul Muenzinger (Stalker #150) says:

Craig, the answer disagrees with an ME lecture (Hathaway) that I found which says the cosine is squared:

Hathaway goes on to say that the wheel rate can be determined experimentally by measuring spring movement vs wheel movement, taking the square, and multiplying by the spring rate, which makes sense. But this would not work if the cosine was not squared.

So I tried to work it out but my math is pretty rusty; I got stuck on the set up of the first equation for Fspring.

In order for the derivation to work, I believe that both Fspring and Yspring have to be the vertical components of the spring force and the spring bracket movement, or the components that are normal to the suspension arm.

I tried renaming the quantities to distinguish the vertical components of spring displacement and force, and the axial components:

Fs – the spring force acting in line with the inclined spring axis

Fsy – the component of spring force normal to the suspension arm at the coilover bracket

X – spring compression measured along the spring axis

Yspring – the vertical displacement of the spring bracket on the suspension arm

Then

Fx = Kspring X

Fsy = Fs cos (a) = Kspring X cos (a) ,

X = Yspring / cos (a)

which gives

Fsy = Kspring [ Yspring / cos (a) ] cos (a)

The cosine drops out! I must be making a high school math error but I don’t see it.

9. Paul Muenzinger (Stalker #150) says:

OK, I found my dumb mistake; please disregard my first attempt to set up derivation.

Define Yspring as vertical but Fspring acting at angle (alpha).

The change in spring length at angle (alpha) in terms of the change in vertical spring bracket position is:

Yspring cos (alpha)

You correctly use this angle correction factor in first equation of your derivation.

But vertical component of the change in spring force on the suspension arm is:

Fspring cos (alpha)

So balancing the moments in equation 4 should include the angle correction factor again:

Fwheel d2 = Fspring d1 cos (alpha)

Then proceeding with the rest of the derivation as before yields:

Kwheel = Kspring (d1/d2)^2 cos ^2 (alpha)

So the cosine should be squared.