Vertical CG on an F1 car?

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xpensive
xpensive
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Vertical CG on an F1 car?

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Since the traction thread, I have wondered about the front/rear distribution of contact-force between tyre and surface.
This should of course be dependent on acceleration, positive or negative, but also on the location of vertical CG.

Does anyone have an idea of where this is and how it changes with fuel-load?
"I spent most of my money on wine and women...I wasted the rest"

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machin
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Re: Vertical CG on an F1 car?

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My simple CAD model shows it is about at the centre of the wheel's (330mm above the ground); probably a bit lower in reality (I've only modelled the major masses accurately).

Note that with my tank shape the CG hardly changes at all... infact it goes up slightly (a couple of mm) when the tank is empty because the tank is much wider/longer at the bottom.

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Jersey Tom
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Re: Vertical CG on an F1 car?

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xpensive wrote:Since the traction thread, I have wondered about the front/rear distribution of contact-force between tyre and surface.
This should of course be dependent on acceleration, positive or negative, but also on the location of vertical CG.

Does anyone have an idea of where this is and how it changes with fuel-load?
Vertical CG? Not quite sure what you mean. Do you mean height of the CG?

I don't have specifics on hand.. but axle height isn't a bad first guess.
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xpensive
xpensive
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Re: Vertical CG on an F1 car?

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Indeed JT, with vertical CG I mean the distance from bottom of the car, or "height" if you prefer.

Now, if CG is level with rear wheel-center, there should be no change to front/rear distribution of contact-force between tyre and surface due to the car's forward accelleration, correct?
"I spent most of my money on wine and women...I wasted the rest"

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Ciro Pabón
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Re: Vertical CG on an F1 car?

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If I understand your question correctly, the answer is no, X. The braking/accelerating force happens at the contact between tyre and pavement. For you not to have weight transfer, you have to have the CG at the level of the pavement.

Here you have a simple explanation (please, read the balloons in numerical order):

Image

The red arrow marks the location of the braking force. The orange line is the "lever" of this force. The blue arrow shows the increase in frontal weight. The green line is the lever of the increase in weight. You get the idea: "red arrow" times "orange line" is the same as "blue arrow" times "green line".

I used this text ages ago for the first lesson in highway design, perhaps you might want to read it.:

http://sites.google.com/site/cirospictu ... eering.zip

This is common knowledge (or so I think), so I apologize if you already knew all this and I'm misunderstanding your question. I also apologize for the horrible translation (the lesson and the image text were written originally in spanish around 1994... when my english was even worse).
Ciro

xpensive
xpensive
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Re: Vertical CG on an F1 car?

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Nice display Ciro, seems intellectually credible at de-accelleration with locked fronts in JPM's fashion,
but is it xactly the same thing at accelleration with the traction on the rear-wheels, think about it?
"I spent most of my money on wine and women...I wasted the rest"

scarbs
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Re: Vertical CG on an F1 car?

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I doubt vertical (Z) CofG is as high as the wheel centres. No Tech director has ever given me much of an insight into its position. Some vague reference to being around the crankshaft centre height in the early noughties, but probably much lower now.

Some assumptions (Ignoring fuel load, ride height which is a few mm at most and all but zero at speed). The wheel centres are at Z330mm, less than half way up the car, the cockpit opening at just Z550mm. Excluding the roll structure there is precious little structure above this point and lots of heavy bottomed parts and ballast at zero height. A monocoque being around 80Kg with probably half that again in ballast at zero height.
The engine (100kg) and gearbox (30Kg) total height is no more than 300mm. Assuming the gearbox CofG is some way in between the gear shafts, the high diff being offset by the ballast low down (input shaft at Crank centre line Z58mm and output shaft 85mm above that) giving an approximate CofG height of Z100mm, then the engine CofG is mandated in the rules at Z165mm.

I would say the dry CofG is somewhere between 100mm & 165mm (say 135mm..?). The 100Kg of fuel over the 600Kg chassis would have an impact (with its own CofG at approx Z300mm), would only raise the total CofG by 15-20%

marcush.
marcush.
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Re: Vertical CG on an F1 car?

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scarbs wrote:I would say the dry CofG is somewhere between 100mm & 165mm (say 135mm..?). The 100Kg of fuel over the 600Kg chassis would have an impact (with its own CofG at approx Z300mm), would only raise the total CofG by 15-20%
I doubt vertical (Z) CofG is as high as the wheel centres. No Tech director has ever given me much of an insight into its position. Some vague reference to being around the crankshaft centre height in the early noughties, but probably much lower now.

Some assumptions (Ignoring fuel load, ride height which is a few mm at most and all but zero at speed). The wheel centres are at Z330mm, less than half way up the car, the cockpit opening at just Z550mm. Excluding the roll structure there is precious little structure above this point and lots of heavy bottomed parts and ballast at zero height. A monocoque being around 80Kg with probably half that again in ballast at zero height.
The engine (100kg) and gearbox (30Kg) total height is no more than 300mm. Assuming the gearbox CofG is some way in between the gear shafts, the high diff being offset by the ballast low down (input shaft at Crank centre line Z58mm and output shaft 85mm above that) giving an approximate CofG height of Z100mm, then the engine CofG is mandated in the rules at Z165mm.

I would say the dry CofG is somewhere between 100mm & 165mm (say 135mm..?). The 100Kg of fuel over the 600Kg chassis would have an impact (with its own CofG at approx Z300mm), would only raise the total CofG by 15-20%
only fuel loads above 135mm height would rise the CG.So anything more than around 30%of total fuel capacity filled into the car would seriously affect the cornering abilty of the car whereas everything below this would actually lower CG height but at the expense of added weight.So to my mind the best design for the fuelcell would use the maximum allowed planeview area to keep the maximum amount of fuel under or close to those 135mm...to optimise tyre use.This is in my view one of the areas where you have to decide a compromise aero vs mechanical in the 2010
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xpensive
xpensive
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Re: Vertical CG on an F1 car?

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I had to require assistance from a good engineering friend of mine in order to formulate an intellectually credible model
for how to estimate the re-distribution of vertical contact-force between tyre and surface during accelleration.

Imagine the rear wheel-center as the pivot point of moments, no wheel-spin involved.
- Propulsion Force from tyres is acting at 0.33 m, trying to raise the front of the car.
- Reaction Force from CG is equal in amount, but acting at (0.33 m - CG), trying to lower the front of the car.
- Total moment is as a concequence Force times (0.33 - (0.33-CG) = Force times CG.

In other words, Ciro was correct, credits due, moderator.

- If Force is 20 kN (500 kW at 25 m/s or 90 km/h) and CG is 0.150 m, Total moment is 20 000 * 0.150 = 3000 Nm.
- A wheel-base of 2.5 m (?), means that front wheels will lose 3000/2.5 = 1200 N due to the accelleration.
- Rear wheels will obviously gain the same 1200 N.

Please note that the total mass of the car is irrellevant here.
"I spent most of my money on wine and women...I wasted the rest"

Carlos
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Re: Vertical CG on an F1 car?

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OT but related, a copy of this would be interesting:
Racing Car Design Using Knowledge Aided Engineering
http://journals.cambridge.org/action/di ... &aid=59051#

xpensive
xpensive
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Re: Vertical CG on an F1 car?

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When I put this together the other day, I was hoping for comments from Ciro and scarbs, where are you guys?
What is the wheelbase and static weight distribution, front/rear of a modern F1 car anyway?

I had to require assistance from a good engineering friend of mine in order to formulate an intellectually credible model
for how to estimate the re-distribution of vertical contact-force between tyre and surface during accelleration.

Imagine the rear wheel-center as the pivot point of moments, no wheel-spin involved.
- Propulsion Force from tyres is acting at 0.33 m, trying to raise the front of the car.
- Reaction Force from CG is equal in amount, but acting at (0.33 m - CG), trying to lower the front of the car.
- Total moment is as a concequence Force times (0.33 - (0.33-CG) = Force times CG.

In other words, Ciro was correct, credits due, moderator.

- If Force is 20 kN (500 kW at 25 m/s or 90 km/h) and CG is 0.150 m, Total moment is 20 000 * 0.150 = 3000 Nm.
- A wheel-base of 2.5 m (?), means that front wheels will lose 3000/2.5 = 1200 N due to the accelleration.
- Rear wheels will obviously gain the same 1200 N.

Please note that the total mass of the car is irrellevant here.
"I spent most of my money on wine and women...I wasted the rest"

DaveW
DaveW
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Re: Vertical CG on an F1 car?

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A simple "thought experiment":

Suppose a vehicle has its rear wheels driven directly via drive shafts by an electric motor attached to the chassis. Suppose that the electric motor is constrained laterally, vertically and longitudinally, but is allowed to rotate about the drive shaft axis (I know, but this is a thought experiment). What will happen when a current is applied to the motor?

My mind suggests that the wheels will start to rotate in one direction, whilst the motor will start to rotate in the other. If that is correct, then it suggests that the act of constraining the motor's rotation relative to the chassis will cause a torque to be applied to the chassis, which must be reacted at the contact patches. The reaction loads appear to be independent of the height of the centre of gravity, & hence must be independent of changes in the contact patch load distribution caused by accelerating the chassis.

What is wrong with my logic?

BTW my logic implies that a fundamental difference exists between outboard & inboard brake installations (I think).

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Ciro Pabón
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Re: Vertical CG on an F1 car?

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DaveW, if I understand correctly your thought experiment, the wheels would rotate, yes, but very little. The relationship between the rpms of the wheels and the rpms of the engine would be the same as the relationship between the torsional moment of inertia of the wheels vs the one of the engine.

Simple: if the engine (as I pressume) is much "lighter" than the car, the engine will turn, but the wheels will turn by a very small amount.

Actually, when I say "lighter", I'm referring to torsional inertia moments, not to real weights, but I hope you get the idea: if an ant jumps forward from the back of an elephant, the elephant moves backwards, but how much? ;)
Ciro

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ringo
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Re: Vertical CG on an F1 car?

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DaveW wrote:A simple "thought experiment":

Suppose a vehicle has its rear wheels driven directly via drive shafts by an electric motor attached to the chassis. Suppose that the electric motor is constrained laterally, vertically and longitudinally, but is allowed to rotate about the drive shaft axis (I know, but this is a thought experiment). What will happen when a current is applied to the motor?

My mind suggests that the wheels will start to rotate in one direction, whilst the motor will start to rotate in the other. If that is correct, then it suggests that the act of constraining the motor's rotation relative to the chassis will cause a torque to be applied to the chassis, which must be reacted at the contact patches. The reaction loads appear to be independent of the height of the centre of gravity, & hence must be independent of changes in the contact patch load distribution caused by accelerating the chassis.

What is wrong with my logic?

BTW my logic implies that a fundamental difference exists between outboard & inboard brake installations (I think).
This sounds like sound logic for a rigid body. The COG height won't affect the reaction force. However i think when the vehicle pitches from the reaction (non rigid body, with suspension movement) then the height would be a factor.
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DaveW
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Re: Vertical CG on an F1 car?

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Ciro, ringo,

My "thought experiment" was intended to suggest that load transfer would happen when a conventional vehicle was accelerated even if the the suspension was locked & the c.g. was at the ground plane. I'm not completely sure of my logic, hence the post. If true, then an additional transfer component should be added to Ciro's very clear diagram illustrating the "braking" case.