Drag & mass calculations, am i missing something?

[101m4n]

I have noticed something rather peculiar things whilst mucking around with various plane designs, particularly to do with the mass of a plane affecting drag forces, which is rather peculiar.

So earlier today I devised a simple experiment to clear things up, my plan was to fly two different planes one with a single fl-t500 fuel tank (liquid rocket fuel) and another with a single mk1 aviation tank.

as seen here:

Now, these two craft have identical aerodynamic properties, supposedly. The only difference between them is mass, as the fl-t500 fuel tank has a mass of 2.5 whilst the mk1 has a mass of 1.2. note that the drag ratings for these parts are also identical.

my method: (note that mech-jeb was used for accuracy)

- set heading of 90.4 degrees (runway is aligned to 90.4 for some reason)

- set a pitch of 15 degrees

- throttle up to 2 divisions below maximum thrust

- fire engines

- record speed at 10,000 meters

- rinse and repeat.

Now for a little theory, if lift is proportional to the wings attack angle relative to it\'s velocity vector (which i believe to be the case currently) and the vertical and horizontal components of force in the form of thrust from the engines are also the same (which they are, as pitch and thrust setting were kept constant), then one would expect to find that the two identical aircraft climb through the same altitude at the same velocity, regardless of differences in mass. however this does not appear to be the case as seen below:

the single mk1 craft: (average 473m/s at 10,000 metres) 5 repeats

the single fl-t500 craft: (average 407m/s at 10,000 metres) 5 repeats

as you can see, there is a very noticeable difference between the two crafts speeds at 10,000 meters, the surface info panel was also included as there may be something that i missed there.

either way, this would seem to imply that the drag force on an aircraft is proportional to it\'s mass in some way, which is not terribly realistic to say the least :S, if i am missing something obvious, or there is something horribly wrong with my method (entirely possible), then please do point it out. Else i am afraid that this may be something for the developers to explain

[softweir]

Yes, there appear to be some very.... interesting.... terms in the drag calculation that have sparked a great deal of discussion over the months!

The current drag system is a placeholder - just about realistic enough that the game doesn\'t fail spectacularly, but no more than that. I gather that the intent is to leave things as they are until the team feel brave enough to replace the drag system with something very much better, including proper wing-lift dynamics. (Currently, stall is not simulated. At all.)

[Shadownailshot]

Your vertical speeds are also WAY different. Probably due to the mass though. That may cause for a variation in the actual speed readout. I don\'t feel like doing the numbers, but your average speed may actually be the same. Or at least closer than it looks at a first glance.

[101m4n]

(Currently, stall is not simulated. At all.)

haha stalls... the number of times that i have simply drifted the full length of the runway...

fair enough i guess though, best alpha i have seen for a long while not been playing very long but it has proven to be great fun despite the few shortcomings, excellent community also

Your vertical speeds are also WAY different. Probably due to the mass though. That may cause for a variation in the actual speed readout. I don\'t feel like doing the numbers, but your average speed may actually be the same. Or at least closer than it looks at a first glance.

so they are... hadn\'t noticed that

ground-speed is what i was reading however, and as the two craft have the same forward component of thrust acting on them (same setting altitude and pitch) so they should still be travelling at similar forward velocities.

but this is besides the point, if as sofweir says, the simulation algorithms are liable to change in the not too distant future, then im not at all worried by this

thanks anyway

[Kosmo-not]

The current drag model in KSP is:

drag_force = 0.5*maximum_drag*density*mass*velocity^2

[Zephram Kerman]

The current drag model in KSP is:

drag_force = 0.5*maximum_drag*density*mass*velocity^2

What Kosmo-not means is the KSP drag model includes an extra mass term that does not appear in real world physics.

For example, a full fuel tank has more drag than an empty one. So if you decouple both tanks in the atmosphere, the full tank falls more slowly than the empty one. In the real world, it would be the other way around.

I think it\'s awesome that you noticed this independently, and proved it scientifically using a different method.

[Midnight Star]

What Kosmo-not means is the KSP drag model includes an extra mass term that does not appear in real world physics.

For example, a full fuel tank has more drag than an empty one. So if you decouple both tanks in the atmosphere, the full tank falls more slowly than the empty one. In the real world, it would be the other way around.

I think it\'s awesome that you noticed this independently, and proved it scientifically using a different method.

In the real world, both tanks would fall at the same rate until they reached terminal velocity :3 in which case, the full tank would have a higher terminal velocity and would fall faster.

[101m4n]

The current drag model in KSP is:

drag_force = 0.5*maximum_drag*density*mass*velocity^2

right ok then, that explains a lot!

i may try messing with that at some point in the future

[Zephram Kerman]

In the real world, both tanks would fall at the same rate until they reached terminal velocity :3 in which case, the full tank would have a higher terminal velocity and would fall faster.

Yes, good point. The way you said it is better. Although, technically, the difference would appear gradually while approaching V_t, blah blah blah. Besides, when is Kerbal stuff ever falling slower than that? So your argument is invalid kerboom. ;D

[closette]

Actually the mass term in the drag equation means that a full and an empty tank should both fall together, accelerating at the same rate and reaching the same terminal speed.

For a falling object the downward acceleration of a falling part in the direction of 'g' is given by

accel = {Net Force}/ mass

= {Weight - Drag}/mass

= (mass x gravity - 1/2 x mass x maximum_drag x density x speed²) / mass

= g - 1/2 x maximum_drag x density x speed²

...so mass cancels out in both terms. Setting a=0 one can solve for the terminal speed:

v_T = Sqrt[2g/(maximum_drag x density)]

Unlike in the real world where if you drop a full and an empty styrofoam cup off a balcony you\'ll easily see the difference in accelerations and terminal speeds.

The atmospheric density on Kerbin for drag purposes is given by 0.009785 * exp(-altitude / 5000 m), so you can see that the local terminal speed at the surface for most parts (with maximum_drag = 0.2) is about 100 m/s for any mass.

Edited August 8, 2012 by closette
included density term