Spring Rate Theory « Mechanical Daydream

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.

Published on 20070218 at 0938 Comments (3)

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3 Comments Leave a comment.

On 20090401 at 1801 Braxton Seraphin Said:

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

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On 20090401 at 1804 Braxton Seraphin Said:

Is that spring supporting an axial or a bending load.

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On 20090401 at 2004 enderw88 Said:

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…

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