danb1990 Posted January 31, 2009 Share Posted January 31, 2009 Right, I've been looking into race car aerodynamics. Using the formula: D=0.5 x (WxHxA) x C x Ad x V where D is downforce in Kg, w is the width of the aerodynamic aid, and H is the height (both in metres), A is the angle of attack in degrees, C is the drag coefficient (constant), Ad is air density in kg per cubic metre, and V is velocity in metres per second. I have already calculated that C is 0.369 for my tourer, Ad is variable by temperature but using a theoretical constant temperature of 20 degrees celsius the value of Ad is 1.2041. And the value of V I am using is a max speed of 25kph (calculated on Gearchart) which is roughly 7 metres per second, which I shall call Vmax Now, from this I have calculated that the front splitter of my R9-R shell (190mm wide by 18mm tall) is producing 0.47kg of downforce at Vmax. The problem arises when I consider the rear wing. Do I measure the height of just the wing, or do I measure from the top of the wing to the ground? Because I have been doing the latter and calculated from this that the rear wing is producing 4.5kg of downforce, which sounds a little too much, but could be believeable. A quick recalculation of D using the approximate height of the wing (30mm) brings up a value of 1.17kg of downforce, which sounds more realistic considering the front is producing 0.47kg. I hope there's someone who can help! Dan Quote Link to comment Share on other sites More sharing options...
joe of loath Posted January 31, 2009 Share Posted January 31, 2009 use the approximate height of the wing. a CD of 0.37 sounds quite high. I think a citroen 2CV is 0.4. I would expect low 3's/high 2's for a TC. oh, and remember, the whole bonnet should create downforce. try with the size of the bonnet and see how realistic the figures are. Quote Link to comment Share on other sites More sharing options...
danb1990 Posted January 31, 2009 Author Share Posted January 31, 2009 (edited) Thanks very much, just one thing I have gone wrong with is that I was multiplying by the velocity, where I should have been multiplying by (velocity) squared. The only problem is that this gives an approximation of the rear wing downforce to be 8.15kg, which sounds not very realistic. A quick recalculation of the front splitter downforce gives 3.35kg. This sounds less realistic than my original calculations which had a mistake in them? Also, I am trying to experiment with extra add ons (I know not BRCA national legal but only racing at club level where they are not so fussy), like extra winglets and sprouty things coming out, a 1:1 motorsport example being the nosecone horns on the 2008 McLaren at Brazil: Will post calculation results later Dan PS: Cheers for the tip on the bonnet, will take a look later. Also, I agree that the drag coefficient sounds rather high, however a Bugatti Veyron has a coefficient of 0.36, so I'm not sure. I googled it to find a formula for drag coefficients, but couldn't find one, however an Excel spreadsheet came up where you put in a value of Vmax in kph, the mass of vehicle, then let off the throttle (no brakes) and estimate the speed at 10 second intervals, which was extremely difficult, but eventually found me a coefficient of 0.36. I don't trust it completely but I shall have to keep using it unless I find a formula I can use, or somebody out there knows the exact coefficient of an R9-R shell? I shall give it a go with something in the range you suggested and come back to you with that. Edited January 31, 2009 by danb1990 Quote Link to comment Share on other sites More sharing options...
rob_b88 Posted January 31, 2009 Share Posted January 31, 2009 (edited) If you are calculating down force (or -ve lift), you should be interested in the lift co-efficient, not the drag co-efficient. These values are effectively the aerodynamic cleanliness of the model, so if you add extra spoilers etc, they will increase accordingly. Also the angle of attack can be taken out of the equation, as that will only affect the lift coefficient. i.e higher AoA, higher lift co-efficient (and drag co-efficient). Also if you are trying to calculate the lift of a specific wing, you will need the specific lift/drag co-efficient for that wing. The overall lift/drag co-efficient of the car is going to give the wrong results. Edited January 31, 2009 by rob_b88 Quote Link to comment Share on other sites More sharing options...
danb1990 Posted February 1, 2009 Author Share Posted February 1, 2009 If you are calculating down force (or -ve lift), you should be interested in the lift co-efficient, not the drag co-efficient. These values are effectively the aerodynamic cleanliness of the model, so if you add extra spoilers etc, they will increase accordingly. Also the angle of attack can be taken out of the equation, as that will only affect the lift coefficient. i.e higher AoA, higher lift co-efficient (and drag co-efficient). Also if you are trying to calculate the lift of a specific wing, you will need the specific lift/drag co-efficient for that wing. The overall lift/drag co-efficient of the car is going to give the wrong results. Thanks very much, I was perplexed when I got a result of nearly 8kg of downforce from the rear wing! Quote Link to comment Share on other sites More sharing options...
amnesia Posted February 1, 2009 Share Posted February 1, 2009 I would imagine you also get ground effect with low ride-heights and virtually flat chassis bottoms on most touring cars Quote Link to comment Share on other sites More sharing options...
mad-wolfie Posted February 1, 2009 Share Posted February 1, 2009 Right, I've been looking into race car aerodynamics. Using the formula: D=0.5 x (WxHxA) x C x Ad x V where D is downforce in Kg, w is the width of the aerodynamic aid, and H is the height (both in metres), A is the angle of attack in degrees, C is the drag coefficient (constant), Ad is air density in kg per cubic metre, and V is velocity in metres per second..... you lost me at this point. Quote Link to comment Share on other sites More sharing options...
glypo Posted February 1, 2009 Share Posted February 1, 2009 (edited) I can't really figure out your maths to be honest. W*H*A is total rubbish I feel. As are basically most the comments above by everyone (sorry, but it's true). How on earth can there by any generic factor where you can multiple a scalar in this manner with the wing alpha. The basic lift equation, and one that should always be used is: L = 0.5 * rho * (v^2) * cl * s So your S is reference length, typically you can use wing chord or span, just ensure the reference stays the same. Rho is density at ground level. You are kind of right, but again kind of wrong on this. Just pick the value for sea level from the ISA, this is 1.225kg/m^3. Cl is the lift coefficient of the wing. Simple as that. Also make sure velocity is also in SI units (m/s). This is how you work out the wing. To be honest I have no idea how you worked out your Cl's before without any kind of computational fluid dynamics. If I was you I would just chuck your wing geomety in a basic Kutta Joukowski solver, or one of panel methods that has the NACA 4xxx aerofoils already built in for basic wing design and analysis. Edited February 1, 2009 by glypo Quote Link to comment Share on other sites More sharing options...
joe of loath Posted February 1, 2009 Share Posted February 1, 2009 I can't really figure out your maths to be honest. W*H*A is total rubbish I feel. As are basically most the comments above by everyone (sorry, but it's true). How on earth can there by any generic factor where you can multiple a scalar in this manner with the wing alpha. The basic lift equation, and one that should always be used is: L = 0.5 * rho * (v^2) * cl * s So your S is reference length, typically you can use wing chord or span, just ensure the reference stays the same. Rho is density at ground level. You are kind of right, but again kind of wrong on this. Just pick the value for sea level from the ISA, this is 1.225kg/m^3. Cl is the lift coefficient of the wing. Simple as that. Also make sure velocity is also in SI units (m/s). This is how you work out the wing. To be honest I have no idea how you worked out your Cl's before without any kind of computational fluid dynamics. If I was you I would just chuck your wing geomety in a basic Kutta Joukowski solver, or one of panel methods that has the NACA 4xxx aerofoils already built in for basic wing design and analysis. this guy knows his stuff Quote Link to comment Share on other sites More sharing options...
rob_b88 Posted February 1, 2009 Share Posted February 1, 2009 ... The basic lift equation, and one that should always be used is: L = 0.5 * rho * (v^2) * cl * s So your S is reference length, typically you can use wing chord or span, just ensure the reference stays the same. Rho is density at ground level. You are kind of right, but again kind of wrong on this. Just pick the value for sea level from the ISA, this is 1.225kg/m^3. Cl is the lift coefficient of the wing. Simple as that. Also make sure velocity is also in SI units (m/s). This is how you work out the wing. To be honest I have no idea how you worked out your Cl's before without any kind of computational fluid dynamics. If I was you I would just chuck your wing geomety in a basic Kutta Joukowski solver, or one of panel methods that has the NACA 4xxx aerofoils already built in for basic wing design and analysis. Firstly S is not a length, it is an area. You are working with a cross-section, hence one of the dimensions is assumed to be unitry. Secondly, rear wings on RC cars don't conform to any NACA aero foil shapes. They do not rely on the airflow around both surfaces, only the upper surface. They tend to be more ______/ shape. They work by deflecting the air over the upper surface upwards, thus pushing the rear down. In his second post danb1990 realised that he should be using the square of velocity, thus you have just quoted pretty much the same equation. You can use sea level ISA density (1.225kg/m^3)... as long as you at sea-level and it is standard ISA sea-level temperature and pressure... Otherwise the density will be different. Higher temperature or pressure, higher density. You are right in that it is not possible to assume the Cl (and Cd) values. These can only accurately be calculated from wind-tunnel testing, or complex CFD modelling. Quote Link to comment Share on other sites More sharing options...
glypo Posted February 7, 2009 Share Posted February 7, 2009 (edited) Firstly S is not a length, it is an area. You are working with a cross-section, hence one of the dimensions is assumed to be unitry. Secondly, rear wings on RC cars don't conform to any NACA aero foil shapes. They do not rely on the airflow around both surfaces, only the upper surface. They tend to be more ______/ shape. They work by deflecting the air over the upper surface upwards, thus pushing the rear down. In his second post danb1990 realised that he should be using the square of velocity, thus you have just quoted pretty much the same equation. You can use sea level ISA density (1.225kg/m^3)... as long as you at sea-level and it is standard ISA sea-level temperature and pressure... Otherwise the density will be different. Higher temperature or pressure, higher density. You are right in that it is not possible to assume the Cl (and Cd) values. These can only accurately be calculated from wind-tunnel testing, or complex CFD modelling. You are spot on about the S, not sure why I made such a bad error. Looks like I essentially joined two sentences together. But yes the reference area applies for the wing, and rest of the sentence is fine. Trust me you don't use cross sectional area for a wing for reference! Typically you can use planform area, as I suggest in my post (use wing span and wing chord) or if you can be bothered use a wetted area. It is essential you keep reference lengths the same though for all the calculations, or you need to non-dimensionalise all your values when transferring between calculations. And no, we have not quoted the equation, for the very reasons mentioned above. For a non-lifting body such as a fuselage, or a car body, people typically use frontal cross-section. Thus if you look for empirical cl/cd's, as I assume was done in this case, you need to keep reference the same hence by point. However a lifting surface you always tend to look at planform or wetted area, thus empirical data will also be based on this, thus you need to use the standard lift equation I stated above. As for ISA, there is a reason in aerodynamics we use it. There is no such thing as standard! Simply assuming ISA at sea level makes life infinitely easier. ISA was set based on North American and European averages so it's as good as it gets for an average day. Hence everyone else uses it, and if you pull from dimensionalised empirical data you can almost be assured for race-car that ISA was set at sea level, thus non-dimensionalise, transfer to the rc car and then dimensionalise again but with the new values. As for CFD being complicated, it really depends on methods. Sure you can run an Euler, Reynolds-Averages Navier-Stokes..... and they take hours or days even on the big cluster machines at work. However the basic methods I stated don't require griding or computation, just plug in some numbers and get a result. It won't be totally accurate, but it's not as if an aircraft is being certified. Being in the ball-park I imagine would be fine. As for wings being used in RC, I have never seen a proper wing being used in RC so your argument about NACA sections falls over there anyway. If it was me, I wouldn't even use an aerofoil and just do flat plate theory, much easier. Typically you get spoilers on RC cars, so if this guy is trying to work out aerodynamics of a spoiler he is wasting his time using the lift equation. However judging by the fact he has gone to some effort, and done quite well minus a few mistakes, I assume he would be using a custom/special wing of some kind. Edited February 7, 2009 by glypo Quote Link to comment Share on other sites More sharing options...
tt-01 mamba Posted February 7, 2009 Share Posted February 7, 2009 Do the walls of texts say that a standrd touring car can drive on my ceiling?? Thanks Quote Link to comment Share on other sites More sharing options...
joe of loath Posted February 8, 2009 Share Posted February 8, 2009 Do the walls of texts say that a standrd touring car can drive on my ceiling?? Thanks if you get a good enough run up, and the right shaped wall to drive it up (like a giant wall ride, only all the way to the ceiling) it might! Quote Link to comment Share on other sites More sharing options...
danb1990 Posted February 9, 2009 Author Share Posted February 9, 2009 Do the walls of texts say that a standrd touring car can drive on my ceiling?? Thanks Thou shalt not pay attention to James Allen. But as Joe says, go quickly enough up a wall with the perfect angle and it is feasable Quote Link to comment Share on other sites More sharing options...
glypo Posted February 9, 2009 Share Posted February 9, 2009 Yup, but that's basically looping the loop. F1 cars though, probably have more than enough force to drive on a ceiling. Quote Link to comment Share on other sites More sharing options...
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