John - how about you get someone to look at your digital levels while you put more and more of your own weight out on the end? Just to get that qualitative look before you start quantifying.

Should give you an idea of how things are going, though the question you raise is a good one, since you have a fulcrum in the middle of chassis.

It does not matter where the supports are, torque applied to chassis will result in the same rotation about the centroid axis.

You need to think of using a torque wrench (or any wrench). If you have a deep socket or extension you need to apply a good amount of pushing force on the socket to counteract your other hand pulling on the handle. If you apply a force of 1 pound to the end of a wrench the wrench just moves linearly, falling off the nut. If you apply a pull of 1 pound at the handle and a push of 1 pound at the socket and the handle is 1 foot long, then you are applying a couple (two opposing forces separated by a distance) of magnitude 1 foot-pound of torque.

Firstly, we assume the chassis is close enough to call bilaterally symmetric (left and right are mirrored). Now, a lever arm of length L (from the center of the chassis to the weight) with weight W at the end is added. You have two mounting points for the lever on the chassis, and in this case the mounting points are also bilaterally symmetric. They are separated by a distance D (so the total length of your beam is L+D/2).

This is a simple statics problem. We want the sum of the forces and the sum of the moments acting on the lever to be zero. We have a moment about the end of the lever (at the chassis centerline) L x W and a force W. Therefore, the two reaction forces at the chassis (call them R1 for the left side and R2 for the right) must cancel the applied force and moment.

Now, so that I only need to type scalars we assume that a tension force in the bolt is a positive force, and a positive moment is one that turns the end of the lever towards the ground.

R1 + R2 = W

LW = R1*D/2 - R2*D/2

Now, some algebra

R2 = W - R1
LW = (R1-R2)*D/2
LW = (R1-(W-R1))*D/2
LW = (R1-W+R1)*D/2
LW = 2*R1*D/2 - W*D/2
LW + W*D/2 = R2*D

W*(L/D+0.5) = R1 This is a positive quantity as I have defined my vectors, so the left side bolt is in tension with a force significantly greater than the applied weight

R2 = W-R1

R1 > W, therefore R2 is negative and the right side bolted joint is in compression.

The bolt on the side where the weight is is doing most of the work.

There are quicker ways to get this answer but this is a little more to the book.

Even more basic than that, I was just pointing out that if he only gets a deflection equal to one unit of the resolution of the device, then he has +/- 50% uncertainty (error bars). That's hardly measuring something. 2 units he has +/- 25%, 3 units +/- 16.7%, 4 units +/- 12.5%, etc. So the bigger the deflection, the smaller the uncertainty.

About 1.2M GBP.
Or you can rent time on one. Morse Measurements

The reinforcing cross member is held to the chassis with an M12-8.8 bolt. The first 8 indicates that the tensile strength of the steel is 800 Newtons per square millimeter. That's 4568 psi, or 20,340 pounds tensile strength.

So I don't think you will break the bolt.

A little humor!
 

Hey Boss, how much torque do you want me to apply?

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You are correct that an 8.8 will have an ultimate tensile strength of at least 800 MPa (Mega Pascals = Pascals * 10^6 = 10^6 N/m^2 = N/mm^2). The yield will be 80% of that.

However, 800 MPa is 116,000 psi. 6061-T6 aluminum is around 30 ksi (30,000 psi). 4130 steel (normalized) is about 63 ksi. 4.5 ksi is about the strength of cheese :).

The minor diameter of an M12 bolt is 9.6 mm minimum, which gives an area of 72.4mm^2, corresponding to an ultimate tensile load of 13,476 pounds. You want to stay well clear of yield however, which means more like 6,500 pounds.

I knew I should have found one of those charts which shows strength by bolt size and grade directly, which would have kept me from overlooking the minor diameter. And maybe kept me from confusing UTS with the elastic limit.

But 200 pounds (a guy sitting on the lever) would need 32.5 feet to apply a force of 6,500 lbs, would it not?

A lot of force may be necessary. Remember the photo, in Puhn I think, of a bunch of guys standing on the end of a rather long lever to do this kind of measurement?

If the lever arm was 32.5 feet (not 32.5 feet from weight to opposite side crossmember mount) then there would be 6500 ft-lbs of moment on there, so I would expect the chassis to deflect .5 degrees or less based on the stiffnesses given in post #34. The tension force in one bolt would be about 3350 pounds (I assumed a 24" bolt spacing, I fon't know what it really is). The other bolted joint would have 3150 pounds of compressive force in it, so no bolt necessary.

Hi guys,
Do keep discussing, just letting you know im away on holiday without access to the car for 10 days. I will resume next saturday.

Regarding the discussion on the steering rack bolts, isnt it a bit of a moot point? If it takes more force to twist the chassis than the steering rack bolts can withstand, and the bolts didnt break or pull before, then the car wasnt being subjected to enough twisting force for it to do so? So it doesnt matter? I guess it matters if my experiment subjects the car to more force than it encounters while driving and i were to break something. But that is an interesting question for me...if i have to subject the car to so much force that it exceeds normal use to get a measurable deflection then my results dont really matter for the purposes of seeing if my chassis mods impact handling?

The rack mount does not experience much force in tension or compression during operation but does react lateral loading. Its design was probably based upon loads induced by bumping a curb or bouncing through a chuck hole and not from normal loads due to tire friction with the road such as from braking or cornering. The large bolt is simply large due to the requirement of the lug welded into the chassis needing enough shear strength for the lateral loads. As such, the lug is probably adequate for reasonable tension/compression loading per your requirements.

Based upon your setup I calculate:

• 90 pounds generates 590 ft-lbs of torsion to chassis
• assuming torsional spring rate of 15k ft-lbs/degree your chassis twisted 0.04 degrees
• tension load on cross bar bolt (closet to the applied weight) is 298 lbs
• compressive load on opposite mount is 208 lbs
• deflection of your loading bar is 2.25" assuming 1 3/4" x 1 3/4" x 3/16" angle
• compressive stress in leg of angle is 40.5ksi which is 80% of yield strength of your angle

You just blew my mind.

Before I left I did bolt the crossbar back on and removed the ruler/placed a bolt on the end of the lever bar so I can just stack the weights on the end without them slipping off. I am going to take some collective advice here and;

- I took the straps off the torsion bars and will hook the body directly to the lift arms - no bushings there.

- Will use the bar bolted to the steering member bolts only for loading with a bunch more brake discs.

- Will measure the change in angle across the rack rather than the change in height of the end of the bar

Still not intending this to be lab-level data but I am listening to suggestions here.

I als realize earlier I used the term "gravity-based" in reference to my dial indicators. What I should have said is that they are not spring-loaded. I have typically used them in a vertical, upright position such that gravity pulls the measuring pin down rather than any spring.

I've never seen dial indicators without some mechanism to apply some tension to the arm/plunger. That would require a precise low friction mechanism to accurately measure movement. And other than used in a vertical axis, it wouldn't be accurate if used to measure movement at any other angle (e.g. horizontal and thereabouts). You sure the plunger only uses its weight to maintain contact with the part to measure?

Sherwood

Yep. One I got from Pelican about 10 years ago when I did my first engine build. I forget where I got the other.

Maybe they are just broken...

Forgot to add deflection at end of angle due to chassis twist which is an additional 0.06". Note that deflection of angle is 2 1/4" and deflection due to chassis twist is 1/16"; therefore, the angle as a loading beam is too limber and is close to yielding/buckling already.

Recommendations:

• support front with digital scales under loading beam
• use 2" x 3" x 3/16" wall rectangular tube to load chassis; orient so 3" face is vertical
• create anchor point in floor and use turnbuckle to load up the beam
• attachments to beam to be through the 2" wide wall which would be horizontal
• place digital level on cross member between rear shock mounts
• place digital level on horizontal member in front of gas tank

Dial indicators will record movement at the strapped mounting points as twist which is an error. Also, if dial indicators are not isolated from the supporting structure then they will move with the supports which is an error, they must be mounted to the shop floor to help assure they are as isolated as possible from motion of the chassis and its supports.

Paul
His goal is not to that level of precision.

For the stiffer beam suggested he'd have to conjure up a way of mounting the beam. With horizontal bolting he'd need to weld a small tube in to hold the horizontal bolt to replace the strength lost by drilling (though it could be done in the neutral axis to reduce that effect). That bolt could be larger than stock so its max shear would approach max tensile in the current arrangement. And he'd have to weld up some kind of fitting which would bolt into the stock location, and then stick down on both sides of the beam for the cross bolt.

He has said he doesn't want to come up with floor stands for dial indicators.

I am pretty sure he isn't going to want to install something on his garage floor for a one off experiment, and getting a pull out strength for some kind of drill a hole and insert a fastener system may not even be possible, as those are usually set for shear. I was happy enough using old 1/16" bolts climbing the Nose of El Cap, but I was very careful not to place an outward pull on them 30 years ago.

And won't digital levels, placed as suggested, be about as good as dial indicators? Perhaps we are not concerned with deflections of less than 0.1 degrees? Yes, given the fairly modest weights he proposes using, or even just 200 lbs of largish male, if he could be very precise then extrapolation would be more valid, I think.

The levels will compensate for the non-rigid attachment at the rear, won't they? Wasn't that your idea earlier?

What he may prove to himself is that the chassis deflection he records is so far down in the noise that it won't tell him if various strengthening measures he anticipates taking are doing much.

But it looks like a worthwhile endeavor, and he's pretty much got it all set up. Nothing much lost if his angle arm buckles.

Right.

Well, the thread said measurement, not comparison.

I still think that he will have a hard time getting even comparative data with a digital level. He'd need to put one on the lift to measure the deflection of his fixture, so then he has twice as much uncertainty. I just don't think .1 degree resolution is enough. I think a plumb bob and pointer would be better. Fix a non-load bearing stiff beam to the chassis and a string with weight at the same spot, then you can either do some trigonometry to find the angle by measuring the much larger distance that the end of the beam is from the plumb bob. You could even use a laser for the beam and measure on the ground to the plumb bob.

I think the simplest method is a beam that is perpendicular to the loading beam that is a few feet long with a plumb bob at the end, as well as one where it meets the chassis. Measure the distance between the two attachment points of the strings as the beam length, then measure the distance between the parallel strings after the chassis is loaded. Then do sine^-1 of the distance/beam length. You will have to account for any offsets between the strings when the chassis is unloaded (they should be inline with each other in the unloaded condition).

Edit: nevermind. The separation of the strings for .1 degree with a 3 foot beam is only about 1/16th of an inch. Not much.

That was the title, I think. When you read his first post, I read the intent to be comparison. And the title warned us (pseudo science). Who knows, we will have to await results.

Burgomeister once posted the results of his testing of front shock tower movement (AKA, how much do I really need a brace). His results persuaded him, as a full on automotive engineer, that they really don't flex much even though he was quite aware of how his methodology fell short of what he'd have done at GM or wherever. And of people's anecdotes about paint worn off the sides of hoods and the like.

This is now getting down to some real science, though. Flieger and Paul Abbott (who also sounds a lot like an engineer) have made predictions based on engineering science. Will we discover if the sun distorted the starlight at total eclipse? Oh, wrong theories being tested.

I got back and resumed and revised the baseline measurement based on many but not all the suggestions made here. I know it could be done with more precision, but this is as much time as I'm willing to invest into it for now. On with the actual modifications soon.

I separated the loaded lever from the measurement process, and I am using degrees rather than distance the beam was deflecting.

I als strapped the chassis directly to the rear lift support arms via a spring arms mount bolt.

The front still balanced at 53 kgs. I zero'd the digital level across the steering mounts, then put 40kgs at the end of the 2m lever. THe car visibly twists, and I got a -0.3 degrees of twist across the steering arm mounts with the weights on.

I then took the weights off and checked that I got the original baseline setting the same and I did.

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more

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John - this is in the best tradition of shad tree race car engineering!

Did you put the level on the rear end too, to verify that it didn't move despite your ingenious attachment system?

The photos of the lift pads are to show the "visible" part?

40Kg @ 2M = 80 Kg/M = 579 lbs/F if I did the conversions correctly. So 173 lbs/ft per 0.1 degree, or 1730 lbs/ft per degree? Or is torque the wrong force? How does this translate into the units Porsche uses for chassis torsional rigidity?

Springs are normally specified in pounds per inch. Torsion bars, which also work as springs do, would usually be specified in degrees twist per unit of torque, and to translate to pounds/inch need to have A arm geometry factored in. You don't see those figures in torsion bar sales catalogs - just their diameter.

But if we take 200 pounds per inch of suspension movement as a guestimate (I think that's well above stock), what does the 0.3 degree value measured (if we ignore factors like measurement error, and how the level rounds up or down) mean as a spring rate equivalent? You'd have to know the distance from the torsion bar center to the ball joint center or some slightly more accurate measure, like the tire centerline on the axle). I don't know that, and it is Saturday morning.

Next in the analysis might be to learn what forces act more or less routinely on the chassis. For instance, if you can corner at 1.2G (not out of reach with good tires), how much torque does that put on the front? Since the rear also leans in a turn, would you also need to know what the differential is?

But that exceeds John's goal, I think, which was to learn if he has stiffened the chassis after doing what he plans to do?

I get 1925 ft-lbs/deg +14%/-20% using his .3 degree deflection. That means between 2300 and 1650 ft-lbs/deg. Seems a bit low compared to what we were expecting based on the 964.

2300 ft-lbs/deg = 3126 Nm/deg
1925 ft-lbs/deg = 2616 Nm/deg
1650 ft-lbs/deg = 2243 Nm/deg

You need to read chassis at back and at front and subtract the two readings, strapping to your lift is not a rigid mounting and deflects during your loading process. Also, better to place your level on the structure and not on your beam, your beam assumes a curved shape during loading and if your level isn't placed in the same location then it will measure slope of the deflected beam in addition to the chassis twist.

The figure reported earlier was for the 993 at 24,000 lbs/ft/degree. An order of magnitude stiffer?

Keep in mind this car has some nice rust damage along the rocker panels and I suspect kidney bowls.

This would be fun to test on a Targa.
My friends say that they can see my car twist in the roof line when I auto-x.....

Doesn't sound right, as the 996 was much stiffer, and quoted as just over 20,000 nm/deg. What was the source?

Here's what I'd previously backed out:

'88 911 7000 nm/deg (measurement by burgermeister)
964 11500 nm/deg (993 was 20% stiffer)
993 13900 nm/deg (996 was 45% stiffer)
996 20120 nm/deg (Excellence was Expected p.1381)
997 33000 nm/deg
991 40000 nm/deg (porsche used 20% torsionally stiffer as a design target, has claimed "up to 25% stiffer")

Early car chassis had less steel to begin with, and if there was any rust... Short way of saying I don't think this is an order of magnitude out, and in fact it would be close to what I'd guess. Unfortunately the measurement resolution doesn't look like it will be enough to divine subtle differences...

I know you're done, but... laser pointer strapped to the chassis pointing at the wall. Mark the spot with an without load, measure the distance between the spots... If you want to get really fancy, strap a second pointer at the back of the car to take out the movement of lift.

Laser pointer accuracy and measuring distance? My el cheapo construction laser pointer is accurate to maybe 1/16" at around 20 feet. Could laser resolution slant end-calcs significantly?

Sherwood

You measure from the center of the spot, but 1/16" at 20 feet is .015 degrees. Order of magnitude better resolution...

Two thumbs up for two laser pointers. Accuracy isn't a problem since it is motion differential between beginning dot location and ending dot location and not pointing accuracy.

Donohue in The Unfair Advantage reported the 73RS coupes had stiffness of "over 2000 ft-lb/degree" without a cage.

In post #34 I stated 993 torsional stiffness as 24,000 but it was just a grab from a quick internet search trying to put some kind of number to the question at hand.

Some progress....

In another thread I documented how I cut out my old sunroof and made a patch from the old sliding portion. Here's what it looks like posing with the 2 products I used...

http://forums.pelicanparts.com/uploa...1415595405.jpg


I also mocked up a set of brackets I am thinking about welding in around what I am guessing are weaker areas prone to flex. Interested in comments about these.

I also got myself a set of dimple dies and a press and played around a bit in anticipation of making the brackets.

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also a knee bar with gussets and a bar across the sills behind the seats, with gussets to the sills and the center tunnel.

also moving the shifter up 3" and back 3"

Mine is a coupe.

I don't think the top B pillar braces will do much. Seems to me that the arch of the roof is one of the strongest/stiffest parts of the shell and it forms a sort of triangle with the A pillar so there isn't much load going through the B pillar, at least in shear. So I think the windscreen pillar brace will do much more than the B pillar corner gussets.

The bottom gussets for the lock post should be worth their weight. They would be approximating a diagonal door bar from a roll cage.

Hmm, I think these gussets would be horrible, put in a cage instead.

If you don't like that idea, I have another one that I have been playing with in my mind:
Make an extra layer of sheet metal to cover the sills, a-pillar down to the sills, b-pillar down to the sills, roof line all along the c-pillar to the shelf, over the windscreen. Make lots of dimple holes and bevel the edges so you get a nice distance to the existing metal and weld it in.
Sort of an integrated roll cage.

I have been borrowing ideas from the other caged cars I have here, incuding an identical 83 with a Bond full weldin cage.

I don't want a cage as I wont be wearing a helmet often in this car.

I am going down the road of what you suggested, though I don't plan on completely "reskinning" the posts and sills. Just reinforcing and gusseting.