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9 Jan 2013, 20:58 (Ref:3187027) | #101 | ||
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Yep, a car traveling at 3.00x10^8m/s would be so heavy it wouldn't be able to move without an astronomically powerful engine. It'd be pretty hard to see the races happening too. E=mc^2 and all...
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9 Jan 2013, 21:06 (Ref:3187033) | #102 | |||
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The lighter car with downforce will therefore be able to accelerate both laterally and longitudinally than the heavier car without downforce. It is this principle that makes a 500kg FRenault faster than a 1300kg GT car. |
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10 Jan 2013, 13:44 (Ref:3187296) | #103 | ||
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Another example to show my meaning, this time two 500kg cars, one with 100kg downforce and the other with 0kg. Using the same F=ma logic the acceleration would be the same for both cars. I would suggest that F must increase, by applying more vertical load to the tyres, you achieve more max grip from them. This is why squat can aid acceleration. For overall simplification I would suggest that the original 500kg car makes its 100kg downforce at its max cornering speed. Simplistic I know but it is very easy to over complicate. |
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10 Jan 2013, 15:35 (Ref:3187316) | #104 | |||||
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10 Jan 2013, 15:44 (Ref:3187318) | #105 | |
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To a first approximation, the 500kg car will be about 9.5% faster. (square root of 600/500)
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10 Jan 2013, 16:31 (Ref:3187333) | #106 | |
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10 Jan 2013, 17:25 (Ref:3187344) | #107 | |||
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10 Jan 2013, 19:37 (Ref:3187383) | #108 | |||||||
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I made an assumptions about the example given. The 100kg (or 1000N) of downforce was at the particular corner in question. My answer differs from barnettracing two posts later as this user assumed the downforce was a maximum downforce. Quote:
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In the first example, the veritcal load in the example is the same for the downforce car (500+100) and the non downforce car (600). I thought keeping the vertical load/grip the same was the point of the example you gave? Quote:
I'm sure you can't be, but the 100kg of downforce can't be considered as m. Said just in case. Also for clarification we should just be considering acceleration and force as vectors. Acceleration is the rate of change of velocity. In this case it is the change of direction, not magnitude of velocity. Again said just in case. Quote:
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10 Jan 2013, 19:43 (Ref:3187384) | #109 | ||
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As a random comment I wouldn't really describe F=ma as "logic" more a law of physics. Newton's second law of motion.
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10 Jan 2013, 19:53 (Ref:3187388) | #110 | |||
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A simple analogy: Put a 1 kg bag of sugar on your kitchen scales; the scales will show a weight of 1kg. Now press down on the bag of sugar until the scales indicate 1.5kg. Now how much does the bag of sugar weigh? The scales are actually adding two forces - the force exerted by the mass of the bag under the influence of gravity, & the force exerted by your downward pressure. Maybe a little bit pedantic, but downforce should be quoted in Newtons, not Kilograms - that would actually remove a lot of the confusion. |
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10 Jan 2013, 20:07 (Ref:3187396) | #111 | ||
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Yes, while using kg for forces is often done it doesn't aid understanding.
Good analogy about the scales measuring the total force. I am quite enjoying this conversation, so while we are at it. I would consider 'weight' a force. It is the force of a mass caused by a gravitational field. Ie the weight of an object is less on the Moon than on the Earth, but the mass of the object is the same. Of course downforce does not add weight! |
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10 Jan 2013, 20:14 (Ref:3187399) | #112 | |||
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Therefore, if you measured the weight of the car when at vMax it would weigh more, but it wouldn't have an increased mass (ignoring the minute increase caused by its extra kE). Because it would weigh less than the 600kg car, it would be able to accelerate faster for the same amount of force provided (engine power) and it would also be able to change direction quicker. (I know I'm not really arguing with you now David). However, the extra downward force created by the wings would provide the car with even more 'grip' as it is effectively being squashed into the tarmac. For this reason, the 500kg car with '100kg' downforce will be quicker than the 600kg car. It will accelerate (and decelerate) faster and will be able to change direction quicker. Furthermore, the faster it goes, the more downforce it generates, enabling it to go faster, enabling more downforce etc. At this point, the limiting factor becomes the force that the tyres are able to create through their chemical interaction with the tarmac. Again, this is all off the top of my head, but I believe my reasoning is less flawed than last night. And yes, forces should be in Newtons (N) rather than kilograms (kg). |
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10 Jan 2013, 20:24 (Ref:3187402) | #113 | ||
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Look at us all trying to say the same thing in different ways!
On the go quicker, more downforce, means you can go quicker, more downforce, means... I don't think this ever over rewards. For example I don't think there is ever a situation where a corner isn't possible at 100mph, but is possible at 110mph! Don't tell Gilles that though |
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10 Jan 2013, 23:22 (Ref:3187485) | #114 | |
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Good discussion.
Regarding earlier posts I think I got confused as everybody is suggesting different meanings for "F" Any excuse to post this When you resolve all the vertical forces it doesnt matter whether the 600kg (or 6,000N if you prefer) comes from mass or downforce, the vertical loads on to the tyres will all be the same. Therefore the friction generated by the tyres will be the same. If all the forces are the same then the car is going at the same speed. Using the earlier reference to pushing down on a sugar bag, I dont see why it matters whether you add extra sugar to the bag or push down on it, as long as the vertical load sums to the same figure, it will require the same force to push it along the floor. I feel like I am missing something obvious because nobody else agrees |
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10 Jan 2013, 23:37 (Ref:3187487) | #115 | |||
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The force is the same. Yes. The mass is lower for the downforce car in this example. 500kg v 600kg. So the change of direction is quicker for the lower mass car. The rate of change of direction is the acceleration. It can have a higher corner speed (the magnitude of velocity) as it changes direction quicker. 500xa1=F1=F2=600xa2 therefore a1>a2 (Sorry can't do proper subscripts) Cool old diagram Last edited by Adam43; 10 Jan 2013 at 23:44. |
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11 Jan 2013, 00:10 (Ref:3187498) | #116 | |||
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It may also be that vintage racers, and their crews, don't have the right height, rake, etc. right, and a car property set up by a professional team wouldn't experience that. Anybody know? |
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11 Jan 2013, 00:22 (Ref:3187499) | #117 | |
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Once you have figured out your downforce and the ability of the tyres to resist sideways motion i.e. Centripetal Force
The magnitude of the centripetal force on an object of mass m moving at tangential speed v along a path with radius of curvature r is:[5] where is the centripetal acceleration. The direction of the force is toward the center of the circle in which the object is moving, or the osculating circle, the circle that best fits the local path of the object, if the path is not circular.[6] The speed in the formula is squared, so twice the speed needs four times the force. The inverse relationship with the radius of curvature shows that half the radial distance requires twice the force. This force is also sometimes written in terms of the angular velocity ω of the object about the center of the circle: from http://en.wikipedia.org/wiki/Centripetal_force (Good instructive article) Downforce generates grip, but does not add to mass, so all the grip is available to overcome the forces generated by a smaller mass in the above equation. |
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11 Jan 2013, 13:41 (Ref:3187729) | #118 | |
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I see now
A plane has so much lift that it flies but the kinetic energy is a property of the mass of the object so it still takes force to accelerate it. duh. lol. |
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11 Jan 2013, 15:03 (Ref:3187767) | #119 | |
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So, can lap times actually be maintained with half the downforce?
It seems as though it's not an easy question to answer. Not even if you only apply it to the current 750 bhp F1 cars. There's also the question of refueling and tyre changes, which could have a significant effect on an F1 car being able to 'maintain' lap times over the course of a race distance. |
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11 Jan 2013, 18:13 (Ref:3187836) | #120 | |||
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12 Jan 2013, 00:26 (Ref:3187983) | #121 | ||
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Of interest would it be easier to achieve at a slow circuit like Monaco or a high speed circuit like Monza/
Monaco, more torque less wieght out of slow corners. Greater mechanical grip from wider tyres. Less appendages to knock off on kerbs or barriers. Monza, More power, less drag to achieve higher terminal velocity. |
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12 Jan 2013, 01:14 (Ref:3187989) | #122 | |
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Your post above gets us back to here Oldtony. Still a very interesting question in my book. |
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12 Jan 2013, 01:15 (Ref:3187990) | #123 | ||
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It was interesting to note that the HRT cars, despite having engines with competitive bhp figures, were pretty much hopeless everywhere. Having not that much less downforce than their contemporaries, they suffered from every handling flaw under the Sun, whether it be Monaco or Monza. Brake problems and 'locking up', through lack of downforce going into the corners, was also a big issue for them. |
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12 Jan 2013, 01:30 (Ref:3187996) | #124 | ||
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12 Jan 2013, 01:33 (Ref:3187997) | #125 | |
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Those 23 seconds to find don't seem a lot. But what is the downforce of an F3 car compared to an F1 car? Are F3 cars also considered to be too reliant on aero for grip? I don't know.
A 650 bhp, lightened F3 car with half of the downforce of a current F1 car. Sounds interesting. Later versions of the 4 cylinder turbocharged 'Offy' Indy car engines produced 800 bhp. One of those in an F3 car would be interesting just by itself. Also very interesting to drive, I shouldn't wonder! |
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