jackssmirkingrevenge wrote:Not disputing that at all, especially with a streamlined dart as opposed to the lumpy 45 slug, it's the 3º angle that I have a problem with.
Well, you've read what I had to say about the maths of a 3 degree angle equating to about 500 metres. That method to an approximation was simplified, but it doesn't cut too many corners - could you tell me where you think I went wrong with the calculations? Drag, bullet drop, or the rise from the increased angle?
Because I'm a bit bored, I'll go over it again, without omitting the information I did before.
Drag force is -<sup>1</sup>/<sub>2</sub>*drag coefficient*air density*area*velocity<sup>2</sup>
The minus sign means it's against the direction of movement. I ommitted it earlier for writing the forces, but I did keep it in the equations. If I put the various numbers into that:
Air Density: 1.2 kgm<sup>3</sup>
Drag Coefficient: 0.1 (but, if you prefer, 0.2)
Area: 0.56 cm<sup>2</sup>
Velocity: 240 m/s
You need to remember to put the area in square metres, but it's in square centimetres for the moment because it looks nicer. Output from the equation is in Newtons
All those numbers condense to -0.2 N or -0.4N depending on your drag co-efficient (or C<sub>D</sub>). I will use -0.2N in the calculations, but you can return and use -0.4 if you want.
To get the accleration, we divide by the mass, because:
F = m*a
and therefore: a = <sup>F</sup>/<sub>m</sub>
M needs to be in kilos, which is 0.016kg for this dart design.
So, a is -12 m/s<sup>2</sup>
Now we know that, we need to find time in the air so drop due to gravity can be found. For that, we need:
s = u*t + <sup>1</sup>/<sub>2</sub>a*t<sup>2</sup>
Or, with the values: 500 = 240*t + <sup>1</sup>/<sub>2</sub>-12*t<sup>2</sup>
As I said before, I am assuming drag is always it's maximum value to simplify it a bit.
We need that in terms of t (time) so we simplify to:
- 3*t<sup>2</sup> + 120*t - 250 = 0
Which is of the form ax<sup>2</sup> + bx + c = 0
So using the quadratic formula to find the values that solve that:
I find that the solutions are: 2.20487, and 37.79513
The 37.79513 is a result of the fact I assumed the drag force was constant. If it were, the round would slow, then could actually turn round and hit the target again. Because of that assumption, it's ignored now.
To find the drop, I'll need s = u*t + <sup>1</sup>/<sub>2</sub>a*t<sup>2</sup> again.
In this case, u is zero, the round is not initially falling. a is 9.81m/s<sup>2</sup>, acceleration in free fall.
S is therefore 23.845m. This is the drop of the round. The launch angle will therefore need to be sufficient to counter this drop.
Dividing 23.845 by 500, I get 0.04769. The tan<sup>-1</sup> of this number will give our angle.
Tan<sup>-1</sup> (0.04769) is 2.73<sup>o</sup>.
The number at the end is actually under the 3 degrees I was stating. Unless you can find a flaw in my logic or maths, I have to say, I think it's right.
EDIT: Damned hotlink blocking images. Hopefully it will work this time.
