Race car suspension Class

Ranger Mike

Many private messages on this weight question-
Let us look at weight (mass) and how it effect a race car-

UNSPRUNG Weight -Everyone parrots “ Unstrung weight is bad”..but why? Because we can not control it. This is why we go with light weight wheels, tires, turn shock absorbers upside down. Sprung weight must be placed where we can have maximum control of it. Ballast is the biggest problem so we minimize its height and mount it at the polar moment to minimize its effect during cornering.

Sprung weight – Weight we can “ control” as far as that thinking goes. During this discussion we focus on Ballast or weight we need to meat the governing rules of the race track for minimum weight and percent left side weight and percent rear weight. So where do we place this ballast weight?

Simple moment of inertia is used to estimate resistance to rotation.. It is analog to inertial mass, but for rotation instead of linear displacement. Moment of inertia: Kg*m2

The area moment of inertia or second moment of inertia, is used to estimate resistance to bending. Think of a beam when you want to know how much will it distort under load. It’s a measure of the resistance of a section of a solid to perpendicular loads. Area moment of inertia: m4

Polar Moment of inertia is a quantity you use to estimate resistance to angular torsion. It’s a measure of how much resistance to twisting around an axis has a section.
Polar moment of inertia: m4

Polar Moment is the center of all forces in a race car. This is the Point about which the car pivots during weight transfer ( cornering). This point will move the least amount during this action. Indy cars have low polar moment of inertia. Sportsman Saturday Night special have HIGH Polar Moment of Inertia. Look at the attached illustration. Indy car has Low Polar Moment of Inertia as all the Mass is concentrated as close to the CG as possible. The door slammer has a V 8 mounted up high ( usually with minimum height requirement), has a big old battery mounted up high and a fuel cell mounted past the axel ( outside the wheel base). High Polar Moment does not have to do with height. High means it takes a lot of force to change the DIRECTION of Mass. Think of a 50 pound fly wheel vs. a 10 pound flywheel. Once the Mass rotates ( as going into a corner) it wants to keep rotating. Think of a bowling ball and a volleyball . The volleyball is easy to get rolling and easy to change directions..not so with the bowling ball.

So why is it important to know about the Polar Moment? We want to build the car with the desired percent of left side weight, front to rear weight. Only when we achieve this do we want to add weight to meet the minimum weight requirement when we scale. We want to add this Mass or weight at the Polar Moment as this is the point where it will move “ the least” when cornering.

Think of the old teeter totter at the local school. If two people of equal weight sit on it at the same position from the middle pivot, no teetering or movement. The center pivot is the polar moment. Now bounce up and down and the teeter totter moves but notice the center pivot has the “ least movement” relative to the rest of the board.

So if I go get the neighbors fat kid and put him on the right side of the teeter totter opposite me, the teeter totter will tilt in my favor. I have to scoot toward the center pivot point to “ equal up” the weight distribution. In fact, an observer looking at this mess from the side would see my weight was shifted 10% bias toward the center pivot point. If we were to measure the distance between the seat positions ( think track width) and measure the center pivot point from my seat position we would see a 60% left side weight bias.
If we then try the same drill but at 90 degrees ( think wheel base ) and the center pivot point – we have front to rear weight bias.

All polar moment is the intersection of these two points AT CENTER OF GRAVITY HEIGHT.

By the way, most door slammers run 18 inch CG Height and super late model cars with dry sump oil pans and aluminum heads get as low as 16” CG Height. There is a method you can use to measure exact CG on your car with wheel scales and a floor jack,

You can run up to 70% left side weight on asphalt cars but 55% is max on dirt, and 52% rear is good for asphalt and 55% max rear wt. on dirt..
see good article on this at

http://content.yudu.com/Library/A1vg...sources/33.htm

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Ranger Mike

Ok I been getting a lot of PM about finding Center of Gravity.

If you do not have access to a set of wheel scales you can rough estimate the CG by measuring the distance form the pavement to the center line of the cam shaft on the typical V8 door slammer or V6 for that matter..even OHC 4 cylinder will get you close.

Use your local Landmark or Farmer s Coop grain elevator scale. Measure your wheel base. Mark the mid point of the wheel base on your car with a piece of masking tape. Drive on the scale so the front tires are on the scale and the tape is aligned over the entrance edge of the scale. Record the weight. Drive off the scales to the point the masking tape is over the exit edge of the scales and only the rear tires are on the scales. Take a reading. You now have the Front vs. Rear weight. Simply divide the front weight by the total to get % front weight. Then multiply the wheel base by this percentage.
Say we have 3000 pound door slammer with 104” wheel base. we scale it as noted above and get 1590 front weight, 1410 rear weight. 1590 / 3000 = .53 or 53%.
104 inches x .53 = 55.12 inch so the COG is located 55 and 1/8 inch forward of the REAR axle.

This will give you COG in one axis or 1 dimension...front to rear.



If you have wheel scales you will be able to measure this a lot more accurately but the same math is involved.
With the wheel scales you can calculate the COG in two dimensions. We get Front to Rear as noted above and we can calculate left to right % by substitution Track width for wheel base.
Same 3000 # car that has 60 inch track width. Our wheel scales show 55% left side weight. This means COG is .55 x 60 = 33 inch to the left or 3 inch toward the left side from the center point of the track width and 55 1/8 inch forward of the rear axel. Better but his does not give us the true COG which is a 3 dimensional point.

To measure the COG most accurately, we need to prepare the car as race ready..tires properly inflated, full fuel load, driver ( or substitute weight of driver PROPERLY DISTRIBUTED). Don’t just throw in a few sacks corn that total the drivers weight, you need to replicate the weight of torso , helmet and legs as close as possible.
Note: You must replace the shock absorbers ( dampers) with solid links to replicate the race ready ride height. These solid bars will permit the car to be raised without collapsing as the shock would do under load.

the following is from Longacre who make a fine series of wheel scales. http://www.longacreracing.com/articles/art.asp?ARTID=22

To find the 3-D COG height we need to use a little trig. Specifically, we are using the Law of Tangents, and the Pythagorean Theorem. We use the wheelbase in place of the Hypotenuse and we will use 10 inches for the short leg of the right triangle since we intend to raise the car 10 inches.
a2 + b2 = c2



Center of Gravity Height Formula

COH = WB x FWc
TW x Tan q

Center of Gravity Height Formula

Definition of Variables

CGH - Center of Gravity Height
WB - Wheelbase (inches)
TW - Total weight
FW1 - Front weight LEVEL
FW2 - Front weight RAISED
FWc - FW2 - FW1 (change in weights)
HT - Height raised (inches)
Adj - Adjacent side (see below)
Tan q - Tangent of angle (see below)
CLF - Left Front tire circumference
CRF - Right Front tire circumference
C - (CLF + CRF) / 2 (average circumference)
r - Axle Height


The center of gravity height is found using the rules of trigonometry and right triangles. Specifically, we are using the Law of Tangents, and the Pythagorean Theorem. The following diagrams are greatly exaggerated for illustration purposes.
Tan q = opposite / adjacent

Tan q = HT / Adj


Pythagorean Theorem


So, in our exercise, when we raise the car 10" we are creating a right triangle with the following properties:
Hypotenuse = Wheelbase = c
Opposite = Height = b
Adjacent = a
C = 2 p r ( r is axle height of 10 inches)
Therefore using the Pythagorean Theorem:

Adj = square root of (WB2-HT2)


Once we know the value of the adjacent side of our triangle we solve for the tangent of q using:
Tan q = HT / Adj

Ok, now that we know the tangent of the angle we can calculate the center of gravity height based on our weight measurements using the following formula:

COH = WB x FWc
TW x Tan q
WB is the wheelbase
FWc is the change in front wheel weights
TW is the total weight
Tan q is the tangent calculated above
This calculates the Center of Gravity Height from the axle height.

To find the CGH from the ground, you must add your axle height to the above calculation. You can measure your axle height or calculate it using the average of your two front tire sizes and the formula for the circumference of a circle.

C = 2 p r ( r is axle height of 10 inches)

C is the average circumference found by adding the LF and RF sizes and dividing by 2.
p approximately equals 3.1416
r is your axle height
For example: Your LF is 85.5" and your RF is 87". Your average circumference is (85.5 + 87) / 2 = 86.25". Your axle height is (86.25 / 2) / 3.1416 = 13.727". Add this number to the CGH to find the center of gravity height in relation to the ground.



Now you have your true 3-D COG.

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Ranger Mike

I apologize for many " errors' in the above formulas used in calculating 3D COG.....I wrote the post in Windows messanger because of spell check..when i copied and pasted it some symbols translated to what ever software is used on this forum and the letters changed to th edefault format...i recommend you click on

http://www.longacreracing.com/articles/art.asp?ARTID=22

for details

Rahul jeta

although all your methods was on the mark,but since i am learner,i do have difficulty in understanding this second point.
2. draw the ground line ,vehicle center line and center of the left and right tire contact patches. Determine where the outer lower control arm ball joints (BJ) are located by bolting the upper and lower control arms to the spindle and bolting the spindle on the wheel to be used...some round track cars have different wheel offsets so be careful. mark these BJ centers on the drawing.

my question is..how will we determine the outer lower ball joints..?
i want to design a suspensiom system,and i wann know where to start from?
and how will we locate the pivot joints and how will we determine the lenth and angle of the arm..?
I am using double wishbone

Ranger Mike

Welcome and great questions...
Please note on Page on page 6 of this post..shows typical ball joint. To do a good job you can contact the manufacturers (Moog) being one), and have them send you diagram of exact pivot point. Another alternative is to find and old ball joint of same type you intend to use and cut it to determine the center of the ball stud.
You have a huge question that has many possible answers.
Will your suspension be:
Front wheel drive, front engine
rear wheel drive rear engine
rear wheel drive front engine
front wheel drive rear engine...??
What will vehicle weight be?
What is wheel base and wheel track width?
Where are the heavy components like fuel cell, transmission, differential, battery, driver going to be located?
What size tires will be used and what are wheel specs?
What is desired ride height to be?
Do we have a roll cage ? is the vehicle a hard top of convertible?

The short answer is once you figure out where the 4 tires will be you can start to connect them to the chassis and locate the engine, transmission etc..
finally, after the heavy bits mounting locations are finalized we work backwards to tweak the wishbones and mount points to get the roll centers we want.
very short answer...

Rahul jeta

Thanks for the reply..:)
Currently I am working on a hybrid trike,it doesnt have any engine.
its a tadpole having rear wheel drive.the track width is 45" and the wheelbase is 90.

latemodel18

Ranger Mike I have been going thru this whole forum trying to learn about this kind of setup and I am enjoying it very much. I was hoping you might be able to look at some of my numbers and setup and be able to give me a few pointers. We are running a template body tubular chassis on a hosier 970r 9" tire on an 8" rim, Our springs are 175 across the front and Lr-200, RR-350 with a medium rubber and 1.375 sway bar, panhard bar is 8.875 left side axle and 10.125 right side chassis. 2736lbs without driver, left side is 59.9- cross 54.4 before preloading the bar and rear 51.2 wheel weights are lf-673, rf-689, lr 918, rr-456, trailing arms are at 3 degrees uphill on the left and 1 degree on the right, top link is centered and alot of downhill angle,{sorry forgot to write that down before I left the shop} and 2-7/8 stagger with a Detroit locker on a medium banked 3/8 mile asphalt. At the track last weekend we ended up starting off with 6 rounds of bar in the car to try even out the nose but as testing continued the driver said he was tight thru the corner and loose off so we bumped the RR up to a 375 with medium rubber took 1-1/2 out of the bar and put 1 round of cross in evenly and he liked the car much better but still was not able to pull it to the bottom at the apex when he needed to, still a little snug. I might be wrong but I think I could be crutching the car with to much bar and maybe need to up the rr more. I done some measurements today and plugged them into my roll center program and I tried to attach it for you to see but no luck. It shows RCH-1.7 and RCL-2.9 at static and 2" dive it moves to RCH-4 and RCL-7. Not real sure where the roll center should be with this setup as I am only into this 3 weeks as a first timer working on race car setups. I am sure I am still off on understanding all this but what I have learned has pretty much been from reading on this site. Thanks again and really enjoy it.
Rod

Ranger Mike

Thanks you fro the kind words and welcome. You have a good grasp of what is going on with the set up.

I had to re-read the part about springs...350 right rear...whoa...
just from what you tell me I think you have really twisted the car to make it work half way..
When you preload the ARB you take away the purpose of the bar...go as light as you can on front springs to keep the nose as low as possible until you hit the corner where the ARB will contribute to roll control. When you preload the bar you are really stiffening up the total front spring rate. I much prefer going up on the front spring rate and keeping the bar neutral.

no way would I run that much preload on the ARB and have my rear springs stiffer than the fronts.

you have a good initial set up regarding weight percentages..almost ideal in fact.
without knowing the A-Arm motion rates and the like a ball park setup on coil overs
350 # on lf and rf, 225 on lr, rr, and 220# ARB

My main question is the Detroit locker situation. Is the driver using a lot of trail braking on turn entry?
If he is, the locker outside ratchet will not sense the power off and will stay locked up, the differential will not work and the car will push.
Did you remove the holdout ring on each side to provide instant lock up of the cogs on the driven assembly.
Sounds like you are running about the same stagger as the folks running a spool and this is not letting the differential do what it is supposed to do.
can you calculate the ARB spring rate.

Please confirm the Roll Center static location.. it is to the left in static...correct?
Also where does the RC end up on dive..to the left?
Looks like you have classic push going in loose off caused by RC being too much located to the left , not providing enough leverage to stick the right front going in...the current twisted chassis setup is a result of band-aids applied at the track to fix the car...

latemodel18

yes Mike the RC is 2.9 to the left at static and goes to 7 left at 2" of dive and the RC height is 1.7 at static and then goes to 4
Thanks
Rod

Ranger Mike

If you are running a race track that could use the big bar soft spring set (BBSS) up...like half mile or longer track...then there would be some justification of having the Roll Center on the left side. The BBSS setup uses aero down force to plant the right front tire in a turn. You run super soft springs to lower the nose and seal off the sides to create downforce. With this set up if you have too much force on the right frtont tire you need to take some of it away by moving the RC to the left until you get the proper amount..tire temps and handling, lap times..etc...will tell you this.
On short tracks this is not a good idea as your top speed just is not there. So on short tracks you maintain the RC offset to the right as is discussed in previous pages on this post.

Because the BBSS set up is the current fad, everyone is going to it with out understanding why it is used and where it is used. So copying a set up used by the hot dogs on a long track , and trying to make it work on a short track will have the following results.

Aero will not be enough to download the right front tire so instead of turning at corner entry it will push. Since you have not transferred enough rear weight and cross weight to the right front tire because the RC is biased to the left, the left front tire have a lot of front end weight staying there thru the middle of the corner. Since the left front spring is so weak it is not able to transfer weight to the right rear tire on corner exit. The car is loose off.
Add to this factor the Detroit locker variable and you end up chasing your tail at the track.
The plus side is the fact you got the weights where they should be so the basic package is there..