Forwarded message:
>From phile Tue Apr 26 15:19:11 1994
From: phile (Philip J Ethier)
Message-Id: <9404262007.AA01853@pwcs.stpaul.gov>
Subject: Re: MGB Roadster vs GT rear springs
To: british-cars@autox.team.net
Date: Tue, 26 Apr 1994 15:07:09 -0500 (CDT)
Cc: sfisher@megatest.com, garyb@iotek.ns.ca
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Scott Fisher writes >
garyb@iotek.ns.ca writes > ~
> ~ It will roll more in the corners not less, do to the higher rear end.
>
> If the springs are stiffer, it will ROLL less no matter how high you jack
> the rear end.
Maybe. Depends on how much stiffer vrs. how much you raise the center of
mass.
> The spring rate determines the roll rate; if it takes 125
> pounds to deflect the spring 1 inch, it doesn't matter whether that inch
> is at stock ride height or four feet off the ground
It matters to body roll in cornering. I'll pull some numbers out of the
air for you:
Vehicle track (tread) = T = 50 inches
Stock center of mass vertical distance from road = C = 17 inches
Modified center of mass vertical distance from road = C + 3 = 20 inches
(more realistic than four feet, I wager)
Vehicle mass = M = 2000 pounds
Lateral acceleration = G = .8
I am going to neglect the front/rear effects and just take the totals for
each side of the car. Assuming the right/left weight distribution at rest
is even, there is 1000 pounds on each side. We will further assume that
Scott's 125 pound/inch figure is at the tire (which we all know it really
isn't, just check out my Europa calculations sometime). The virtual
spring rate is 250 pound/inch because we have two tires on that side.
1000 p/i / 250 p = 4 inch deflection on each side. No roll.
Since we know that a solid cube will just start to fall over at one G, we
can determine the accuracy of a formula for weight transfer. Weight
transfer, which I have arbitrarily called X, is the amount of vertical
weight added to the outside edge of the cube. Obviously, an equal amount
is subtracted from the inside edge of the cube. A cube massing 2 pounds,
assumed to be sitting only on two edges, requires 1 pound to be added to
one edge, and 1 pound subtracted from the other edge, to begin to fall
over, as the weight on the light edge is zero and the weight on the heavy
edge is 2 pounds. Say the cube is 2 inches on a side. Therefore the C is
1 inch (the center of mass is in the center of a monolithic cube) and the
T is 2 inches.
Weight transfer = X = M * G * C / T = MGC/T (darn near made a car name!)
= 2 * 1 * 1 / 2 = 1 pound
Now lets take our virtual car at stock height car and put it on a skidpad
pulling .8 g. The weight on the outside tire patches will increase. The
weight on the inside patches will decrease by the same amount, because the
total weight must equal the mass (no banking on this skidpad).
X = MGC/T = 2000 pounds * .8 * 17 inches / 50 inches = 544 pounds
This will deflect the outside springs 544p / 250p/i = 2.176 inches from
static and allow the inside springs to relax by the same amount. This is
a lean of 4.97 degrees because tangent of lean = 4.352 / 50
If we raise the car 3 inches:
X = MGC/T = 2000 pounds * .8 * 20 inches / 50 inches = 640 pounds
This will deflect the outside springs 640 p / 250p/i = 2.56 inches from
static and allow the inside springs to relax by the same amount. This is
a lean of 5.85 degrees because tangent of lean = 5.12 / 50
So we see that there will be more roll with the same rate springs and a
higher center of mass. What would happen if we did get the center of mass
up to four feet as Scott suggested (although it was hard to understand him
with his tongue in his cheek)?
X = MGC/T so G = TX / MC = 50 inches * 1000 pounds = 50,000 = .5208
_______________________ ______
2000 pounds * 48 inches 96,000
At a mere .52 gee, the damn thing would fall over. Just before it went,
the inside springs would be unloaded as far as the droop allowed. The
outside springs would be loaded at 1000/250 = 4 inches past static as the
inside tires left the ground.
If you figure the center of mass height and the track dimension, this will
give you a maximum lateral force possible on a flat corner. Nothing can
get you past this; not sticky tires; not sway bars; not welding the damned
suspension solid. This is why race cars are low: To reduce weight
transfer. If you are well inside this envelope, however, you can work on
things to get you closer to it.
> And if you put in springs that
> take 125 pounds to deflect one inch, replacing springs that took 100 pounds
> to deflect one inch, it'll still be 25% stiffer, no matter what the ride
> height is.
Certainly.
> Since Smith tells us that
> weight transfer reduces overall grip on the affected pair of wheels, and
> since this weight transfer in particular moves *off* the inner wheel and
> overloads the outer tire (effectively doing the opposite of a sway bar),
No. This is not the opposite of how a sway bar works. This is EXACTLY how
a sway bar works. It tries to pick up the inside wheel. It adds to the
spring rate of the outside wheel by using some of the spring rate from the
inside wheel. A sway bar reaches its maximum effect when the inside wheel
leaves the ground. The point of a sway bar is to get you to your limit of
lean faster, and to reduce body lean. They may not make the end they are
used on stick better (there are exceptions in cars with poor geometry),
but a sway bar can limit body roll to make the OTHER end of the car work
better. Sway bars can make cars with poor geometry handle much better.
They cannot help cars with excellent geometry much. Such cars are pretty
rare, though. In serious purpose-built race cars, sway bars are used for
fine-tuning the suspension.
> seems that older (1971 and earlier) MGBs can use wider tires at the
> rear than newer (1972 and later) ones can. ... early Bs can run 185-60
> tires with no problem, and 195 to 205 series with some modifications
> to the fender lip. Later Bs, however, rub with 185-series tires, and
> need serious cutting or banging to use anything wider.
Right. My buddy Brian has no problem with 185-60-14 tires on stock
Rostyle rims while autocrossing his 1970 MGB.
> Note that apparently the MGB's rear leaf springs,
> once properly located at any rate, are stiff enough laterally that they
> do not work well with a Panhard rod; Phil Ethier reports an MGB autocrosser
> who has had Panhard rod mounting points rip out of the chassis because
> the lateral motion they require (visualize the geometry) is greater than
> that permitted by the B's rear springs, and something has to give.
Zounds! I've been misquoted! I never said that, and I have not heard it
before. But I believe it is possible. What I DID say is that Brian tells
me that a Panhard rod is unnecessary with MGBs for autocrossing (unlike
the whippy springs in the Midget). He has had no problems traceable to
lateral movement in the leaf springs, even though he is putting enough
gees down to flex the axle shafts and make the brake drums rub on the
backing plates in cornering.
I had no problems with the very solid (heim jointed) Panhard rod in my Midget.
BTW, Brian uses sway bars for his roll stiffness and runs the 1970 at
stock height. The sway bars keep the camber and rear-steer changes down.
> --Scott "Time to fix this car and DRIVE it instead of talking about it"
Fisher
Phil "thanks for making me think this through" Ethier
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