Mitsubishi Lancer Evolution Forums banner
1 - 20 of 21 Posts

·
Registered
Joined
·
86 Posts
Discussion Starter · #1 ·
this is the first thread in a planned long succession of aerodynamic analises on the lancer evolution.
since this is the first, i will make a qualifier. I am an aerospeca engineering student at california state polytechnic university, pomona. i have a heightened sense of what happens aerodynamically, but by no means am i an "expert", but i know sufficiently enough for our purposes, because i have a hunch that not many of us on evotuners are in fact professional racers (in that case, if you would like more specific information contact me privately, as this subject loses coherency and interest among the general public quickly)


The vortex generator (vg, hereafter). some believe that it has no real performance enhancing capability, and is only for looks, extra money spent, etc....


the vg actually does have a function on the evo, thats why its on the car - duh

what does it do? first we must look at some automobile aerodynamics to understand why we have shark teeth on the rear roofline.

to keep it simple i'll use a VERY crude picture (hey, engineers cant draw very well, especially with MSpaint)

speaking of school, i must leave for a physics class.
i will return ~8:30 PST and continue, but this should wet your whistle for now :) (which is interesting, because i have told you nothing yet ROFL )
 

·
Registered
Joined
·
86 Posts
Discussion Starter · #2 · (Edited)
right, so where were we?


ah yes the picture........

without the VG's (bad)

with VG's (good)


the red ares shows high pressure air, the blue shows low pressure air, and the orange shows approximate air stream over vehicle

sorry for the poor quality, but im doing what i can with what i got

especially note the differences in the airflow (orange) over the rear window, and its proximity to the rear window, which can be characterized by the sixe of the low pressure bubble (flow seperation) on the back window
 

·
Registered
Joined
·
86 Posts
Discussion Starter · #3 ·
so what does this show. if you concentrate on the VG section you will notice a difference. one (the one whith the huge VG) lets more orange in closer to the car's back window.

this means 2 things.
1. there is more airflow to the rear wing (if you have one)
2. the low pressure bubble behind the rear window is reduced.

low pressure bubbles (or seperated flow). in short, they are bad for aerodynamics.
think of this: how does a vacuum cleaner work? well it sucks air into it. how does it do that? by pumping air out thereby creating a low pressure, and the sucking is the air rushing in to equalize the pressures (its really that simple ROFL )
so effectively low pressure bubbles "suck" cars back (this is one of the reasons for rear diffusers and vortex generators, among other things)

how the vg reduces low pressure bubbles. the vg is exactly what it is named for. it creates vorticies. if you dont know a vortex is basically a "whirlpool" you might see in a draining bathub/sink, or a river. but this vortex is made in the air (because air is a fluid, just like water is a fluid) the vortex, if you notice, has a big side, and a small side. at the tip of the vg is the small side. as distance from the vg increases, the size increases. what this does, is infringes on the low pressure bubble, and "mixes" higher pressure air, with the top layer of the low pressure bubble, effectively moving the pressure gradient closer to the car than a non-VGed evo

so what the VG does, it reduces drag on the vehicle [by reducing the low pressure on the rear window] and it increases downforce [by allowing more air to pass over the rear wing's airfoil]
however the vg's also do create their own drag, because the frontal area technically would create a high pressure region, but the benefit outweighs the cost in this case


here is mitsu's official technical write-up on the subject (and they have real CFD pictures!!!!)
http://www.mitsubishi-motors.com/corporate/about_us/technology/review/e/pdf/2004/16E_03.pdf
 

·
Registered
Joined
·
86 Posts
Discussion Starter · #4 ·
now, i know you are thinking "well matt, what about the STi's new contraption above the rear window?"

well, that also aims to reduce the low pressure bubble (or tries to keep the flow from separating). but it does it in a little different of a way. see the vg uses a vortex to mix air of different pressures. the STi more forces air back onto the window by airflow directing. there are good and bad sides to this approach.

1. more air can be directed to the back window, therefore much less low pressure bubble = good
2. that wing, in that steep of an angle of attack creates FAT drag. think of it, drag is calculated by dynamic pressure, drag coefficient, and the reference area. what is the refernce area (the area seen by the free stream, namely the area when viewed parallel to the ground) between the vg's and the STi contraption.

well we have little spikes that have maybe 1 in^2 of area per fin * so many fins. the STi has the effective height of the airfoil times by the length.

time for a rough calculation........

say we have 8 fins and the antenna nub (im pretty sure we have 8, if we dont the math is simple enough that you could figure it, and the results will be so different it wont make a real difference anyway)
ok, so 8 fins * 1in^2 + say the nub is 3 in^2 = 8+3 = 11 in^2 of area.

STi -
we'll be generous and say the effective height is only 1/2" (but im pretty sure its more than that) and whats the width? an evo is 40", so we have 40" * .5" = 20 in^2

well 20 is alot bigger than 11, and we were being generous. the way the drag calculation is setup, the reference area is a multiplier, so we multiply out the dynamic pressure and the Cd, then times by the reference area. well we can see the evo's reference area is almost half that of the STi's, so what would be the quick assumption? that the evo's vg creates half the drag of the STi's fin thing. except they dont have the same drag coefficient. the drag coefficent is basically just another multiplier that shows the dragginess of some object, the closer to 1.0 you are the more draggy you are. so what has more dragginess, a flat sheet, or a fin sticking up? you guessed it the fin has a lower Cd.

so im not saying STi's are stupid for running their version of the flow seperation preventer, they are just taking a different approach to the same problem :eek: , after all they can reduce their low pressure bubble much more than we can (they just create drag while removing it)

so thats pretty much it, as far as VG's go. questions? post 'em up.
im sure there are some other things of note in the mitsu tech report, but they would have nearly negligable effects (like they discuss why they chose the fin shape as opposed to the square that you see on most airplanes)
 

·
Registered
Joined
·
86 Posts
Discussion Starter · #5 ·
here are some vg's on a B-1b lancer (its a supersonic bomber used by the air force)
these are the square shaped vg's, as opposed to the fin shaped ones that we have

on underside of horizontal tail

on side of vertical tail


so you can tell people that your car has features that are found on airplanes (also 737's have them on the tops of their main wings, if you're ever in a commercial airliner you can generally see them)
 

·
Registered
Joined
·
86 Posts
Discussion Starter · #7 ·
ya, we'll see. it takes a long time to type all this stuff. but tit takes like 30 seconds to go over all it in my head, its kinda discouraging :)

so what would you like to hear about next? think of it as story time with matt
 

·
Registered
Joined
·
11 Posts
So lemme ask you this , i took the wing off my evo cause i didnt care for the look or the dam thing in my rear view mirior for that matter , anyway would i benefit from a vortex generator without a wing on my car ? or does it only help if you have a wing?
 

·
Registered
Joined
·
86 Posts
Discussion Starter · #9 ·
thaigurlygirl said:
So lemme ask you this , i took the wing off my evo cause i didnt care for the look or the dam thing in my rear view mirior for that matter , anyway would i benefit from a vortex generator without a wing on my car ? or does it only help if you have a wing?
it completely depends on what you want to accomplish.

running no wing with a vg will reduce drag (both by the way i mentioned earlier by creating vorticies, and by removing the wing which has its own drag component)

however, running no wing will result in less downforce (regardless if you have the vg or not) but going from a MR to a RS config would be a greater difference than going from a GSR to RS config because of the vg's

so if you are solely commuting, the by all means go wingless, you might see a .25-.5 mpg increase. but if you are tracking or doing high speed rins through nevada, well....i would leave my wing on

hope this answered your question
 

·
Premium Member
Joined
·
718 Posts
So for those of us driving an RS adding on the VG's would give a possible bump in fuel milage and likely a bump in top speed due to the reduced drag. As an aerodynamics guy you know that drag in the simplest terms is D=Cf*FA*V^2 (D= drag, Cf = coeficent of Friction, FA = Frontal Area, and V^2 = volocity squared). Since the Velocity component of the formula is squared, the faster your normal cruising speed, the more benefit fuel milage wise you would get from a drag reducing item like the VG's. Also, you should see a substantial gain in top speed as well.

Keith
 

·
Registered
Joined
·
86 Posts
Discussion Starter · #11 ·
:thumb: keith.


as your equation shows as your speed increases the drag increases exponentially. thats why there is tons more drag at 80mph than at 60mph.

but if you REALLY wanted to figure out how much faster you could go with a vg'd RS, that is entirely possible (but it would be somewhat lengthy, and furthermore very difficult to type coherently on an online forum (but i'll try, using excel function commands - stay tuned) )

also, not really important to your post, but the drag fcn is D = Cd * (rho)v^2 * Sref. where Cd is the drag (friction) coefficient. (rho) is the density of air (this is tempoerature and altitude specific, among other things). v is speed. and Sref is the reference area, in cars this is the vehicle frontal area. [but what is interesting, we were taught D = Cd*1/2(rho)v^2*Sref, but from experimental data, im leaning towards that equation sans the (1/2). this could however be due to experimental error, and my disinterest in interpolating and temperature correcting my rho value, so we'll stick with the equation sans the 1/2 to remain consistent with real-world data, until i can examine this better]

so your eqn was good for the point you were trying to make, but if you used that to calculate drag (i dunno why anyone would ever be dumb enough to want to do that :rolleyes: oh wait, me) the units would not work out to a force unit [lbs] or [N], and drag is a force

ok here we go:
assume: elevation 1000ft STP, v=60mph
and the Sref i calculated experimentally with my car
evo's stock top speed (without vg) is 160mph at which point the drag is too great and the engine cannot overpower it

D = (.36) * (2.31e-3 [slugs/ft^3] ) (235 [ft/s] )^2 * (20 ft^2)
= 918.5 [lbf] drag (actually you can calculate this from the car's horsepower as well, but thats another day)
918.5 = (.36-.004) * (2.31e-3) (v)^2 * (20)
v = sqrt (918.5 / (.356*.0462))
v = sqrt (55845.4)
v = 236.3 [ft/s]
v = 161 mph

therefore a 1 mph increase on top speed. which isnt that interesting, but what is interesting is the drag at 160 mph. 920 lbs of drag!!!!!!

see how much fun all these theoretical differences are, and the hard calculations to find out that delta v is only 1 mph





found this, kinda explains the whole thing: http://www.edmunds.com/advice/specialreports/articles/106954/article.html
 

·
Premium Member
Joined
·
718 Posts
That is the first time I have seen slugs mentioned since physics class in Nuclear power School in the Navy! :) Yeah, I was just using the simple formula to show that velocity has the major impact on Drag. I didn't take into account temp and altitude since these are not aircraft flying at 36000 ft.... and I am not a dry lake bed racer taking advantage of the high temp for better top speed.

Did you know that the top speed racers do better on a hot day? The loss in engine power from the air being less dense is not as much of a factor in speed as the reduced drag from the lower density of the air when speeds go higher than 400 mph or thereabouts.

Keith
 

·
Registered
Joined
·
86 Posts
Discussion Starter · #13 ·
i'll keep that in mind if i venture out into 400mph territory. but wouldnt forced induction multiply the power loss through less dense air? hmm, 400 mph is into compressible flow-land (oh my).

ah, slugs. the bastard unit of the metric vs. SAE clash (but not as bastardized as the kg/cm^3 on my boost gauge)
 

·
Premium Member
Joined
·
718 Posts
rammsteinmatt said:
i'll keep that in mind if i venture out into 400mph territory. but wouldnt forced induction multiply the power loss through less dense air? hmm, 400 mph is into compressible flow-land (oh my).

ah, slugs. the bastard unit of the metric vs. SAE clash (but not as bastardized as the kg/cm^3 on my boost gauge)
Actually boosted engines lose less power from less dense air as long as they are set up properly. That is why a stock EVO does much better at the 1/4 mile track (trap speed wise) up in Denver than a stock TransAm does up there, but at sea level the TransAm will have a much higher trap speed than an EVO.

Keith
 

·
Registered
Joined
·
86 Posts
Discussion Starter · #16 ·
i'll really try to get the next installment up next week or weekend, but it might be after that, as i have a huge aero project to compile (im the team leader of a bunch of fools too :rolleyes: )

its not a top secret project, so i'll enlighten those that are interested. My 201 class (that would be the 4th class in the major) is to design a transport plane of 250 + 14 people. read: a comercial airliner that holds 250 passengers and 14 crew. also it has to fly 7k nmi (trans pacific flight) and carry an additional 40k lbs of cargo on top of baggage. must comply with federal regs.

im the team leader for the flight control and avionics group. basically everything you see in the cockpit and that makes the plane move in flight, is our business. we need to design functional flow and block diagrams. then provide physical block diagrams (pick physical parts) the important part of this process is: weight, volume, power, and cost. needless to say, i spent the last couple days on honeywell's website finding cool avionics, communication, and navigation equipment. (the total cost of my parts is ~$650k ;) )

so thats where i am for the next week at least, as we proceed into critical design review - CDR (we're so screwed)
 

·
Registered
Joined
·
6 Posts
and now you should use the poor mans dyno to find the drivetrain lose of an evo in stock form. us awd cars can use it very accuratly, unless your like me sitting in a talon whos speedo only goes to 145...

920 lbs of drag requires a force on it of x which = y hp.

it was something based off of w=f*d or something like that. Do you know what Im talking about? too bad Im sure that your max speed in whatever gear is not at the same hp as your peak hp. but whatever. Ill try and find the exact equation and post it.

(btw the reason why Im bringing up all these dead threads is because Im looking for one on evo x's, lol, sorry)
 

·
Registered
Joined
·
86 Posts
Discussion Starter · #18 ·
in class to calcualte horsepower of a jet engine we do: P = T * V(infinity) * (Rho / Rho sl)

otherwise, the power output is: the thrust * freestream velocity * air density ratio. of course tis unit of power is: lb * ft / s. to convert to horsepower, you divide by 550.

since we know that the unit of a horsepower (butchered) is: (lb*ft/s)/550

to calculate the power required for an airplane, we take: P = D * V. or, power required is drag times velocity

therefore, if we take a force, a speed, and divide them by 550 from our car, we would get a horsepower. a force = drag, a velocity = speed, and 550 = 550.

using numbers that i gathered experimentally, we find:

velocity = 72.5 mph = 106.33 ft/s
drag = 158.9 lb

such that, power required at 72.5 mph is: (158.9 * 106.33) / 550 = 30.72 horsepower

of course drag is a function of V^2, so the drag, and therefore Preq increases exponentially. naturally at some point your Preq will be greater than your power available (Pavail), and that defines your top speed (discounting things like governors, road conditions, fuel, etc)

since drag is a second power term (or function, in the next sentance), you need 3 points to find the coefficients of the quadratic eqn. there are also factors that contribute to drag that are a function of V (road-tire resistance, friction in bearings, etc) and also some that are constant regardless of speed (rolling inertia of tires, etc)

theoretically, there may be some factor that is a function ov V^3, but that would be the change of acceleration with respect to time. im not sure how that would factor in, if at all. at any rate it isnt an aerodynamic problem, perhaps i'll ask my dynamics professor on wednesday (i was wondering about this last week)

well i tried to calculate the theoretical top speed due to drag, but something has gone awray, so i'll check it out and report back when i figure it our (i suspect my calculator didnt like my matrix trickery)
 

·
Registered
Joined
·
86 Posts
Discussion Starter · #19 ·
tstkl said:
920 lbs of drag requires a force on it of x which = y hp.
well drag is a force.

unless you mean the car needs to exert some force to cut through the air at 920 lb of drag? kinda like the reverse of the jet engine equation i mentioned above, so like finding the thrust the car produces? or not so much the thrust (i just said that to sound funny, because cars dont really produce thrust) but rather the pure force that the engine puts out. instead of the torque produced per time that a dyno measures (lb*ft/ s)
 

·
Registered
Joined
·
6 Posts
really what Im talking about is just setting up a basic force equation. at this speed, there is a drag coefficent of D over an area A, producing a force F sub D that the car has to overcome. What Im talking about is really basic to you probably. its just an expantion on power=work x distance?( dunno if thats right, but I think if you just look at the equations for power, you can get what Im getting at.)
 
1 - 20 of 21 Posts
Top