which of the black triangles produces less drag?

>>48777141
yes, spoilers work on /tg/

>>48777141
The lower triangle would have more initial resistance, but create less of a wake.
The top triangle would generate an area of turbulence behind it that would form a "cushion" at significant speeds that would serve to reduce the resistance.
Someone wanting to take advantage of drafting would do better keeping close to the top vehicle.

>>48777141
bottom one as it's got a high surface area in contact with the winds.

Unless these are the winds of magic, in which case the pointier one will go faster cuz it's roight sharp like.

File: cones with fixed description.png (11KB, 602x518px) Image search: [Google]

11KB, 602x518px

sorry lads the text was incorrect i fixed it. also, how is this thread /tg/ related?

>>48777231
are you drunk or something?

Neither. Air produces the drag, triangles are the surface it's measured upon.

>>48777141
Upper (sharp edge into the wind) will have a drag coefficient of around 0.5. Lower will have a drag coefficient of around 1.14. These are very, very approximate numbers.

>>48778750
I know literally nothing about drag, would you mind explaining to me how to get such numbers and what exactly they mean?

File: drag-cone.jpg (35KB, 550x438px) Image search: [Google]

35KB, 550x438px

>>48779136
Didn't go through any rigorous aerodynamic analysis, just used past testing information. Though I did just take values for other angle cones, instead of calculating them myself.

>Doing it properly:
Measuring shows them to be 400px long with a 50px radius, so ε is ~7 degrees for the upper one, and ~173 degrees for the lower one. Looking at this handy chart, we can see that this gives the upper a CD (3D drag coefficient) of about 0.25, while the lower has a CD of around 1.4.

The 3d drag coefficient lets us calculate the drag force (D) if we also know the air speed (V), the density of the air (ρ) and the cross-sectional area (S). Specifically, D = CD * S * (0.5 * ρ * V^2).

For both of these, the cross-sectional area will be the same. Assuming a cross-sectional area of 1m^2 and sea level density of 1.225 kg/m^3 to simplify the math, and after converting the 100 km/h to 27.8m/s, we can then calculate the drag to be:
>D(Top) = .25 * 1 * (0.5*1.225*27.8^2) = 120 newtons, or about 27 lbs
>D(bottom) = .25 * 1 * (0.5*1.225*27.8^2) = 660 newtons, or about 150 lbs

The numbers are a bit higher if they're wedges instead of cones, but overall the top is still going to be much better than the bottom. Flow separation is going to be the main cause of the difference. Drag increases drastically after the flow separates. The top shouldn't have to deal with flow separation until the airflow reaches the end of the cone, while the bottom one will pretty much instantly separate.

>>48780784
Derp, should be 1.4 instead of 0.25 in the second greentext equation, but the result is correct. Just forgot to change it in the copy-paste.

>>48777141
Nigga, which way do arrows point?

>>48780784
that's pretty neat, thanks

why is the difference between wedge and cone so huge? is it because a cirular edge for turbulences to start is less horrible than a square one?

>>48780784
as i pointed out several days ago in a 40k thread, the air forces of the imperium aint short of a joke

>>48777156
underrated post

>>48777280
>>48777231

Oh come on, I don't even play Warhammer and I recognize Ork speak.