


four bar geometry spreadsheet
I am in the process of setting up the rear four bar setup in my 49 Ford coupe and am trying to calculate out the best angles and mounting points for the bars. I mocked everything up where my common sense told me it should go for the cleanest setup. Then I found a spreadsheet that someone here put together (triaged) and input all my mounting points. I think the calculations that the worksheet gave me are good numbers. But I would like to get a second opinion from someone before i do any cutting and welding because If I need to change something now is the time. Dose anyone know how I can attatch my modified version of the spreadsheet found here. http://home.earthlink.net/~triaged/F...ulatorV1.5.xls
I guess I can just list out the variables I used in case anyone is bored enough to put them into it to see my results. I think I understand what all this stuff means and from what I've read it seems that my results should be good. however I really want someone elses opinion on this thing. Thanks, Jared Vehicle Specifications: Wheelbase 114.0 Tire Diameter 25.50 CG Height 30.00 Weight 3,200 Upper Links x y z Frame End 19.00 19.38 12.50 Axle End 1.50 8.75 15.25 Vector 20.50 10.63 2.75 Lower Links x y z Frame End 25.25 18.13 10.00 Axle End 0.00 18.00 9.5 Vector 25.25 0.12 .5 AntiSquat 107.68 % RC slope 0.02 in/in ("" = roll understeer; "+" = roll oversteer) RC Height 17.92 in RC Angle 1.27 degrees ("" = roll understeer; "+" = roll oversteer) Instant Center XAxis 36.04 in Instant Center ZAxis 10.21 in
Last edited by jhenry; 01052005 at 07:39 AM. 
Sponsored Links  
Advertisement




You might be intrested to know that I have put out a new version of the calculator. This version (3.0) draws a picture of the suspension for you, and even has a travel function. I think it is a bit more user friendy.
To use the travel function make sure you have the "solver" addin checked and macro security turned down some. http://home.earthlink.net/~triaged/F...arLinkV3.0.zip This is what your suspension looks like in it (sorry for the big pic) There is a large dicussion of all of the versions here http://www.pirate4x4.com/forum/showthread.php?t=204893 You might want to start from the end as that thread has been going for over a year. 


Triaged has given you some good information. I'll only add that, if you want to cancel the driveshaft torque effect (unloading of right rear on launch), now's the time to do it.
I'm assuming a 4.11 axle, so the following will change slightly with a different ratio. But, with that 4.11, if you were to change the right upper front to (19.18, 19.38, 15.79), the left upper front to (18.12, 19.38, 8.71), the right lower front to (24.83, 18.13, 14.13), and the left lower front to (24.82, 18.13, 4.85), you would have equal rear tire loading with any value of driveshaft torque and no squat or rise (100% antisquat). Link lengths would be unchanged from those you're considering. Roll steer effects would be no worse than with a symmetrical setup. Last edited by BillyShope; 01062005 at 02:52 AM. 




Jared, this is just an adaptation of an idea first used by Jaguar in their early CTypes and then, later, by the Chrysler Ramchargers in their first car.
It's based on the idea that the link forces are proportional to the driveshaft torque. So, if you provide an asymmetry in the linkage, you can provide a torque which is always equal and opposite to the driveshaft torque. The Jaguar and Ramcharger designs were based on a 3link with a single upper link offset to the right. (The Ramcharger design used a telescoping upper left link which acted only in braking.) All I've done is apply that principle of asymmetry to the adjustment of a 4link. 


Ok, but where did you come up with the exact new locations for my four bar bar mounts. I think I understand your concepts, but are you just visualising something I'm not or do you have some more magic formulas??
Thanks again Last edited by jhenry; 01062005 at 11:47 AM. 


No, nothing magic. You can make the calculations yourself. Assume an acceleration for the car. A convenient choice would be 1 "g." Now, consider the rear axle assembly and links separately from the rest of the car. Just to pick a number, let's say this assembly weighs 250 pounds. That means the total thrust, at the rear tire patches, would be the total car weight and the total horizontal component of the reaction forces, at the front of the links, would be the total car weight less the weight of the rear axle assembly. For no squat or rise, the vector for the reaction force would be at an angle with a tangent equal to the CG height divided by the wheelbase. The weight transfer would be equal to the total weight...less the rear axle assembly weight...times the CG height and divided by the wheelbase. The trick, now, is to distribute that weight transfer in a way that the driveshaft torque is canceled. The driveshaft torque would be equal to the product of the wheel patch thrust times the effective rear tire radius and divided by the axle ratio. So, all that's left is to determine the right and left vector angles which will result in the product of link spacing and difference in vertical components of the vector forces that will yield a couple equal to the driveshaft torque. Remember, the left and right side horizontal components will be equal. Well, there's a bit more to do. It's now necessary to "aim" those upper and lower links so that lines through them converge on the lines defined by these new vector angles. In this case, I used the same link lengths and a little algebra to determine the new coordinates.


Recent Suspension  Brakes  Steering posts with photos 
Currently Active Users Viewing This Thread: 1 (0 members and 1 guests)  
Thread Tools  

