Thrust / Drag question

[Poposhka]

hi all, i'm new here...

I'm a computer nerd who enjoy getting into trouble making hobby programming projects, and this time i'm working on a crude aerodynamics engine. problem is, my physics background is, shall we say, limited. Although i do enjoy the learning experience.

Here's my problem.
I'm trying to calculate the resulting force for an object, given the thrust of the jet engine, the mass of the object, and the drag coefficient. The values i'm working with, to make it simpler, are:

Weight = 3800 kg
Thrust = 17600 N
Air density = 1.225 kg/m^3

I'm dumbing it down a whole lot, using constants for some things ... so my resulting formula looks something like this:

F(t) = (17600/3800) - (0.5 * 1.225 * D * v(t)^2)

Now my big issue is with the drag coefficient D (i'm assuming the reference area is 1 m^2 for simplicity)...

If this craft is to obtain something even close to it's speed at sea level, which is 870 kph, the drag coefficient needs to be 0.0000129.. and that just seems WAY too small if you compare to some common values. If i'm using the drag coefficient that seems about right for this craft (0.25) it achieves only 5.71 m/s, which is just way off.

What am i doing wrong here?

 

[IRstuff]

I'm surprised that you get anything meaningful at all. You're dividing force by mass, resulting in acceleration in the first term. That's one reason for always checking your units.



TTFN

 

[Poposhka]

OK that's a suspicion i had... but here's my thought process, please explain what i'm assuming wrong..

the airframe weighs 3800 kg, and the thrust of the engine is 17600 N. That means that the engine can accelerate a 3800 kg object at 0.46 m/s^2 ... is that correct? no?

 

[IRstuff]

No, it's 4.6 m/s^2, which is 1/2 g, which means that in the absence of drag, it would reach your desired speed in about 52 seconds.

The drag equation results in Newtons, which has the same units as your thrust.

TTFN

 

[MikeHalloran]

The drag coefficient is a nondimensional way to describe how slippery a shape is, _relative_to_ a blunt shape with the same cross sectional area.

So, a truck-size object with a very low Cd may develop a lot more drag force than a brick- sized object in the shape of a brick.



Mike Halloran
Pembroke Pines, FL, USA

 

[Poposhka]

yes mike i'm assuming the reference area is 1 m^2 for simplicity

 

[Poposhka]

IRstuff, point well taken, i made a bonehead division.. 4.63 is correct.

but nevertheless, the resulting Drag coefficient (assuming a reference area of 1 m^2) is still just way too low, 0.001294 just seems strange when i'm looking at the drag coefficient on an airfoil. I know it's nondimensional, and maybe that's my huge mistake in this whole thing.

 

[GregLocock]

I think your main mistake is that you aren't reading what people are writing.

Cheers

Greg Locock

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[Poposhka]

hmm Greg, care to elaborate? I'm a bit lost at the moment.. I'm just trying to find the 2 vectors thrust and drag at time (t), and subtract them to find the resulting vector.


If thrust (17600 N) / mass (3800 kg) = 4.63 m/s^2, and
drag is: (0.5 * 1.225 * 0.02 * v(t)^2)

Am i incorrect to say that the resulting vector is:
4.63 - (0.5 * 1.225 * 0.02 * v(t)^2) ???

I acknowledge that IRStuff said i'm mixing units when dividing force by mass, but was that b/c i got .463 instead of the correct 4.63?

I thought a thrust of 17600 N means that it can accelerate a mass of 17600 kg at a rate of 1 m/s^2, or a 3800 kg mass at 4.63 m/s^2

So if the output of the drag equation is in Newtons, i.e. m/s^2, i am *not* mixing units, right? wrong?


sorry for being so blissfully ignorant, just hoping someone can alleviate my case of blockhead.

 

[btrueblood]

you wrote:
"If thrust (17600 N) / mass (3800 kg) = 4.63 m/s^2, and
drag is: (0.5 * 1.225 * 0.02 * v(t)^2) "

NO!

Thrust is a force (units of Newtons), not an acceleration (units of m/s2).

Thrust you defined as = 17600 N

Let net thrust = FN = Engine thrust - drag

Then vehicle acceleration = dv/dt = FN/mass

You will "need to" numerically integrate the acceleration, since the FN term is dependent on the velocity^2.

 

[Poposhka]

woohoo, that's what i was missing, thanks btrueblood!

 

[IRstuff]

The drag coefficient is, of course, non-linear with v^2, so if you want any kind of accuracy, you'l need to consult a decent book on ballistics.

A simple test of whether Fn is positive will tell whether you can achieve the speed required. Since there's plenty of excess thrust at the speed specified, you can accelerate to the speed desired. A numerical integration is required only to determine how long it will take. But, as indicated earlier, since the drag coefficient is not constant, the numerical integration is of academic interest without the correct function for drag.

TTFN

 

[rb1957]

two things ...

net force on the plane is Thrust-Drag (atleast in the longitudinal direction). Your Thrust is very simplistic, in that it isn't constant with speed. Your Drag is also very simplistic ...
1) drag is based on the wing area (not the frontal area)
2) as noted above CD is a function of speed

it would be easier to express acceleration as (T-D)/m

good luck

 

[sdsurfer]

i would recommend anderson, low speed aerodyanmics. or go to MIT opencourseware, they have tutorials on this. i think you need to understand the concept of a drag for a vehicle in flight, also what kind of speed u r dealing with? that can change too with the drag