zilspeed
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| posted on 20/9/10 at 08:35 PM |
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G forces
Seeing as I'm a sad git, I have spent some time today sifting through data logging figures.
I'm telling you, give me a mug of tea, some cholestorol and an excel spread sheet full of numbers and I'm in heaven.
Anyway.
We data logged a fairly typical road trip and one of the things which I established was that the G forces in an everyday road car really are pretty
tame.
0.81 G giving it the full beans.
0.52G on a reasonably firm non emergency stop.
0.6G lateral, so cornering force.
How do we reckon this compares to what can be achieved with a lightweight sportscar like one of ours, or say a racing car.
As a reference, it seems an F1 car can acheive pretty much 5G in lateral and braking with 1.45G under acceleration.
Or, is it all a load of old cobblers ?
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lsdweb
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| posted on 20/9/10 at 08:42 PM |
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Poor resolution but some data of my single seater on a hillclimb - here
Wyn
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andyw7de
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| posted on 20/9/10 at 08:46 PM |
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John you really do need to get out more
Speed is just a question of money !
How fast do you want to go ?
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A1
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| posted on 20/9/10 at 08:48 PM |
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if i remember correctly, the ultima gtr can reach 1.6g cornering
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mcerd1
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| posted on 20/9/10 at 08:52 PM |
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quote:
Originally posted by A1
if i remember correctly, the ultima gtr can reach 1.6g cornering
isn't that on a 200 ft diamiter circle (that test the americans love to do)
-
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brianthemagical
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| posted on 20/9/10 at 08:58 PM |
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Assuming no aero loading, even weight distribution and no funny business then the maximum accelerative forces are equivalent to the Cf of the tyre and
the gravity under which the vehicle is operating, thus its Cf G's.
So if the Cf of the tyre is 1, then the vehicle will accelerate at 1 G.
If anyone knows anything about the complex life of a tyres Cf while in use, then they'll know it's very unlikely for this to occur in the
real world.
Down force would increase the above example proportionally, thus it now becomes the ratio between the normal tyre load and the mass of the vehicle,
multiplied by Cf. But as we know the Cf of the tyre will alter, so it's not easy for the average driver to use this for calculations.
Ultimately, the higher the G value you log, the faster the vehicle will corner.
All you need to do is increase the aero loading and it will/should increase the cornering force, if you have the power to move the vehicle fast enough
to realise the downforce to generate the normal loading. Such an amount of downforce will cause drag, which would need power to overcome, thus
bringing the circle of traction into play and making the whole situation well beyond anything low cost.
[Edited on 20/9/10 by brianthemagical]
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zilspeed
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| posted on 20/9/10 at 09:08 PM |
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quote:
Originally posted by andyw7de
John you really do need to get out more
Is the correct answer. 
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craig1410
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| posted on 20/9/10 at 09:36 PM |
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Nissan Skyline's are known to achieve 1.3g in race mode with R888 tyres. I don't think that there would have been any significant aero
loading so it's fair to conclude that the friction co-efficient of the tyres in a dynamic situation can be greater than 1.0
The Lotus Europa used to hold the world record for lateral 'g' if I remember correctly and it was just shy of 1.0g. In fact here is the
Wikipedia article. It was 0.9g and this was on 1960's tyres... http://en.wikipedia.org/wiki/Lotus_Europa
From what I've read, a modern street tyre will produce up to 1.2 and sticky racing tyres even higher. I've seen a wikipedia article
suggesting that 1.7 is possible but there was no citation so take that with a pinch of salt.
The biggest benefit a Locost type car has is a low centre of gravity and relatively short suspension travel so roll angle is kept to a minimum and
suspension geometry remains close to optimum, assuming it was optimal before you entered the corner of course.
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rodgling
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| posted on 20/9/10 at 10:48 PM |
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.8g under acceleration is surprisingly good for an everyday road car - I don't think I've managed anything that high in my Elise (standard
1.8 k-series). The best I've managed according to the iphone is .71 under acceleration, 1.23 under braking, and 1.18 laterally (recorded some
slightly higher lateral numbers, but I think those were with a spin). All on road tyres (Kumho 31s), so much higher numbers should be possible for
braking/cornering.
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BobM
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| posted on 21/9/10 at 06:48 AM |
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From my logs on my FuryBusa at Snett on Sunday:
Accel 0.6g
Braking 1.05g
Lateral 1.3g
Clearly not trying hard enough ...
Not very Locost but very BEC 
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britishtrident
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| posted on 21/9/10 at 06:55 AM |
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Depends what you consider a reasonably firm stop.
On a Locost foot brakes should be able to pull 1g with ease even 1950s Ford could pull 0.85g peak braking decelleration without too much problem,
istr 0.5g is the minium legal requirement.
[I] “ What use our work, Bennet, if we cannot care for those we love? .”
― From BBC TV/Amazon's Ripper Street.
[/I]
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hughpinder
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| posted on 21/9/10 at 07:04 AM |
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The only information I have is for Avon pro 13" cart tyres:
coef of friction (dry) = 2-(0.000185*F), where F is the normal force in newtons on the tyre - say 150kg approx 1500N give Cf of 1.7.
Avon state wet = 0.87-(0.00064*f) = 0.7, hence the big difference between wet and dry grip!
ETA
So for a road car, say 400kg/wheel at the front, Cf 1.25 in the dry, .55 in the wet.
ETA2
Your standard road car does 0-60 in 3.4 sec!!!!!
Regards
Hugh
[Edited on 21/9/10 by hughpinder]
[Edited on 21/9/10 by hughpinder]
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chris mason
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| posted on 21/9/10 at 07:16 AM |
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The best i managed in the Sr2 were;
accel: 0.52 (well it's only a 1600)
decel: 0.91 (was hoping for better than that but the brakes were new and needed bedding in)
Lateral: 1.90 (which i thought wasn't bad, the car really does corner well)
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matt_gsxr
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| posted on 21/9/10 at 09:44 AM |
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quote:
Originally posted by brianthemagical
Assuming no aero loading, even weight distribution and no funny business then the maximum accelerative forces are equivalent to the Cf of the tyre and
the gravity under which the vehicle is operating, thus its Cf G's.
So if the Cf of the tyre is 1, then the vehicle will accelerate at 1 G.
[Edited on 20/9/10 by brianthemagical]
This is true for deceleration where all 4 wheels contribute, and would be true for acceleration if your had 4 wheel drive. But normally not all the
cars weight will be on the rear wheels. Just something to remember.
Matt
[Edited on 21/9/10 by matt_gsxr]
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brianthemagical
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| posted on 21/9/10 at 10:05 AM |
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quote:
Originally posted by matt_gsxr
quote:
Originally posted by brianthemagical
Assuming no aero loading, even weight distribution and no funny business then the maximum accelerative forces are equivalent to the Cf of the tyre and
the gravity under which the vehicle is operating, thus its Cf G's.
So if the Cf of the tyre is 1, then the vehicle will accelerate at 1 G.
[Edited on 20/9/10 by brianthemagical]
This is true for deceleration where all 4 wheels contribute, and would be true for acceleration if your had 4 wheel drive. But normally not all the
cars weight will be on the rear wheels. Just something to remember.
Matt
[Edited on 21/9/10 by matt_gsxr]
That's why i stated if weight dist is even, and how it's also unlikley under normal driving. Under any form of acceleration (lateral and
both pos and neg transverse) there is likely to be some weight transfer, the Cf's become less than optimum and it's a factor of all four
wheels and there avalible force.
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boggle
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| posted on 21/9/10 at 10:20 AM |
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managed 1.1g on acceleration in the scoobydoo drag car with nos....oh what a feeling
i miss those days
never went round corners or braked hard so didnt record the rest 
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phelpsa
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| posted on 21/9/10 at 11:23 AM |
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I managed a 64ft time of 2.4secs at shelsey on the weekend. If my maths is correct that equates to an average acceleration of 6.8m/s^2 or 0.7g across
the first 64ft. Not too shabby considering its uphill on old, hard and cold tyres
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