View Single Post
Wil Ferch Wil Ferch is offline
Registered User
Join Date: Aug 2002
Location: Galivants Ferry, SC
Posts: 10,551 problem...but as I said, there is more.
*IF* you are already near desired ride height, then the spring plate angle that "droops" nearly the desired 3.8" at that point....makes BA's statement correct. Namely, 1 degree is about 7-9 mm ( 1/4-1/3").

Look at this for a better backdrop of where I'm coming from, taken from my technical article on Rennlist:


I think I've got it. According to theory and my faith in math, at least.
I think I have a way to predict proper spring plate angle when setting up a
car with OEM or larger torsion bars.
We first need to agree on some things. A while back, I posted equivalent
spring *rates* vs rear torsion bar sizes, and it also included measurements
I've taken on the spring plate length ( 18.5 inches...from TB centerline, to
wheel centerline). We need these numbers , so for convenience, I will repeat
them here:
torsion bar size ( mm) spring rate ( lb/in)
23 100
24 120
24.1 122
25 140
26 165
27 191
28 221
29 254
30 291
31 332
32 377
33 427
We'll also assume that all torsion bar 911's use a spring plate as
measured here ( 18.5"). Let's also repeat some things published by Porsche in
their Tech books, and see if we can verify our assumptions against them:
- Porsche says that every 1 degree of spring plate angle is worth 8-9 mm
in ride height ( early tech books) , but they change that to 7-9 mm, in
later tech books. Actually , this discrepency makes sense, since it varies by
the severity of the angle. Let's use our "assumption" measurements and see
how we stack up. Lets's try 2 cases...the first one going from 37 to 38
degrees angle....and the second case where we only go from 2 to 3 degrees
angle...a much shallower proposition, but the same net change of one degree.
Let's make life easier by pretending we're looking at the driver's side of
the car ( left side, or US spec for our English friends)...assuming "zero"
degrees is a spring plate horizontal to the right, and normal "droop" would
make the far right end drop "down". This would define a right triangle, with
the spring plate as the longest length ( hypotenuse), and the vertical droop
defined as a function of the "sine" of the droop angle...basic trigonometry:
Case 1:
37 degrees down from horizontal.
vertical droop is... ( sin 37 degrees)( 18.5") = 11.1335"
38 degrees down from horizontal.
vertical droop is ...( sin 38 degrees)( 18.5")= 11.38"
difference for 1 degree is about 0.25" , or 6-7 mm. Close,
but no cigar, compared to the spec books.

Case 2:
Lat's look at a shallow angle , like when a car is nearer to
"ride height"...
2 degrees down from horizontal
vertical droop is...( sin 2 degrees ) ( 18.5) = 0.6456"
3 degrees down from horizontal
vertical droop is....( sin 3 degrees) ( 18.5)= 0.968215"
difference for one degree is about 0.32" , or 8.2 mm. Voila !
It center-punches Porsche's claim in the spec books !
This proves that Porsche had "near ride height" in mind when quoting
these numbers, in other words, assuming very shallow angles or "droops" from
horizontal, and it also verifies our 18.5" spring plate measurement,
otherwise, the math wouldn't work ! Let's go on.
We also need accurate car weight and spring rates for this to work.
Let's first take a typical 1974 911, weighing about 2400 lbs, using standard
23 mm rear torsion bars. From my earlier post, a 23 mm bar is about a 100
lb/in spring rate. Typical weight distribution is 40/60, so the rear weighs
1440 lbs...and *each* rear wheel carries 720 lbs. Spec book says free hanging
droop is 36.5 to 37 degrees down from horizontal. Let's use 36.5 degrees:
find vertical this "x"
x = ( sin 36.5 degrees) ( 18.5)= 11.0042255" droop.
As we put this car on the ground, the "droop" will be lessened
due to the 720 lbs weight we are applying. The amount this droop gets
"lessened", is based on the spring rate ( 100 lb/in for 23 mm bar). So......
720 lbs of weight, pressed against a 100 lb/in torsion bar, moves
the droop back "up" by 7.2 inches. The "net" droop then becomes 11.004" -
7.2" = 3.804" droop , when the car is on ground. This is the "ride height"
droop. The equivalent angle that represents 3.804" of droop , can be found
3.804" = ( sin of angle) ( 18.5"). Solving for "angle" ....the
angle equals 11.866 degrees.
Let's move on to a nice example I saw in the spec books. Porsche
claims that the 85 Carrera ( 2760 lbs weight), 24.1 mm torsion bar ,
requires a free hanging "droop" of 35 degrees, yet the identical 1986 car
with 25 mm torsion bar requires only 32 degrees. Let's see how our methods
check against this:
1985 Car:
torsion bar....24.1 mm ( 122 lbs/in rate)
rear *individual* wheel weight of 840 lbs ( assuming 40/60 split,
2760 total).
droop ( "x" ) = ( sin 35 degrees)( 18.5)= 10.61"
put car on ground, apply 840 lbs.
840 lbs / 122 lbs per inch rate = 6.88" less droop

"net" droop is 10.61" -6.88"= 3.73" droop at ride height. This should
be no surprise. Just like the 1974 example, if the suspension plate arm
length is the same, ride height is determined solely by the resulting "ride
height droop"...we shouldn't be surprised these are nearly the same. Let's
see how the 1986 car stacks up. Seeing a pattern develop, we may expect that
the free hanging droop is less ( it is , 32 degrees with the stiffer
25 mm bar)...and we may expect to see a similar "ride height droop". Let's
see if it works out this way:
1986 car:
torsion bar 25 mm ( 140 lb/in rate)
rear *individual* wheel weight is same...840 lbs
droop ( "x" ) = ( sin 32 degrees)( 18.5)= 9.8"
put car on ground..apply 840 lbs
840 lbs / 140 lbs per inch rate = 6" less droop
"net" droop is 9.8"-6"= 3.8"....droop at ride height. ( equiv angle
is 11.85 degrees using similar method as 1985 case.) As before, the ride
heights are nearly the same in all cases, as you would expect.
Now, here is the real question....How do you determine spring plate
angles for "non-standard" ( read "upgraded" ) applications???? Let's use the
1974 car again as an example, and upgrade it to 26 mm bars, from the original
23 mm bars:
1974 car, weighs 2400 lbs...
maintain ride height ( with new springs)...3.804" ride height droop ,
or 11.866 degrees.
We realize that a 26 mm bar has an equivalent spring rate of 165
lb/in. Working backwards from the earlier examples, we want to "add" droop
suitable for the car's weight and torsion bar size. So...we want to "add" 720
lbs of wheel weight, but this will work against a 26 mm bar ( at 165
lb/in) we get:
720 lb / 165 lbs per inch = 4.3636", of added droop.
We'll add this calculated "droop" to the static ride height "droop", to
get total free hanging droop:
4.3636" + 3.804" = 8.1676" droop.
Using trigonometery again, we get:
sin of "angle" = (8.1676 / 18.5) = 0.441491891
Solving for "angle", we get: 26.2 degrees. This is the free hanging
angle of the new set-up using 26 mm bars.
We can conclude with this table:
Weight bar free-droop angle ( vert drop) ride height angle (
vert drop)
2760 24.1 35 deg ( 10.61") 11.63 deg ( 3.73")
2760 25 32 deg ( 9.8 ") 11.85 deg ( 3.8")
2400 23 36.5 deg (11.004") 11.866 deg ( 3.804")

2400 26 26.2 deg ( 8.1676") 11.866 deg ( 3.804")
From these examples, we can calculate any other combination. We need
, however, vehicle weight, torsion bar equivalent spring rates, and a
scientific calculator or trig tables.
Any caveats? Yes. We are working with very precise , and small
measurements, so accuracy is of utmost importance. Rubber bushings that
deflect or deteriorate, don't help. I also find other tech items that the
Porsche spec books qoute ,as not jibing with these calculations. For example,
in almost all our examples, the static ride height is about a 3.75"-3.8"
droop, as measured from the centerline of the TB, to the centerline of the
wheel. Porsche spec books use various numbers ( USA vs Rest-of-World
settings, etc)...but a typical number quoted is 12 mm, or about a 1/2 inch.
Doesn't make any sense , and this would define a *very* shallow, nearly
horizontal spring plate at ride height, and I just don't see many cars this
way. Maybe they mean to measure from the *bottom* of the exposed torsion bar
tube , ....????
Anyway, this won't be perfect, but it may allow you to use "trial and
error" on paper first, and avoid doing this 4 or 5 times for real. Even if
gets you close enough to do it over only two times, it would be worth it.
If this works, ( anyone who tries it, let me know)...I only ask that
from this point forward it be known as the "Ferch" method of suspension
calculation....hell, we have Planck's constant, Avagodro's number,
Googleplexes, etc...why not a "Ferch" constant ?
Hope it helps.....
Wil Ferch -
Wil Ferch
85 Carrera ( gone, but not forgotten )
Old 12-19-2003, 08:11 AM
  Pelican Parts Catalog | Tech Articles | Promos & Specials    Reply With Quote #12 (permalink)