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By darwin dali
#144268
F1 live:
Wind tunnel testing is limited to 60 per cent (or smaller) scale models, and wind speeds capped at 50 metres per second.


50m/s = 180km/h (112mph). Top speeds are easily over 300km/h. 180km/h is 60% of 300km/h. See the coincidence?
Question for the aero inclined among you: Would 180km/h winds on a 60% model roughly translate to 300km/h winds on a 100% model in terms of data gathering?
User avatar
By Jabberwocky
#144609
I might be a bit slow on the uptake of this really but, if the wind tunnel is 60% the size then the airflow is not going to be completely accurate no matter what you do. unless you pressurised it to 1.6 Bar. so that the airflow would be under that same sort of pressure as it would if it was full size
User avatar
By darwin dali
#144610
I might be a bit slow on the uptake of this really but, if the wind tunnel is 60% the size then the airflow is not going to be completely accurate no matter what you do. unless you pressurised it to 1.6 Bar. so that the airflow would be under that same sort of pressure as it would if it was full size

I doubt they'll scale down the wind tunnel (would defeat the cost savings scheme). What is downsized is the size of the car models they're allowed to use.
User avatar
By scotty
#144616
Trying to figure it out (intriguing question!) now using this equation:

Image

Where:

* D is downforce in newtons
* WS is wingspan in metres
* H is height in metres
* AoA is angle of attack
* F is drag coefficient
* ρ is air density in kg/m³
* V is velocity in m/s


watch this space... unless a better mathematician works it out before me! :P

edit: wait, it doesn't even require figuring out the actual numbers, but in theory you'd have to substitute the full scale figures for those that are exactly 60% in order to get an accurate 60% 'prediction', yet is it even possible for teams to simulate certain levels of air density in a wind tunnel?? Is that even relevant?!

Think i'm out of my depth here, ask me again in a couple of years... :whistling::blush::P
User avatar
By Denthúl
#144618
Just to point out, the 2009 sporting regulations contain the same lines.

h) No wind tunnel testing may be carried out using a scale model which is greater than 60% of full size.
i) No wind tunnel testing may be carried out at a speed exceeding 50 metres/second.


So it wouldn't make a difference to how things are now. :)
User avatar
By darwin dali
#144621
Trying to figure it out (intriguing question!) now using this equation:

Image

Where:

* D is downforce in newtons
* WS is wingspan in metres
* H is height in metres
* AoA is angle of attack
* F is drag coefficient
* ρ is air density in kg/m³
* V is velocity in m/s


watch this space... unless a better mathematician works it out before me! :P


Ok, I plugged in some numbers, i.e., I left everything the same for a 100% and 60% model except scaling down H, WS and of course V (which is obviously squared). In theory you should get about 60% for D in a 60% model, but you don't - it's more like 8%.
User avatar
By Jabberwocky
#144623
that is why I said about increasing the air pressure. but then again I am sure the F1 brains just times it by 12 to get 100% levels
User avatar
By Jabberwocky
#144624
saying that these are not single aerofoils. if thing's are being used to direct air flow to a different area of the car it might work well in 60% world but not at all in 100%. maybe that is where McLaren went wrong at the start of the year
User avatar
By scotty
#144626
Ok, I plugged in some numbers, i.e., I left everything the same for a 100% and 60% model except scaling down H, WS and of course V (which is obviously squared). In theory you should get about 60% for D in a 60% model, but you don't - it's more like 8%.


Yeah, that doesn't make sense. Surely to get a D that is 60% of the figure produced by the full scale model, ALL the figures involved in the equation would have to be 60% too? Hence my earlier edit regarding air density.

I can't see any way that the drag coefficient would correlate with a smaller model and reduced speed either.

saying that these are not single aerofoils. if thing's are being used to direct air flow to a different area of the car it might work well in 60% world but not at all in 100%. maybe that is where McLaren went wrong at the start of the year


Or conversely, any problem that may appear to be small could be exaggerated by 40%+...
User avatar
By darwin dali
#144664
Ok, I plugged in some numbers, i.e., I left everything the same for a 100% and 60% model except scaling down H, WS and of course V (which is obviously squared). In theory you should get about 60% for D in a 60% model, but you don't - it's more like 8%.


Yeah, that doesn't make sense. Surely to get a D that is 60% of the figure produced by the full scale model, ALL the figures involved in the equation would have to be 60% too? Hence my earlier edit regarding air density.

I can't see any way that the drag coefficient would correlate with a smaller model and reduced speed either.

saying that these are not single aerofoils. if thing's are being used to direct air flow to a different area of the car it might work well in 60% world but not at all in 100%. maybe that is where McLaren went wrong at the start of the year


Or conversely, any problem that may appear to be small could be exaggerated by 40%+...

Actually no, not all the figures have to be 60% - note that V is squared!
User avatar
By madbrad
#144667
But some of those forces increase in a logorythmic scale. You can't just make everything 60% and think it comes out the same.

Anyway, doesn't making a 60% model that you were not otherwise going to make INCREASE the cost of being in the sport?
User avatar
By Frosty
#145117
surely no matter what you do the results are still going to be slightly screwed because a low wind speeds of 180kmh the airflow is nearly none compressive but when you get to higher speeds of 300kmh you will get something near to 5% compression right? Or can the compression also be solved by equations?

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