RedRevolver

Hopefully this one is correct. If it is correct, here's to hoping for a sticky.


If there are any errors, let me know. Otherwise, this is to educate people on the calculation of spring frequencies, and become a useful tool for people in the future.



Importance of spring frequncies:

Spring Frequencies are an important number to know for the suspension on any car. When trying to figure out proper spring rates for your race car, each car has differences in their suspension geometry, car weight, etc, making spring rates alone impossible to compare across different vehicles. slight variations in suspension geometry may create the need for a higher spring rate on one car than another, even though the cars may be very similar in every other aspect. Spring frequencies allow you to compare the two cars setups without having to worry about these things. If you know the frequency of your ford taurus and want to make it as stiff of a ride as a corvette, all you need are the two spring frequencies, and then you can work backwards from there to figure out what spring rate you would need on your taurus to make it as stiff as a corvette. While this is quite a far fetched example, the reasoning still stands that it is a great way to compare setups across many different type of race vehicles and if nothing else, is very useful information if you wish to change anything on your suspension setup.


Calculating Spring Frequencies:


Calculating spring frequencies requires a few different measurements and calculations. The first thing your going to need is the sprung weight of the corners of your car. This means to take the weight of the car minus all of the pieces that are being held up by the springs. Mostly this is the weight of the wheels and tires and brake components. There is other stuff like the control arms to take into account, but for the sake of simplicity we'll just take into account these two.

The stock weight of my Honda Del Sol is 2,270 pounds. Each wheel and tire of mine weighs 34 pounds so, 136 pounds for those, and the brake assemblies are about an estimated weight of say 15 pounds on each corner, so another 60 pounds off of the weight, putting my approximate sprung weight at 2,074 pounds. If you don't have a corner weight scales to measure each corner (like me) you can just divide that by four to get the approximated sprung weight on a corner.

While this does add a fair amount of inaccuracy, it is the method that will be the most suitable for most people, and it will still get you relativly close. This method puts one corner of my car at 518.5 pounds. If you can find F/R weight distributions on your car, use those, minus the unsprung weight of two corners and divide it by two.

Next up is the motion ratio of the spring. The leverage of the suspension arm acting on the spring has to be taken into account when doing the wheel rate calculations.

There are two distances that you need for this, and they are pretty easy to measure. The first is the distance between the spring centerline to the lower control arms inner pivot point (d1). The second is the distance from that same pivot point, out to the outer ball joint (d2).


The diagram below is for an A-Arm/Double wishbone suspension and can be applied to a MacPhearson strut as well.




Note: Diagram is take from Eibachs website and modified, all credit for the diagram goes to them.

Now you want to take your measurements and divide them. You want to take the larger number (d2 & d4) and divide it by the smaller (d1 & d3). For example, on my car, d1=10" and d2=14", so my motion ratio is 1.4.

Formula: d2/d1

This is all so that you end up with the proper ratio for calculating the next thing we need, the wheel rate.

The wheel rate figure is the figure that tells you how much of the spring rate is actually acting at the wheel. For example, if your spring rate is 200 lb/in, then to compress the spring one inch, it takes 200 lbs of force. But, if you calculate the wheel rate and it comes out to be something like 160 lb/in, then your wheel will be moving one inch for every 160 lbs of force applied to it, thus giving you a much more useful number for calculating the acutal movement of the vehicle under loads.

To calculate the wheel rate though, we also need to take into account the angle correction factor, to account for the angle that the spring is mounted at. This is a pretty easy thing to figure out, just take the angle of the spring from vertical (A1, either measurement works they are congruent), and then take the angle of the lower control arm from horizontal (A2). Then subtract (A1-A2) and take the Cosine of it, making sure your calculator is in the degrees setting (as opposed to radians). A diagram of how to measure the angles, A1 and A2 is shown below



Note: Diagram is take from Eibachs website and modified, all credit for the diagram goes to them.

Since my front springs are mounted at a 17 degree angle and my LCA is at a 5 degree angle, Cos(17-5), my angle correction factor (ACF) comes out to be .978.

Formula: Cos(A1-A2)

This will be used in our wheel rate calculation now.

The wheel rate calculation, now that you have all of this other stuff, becomes quite easy. All you need is your spring rate (SR) divided by the motion ratio (MR) squared multiplied by the angle correction factor (ACF). The reason the MR is squared is because when calculating the wheel rate, it acts as a lever, multiplying the force applied to the spring.

Thus your formula becomes:

SR
--------------------
MR^2 x ACF

Note: Above formula used directly from eibach with no modifications necessary

Broken down, it is:

SR
-------------------
(d2/d1)^2 x (Cos(A1-A2))

So, based on this, my car, with a front spring rate of 182 lbs/in, and distances 10" and 14", with an angle correction of 12 degrees, the formula becomes:

182
---------------------
(14/10)^2 x (Cos(12))



182
= ----------------
(1.4)^2 x .978



182
= -------------------
1.96 x .978




182
= ------------------
1.917




= 94.95 lbs/in Wheel Rate

Finally, we get to total up all this information into the final calculation for the spring frequency. The spring frequency (SF) is the square root of the wheel rate that we just figured out, divided by the sprung weight on that corner, all multiplied by 187.8.

Formula: SF = 187.8 x (√(WR/Sprung Weight))

Note: Above formula used directly from eibach with no modifications necessary.


So, based upon my car, this would end up being:

187.8 x √ (94.95/518.5) = 187.8 x √ .183 = 187.8 x .42793 =

80.36 Cycles per minute (CPM) suspension frequency. the problem with this is that most people tend to use spring frequencies in Hz not CPM. Luckily for us, this an easy conversion, just divide the CPM by 60. Therefore:

80.36 CPM / 60 = 1.34 Hz

The other option, to get the frequency in Hz, is instead of using 187.8, use 3.133, which is just 187.8/60.

This would be the suspension frequency for the front of my car. For the rear, all I have to do is go back and change the values to the correct ones for my rear suspension.

So, to review, calculating the suspension frequency requires: Corner Sprung weight, spring rate, motion ratio, and angle correction factor.

And now, all the formulas again:

MR = d2/d1
ACF = Cos(A1-A2)
WR = SR/(MR2xACF)
SF = 187.8 x √(WR/Sprung Weight)

xd69

iTrader: (1)

If it wasn't for this example i would still be confused as to the benefit of doing all this calculations.


But now that i know what the purpose of all this is, Good Write Up!!

beanbag

ACF formula is still wrong. What happens if A1=A2?
Why is the MR squared, and the ACF is not?

RedRevolver

Wow I fail. I only thought of a1=a2 at 45 degrees, not any other angle. Though if it was something like a1+a2, which i just threw out as a thought, it would bump my frequency up, and it already seems pretty high as it is.... thats the only thing I can think of, and it sort of makes sense, if both were mounted at 30 degrees then a 60 degree ACF. it'd be just like if the control arm was mounted at 0 and the spring at 60. It makes sense to me, though I'll wait for your input before I change it.

Also, he ACF shouldn't need to be squared, it is just accounting for the angle, as far as I know there wouldn't be a lever arm effect on an angle.

beanbag

Well since this is your homework, I won't derive the answer for you, but I will tell you how you should find it. First, you need to start with the very basic definition of wheel rate, which is defined as change in force at wheel divided by upwards wheel deflection, i.e. the deflection component perpendicular to the ground. So now using the powers of geometry, figure out for a given upwards deflection of the wheel, what is the change in angle of the control arm and hence compression of the shock/spring. This gives a change in force at the spring, now go and lever that back down to the wheel.

PS: Why is the ACF not part of the lever?

PatrickGSR94

iTrader: (1)

Shouldn't the motion ratio on a Mac strut setup be 1, since the hub attaches directly to the bottom of the strut assembly?

RedRevolver

Well it's not my homework anymore, I had to make a "final" copy and present it already. Thats the beauty of school sometimes. You don't have to be right, just have to make it look good and acknowledge the fact that you could be wrong =P Though thats not gonna stop me from from still going and trying to make this thing right.

Anyways back to this. My thinking on the ACF not being part of the lever is that It is just the angle acting at that point. The motion ratio is a force being applied over a distance before acting upon the spring, but the ACF is just a correction at that point.

I'm at work right now so I can't work through the math at the moment, but once I get the force at the spring, how do I work that through to find out how much the ACF is affecting? It'd probably make more sense if I had a calculator right now lol but I'm still kind of confused on the method.

Also, patrick, technically yes the motion ratio on a mac strut like the first pic below would be 1 since it is mounted directly in line with the ball joint. My del sol has a double wishbone like an accord shown in the 2nd picture below.

beanbag

no, due to acf, and definition of wheel rate, which only counts upwards deflection of wheel perpendicular to ground

RedRevolver

yeah sorry i forgot to mention the ACF on the mac strut matters as well.

sackdz

So what did you get on your project?

A few critiques from me:

-Practically speaking, the example goes backwards. Normally you would be taking a target frequency and determining what spring rate is needed (or pickup locations). Although this can still be a useful calculation if you want to know how much your freq changes with a given spring.

-You titled the thread 'Spring frequency Calculator' but really it's the suspension frequency you are determining.

-You made mention of the specific corner weight distribution, but then just divided your total sprung weight by 4. I realize it's an example, but it really is critical to use an accurate weight. At least throw a 60/40 f/r split in there...

-MacPherson m/r is not nessesarily 1.. some designs it can be higher or lower by a small amount. generally speaking for estimation its fine, but if you are developing formulas for applicatioin wouldn't leave it out.