Calculating Aircraft Physics for a Simulation Game

  • Thread starter Thread starter boks
  • Start date Start date
  • Tags Tags
    Aircraft Physics
AI Thread Summary
The discussion focuses on developing realistic aircraft physics for a simulation game, specifically addressing challenges with drag and rolling resistance calculations. The user is confused about the distinction between mass and weight, noting that the aircraft's weight is given in kilograms but should be converted to Newtons for force calculations. They correctly identify rolling resistance as the relevant force for moving the aircraft on the ground and provide formulas for calculating drag and rolling resistance. Additionally, they seek clarification on how to determine the net force acting on the aircraft during takeoff, emphasizing the need for accurate calculations that consider varying speeds. Overall, the conversation highlights the complexities of simulating aircraft physics and the importance of understanding fundamental principles.
boks
Messages
1
Reaction score
0
I’m attempting to write an aircraft simulation game with believable physics, although it doesn’t have to be precise but I’m having a bit of trouble getting it off the ground so to speak.

When I plug numbers into the drag and rolling resistance formulas, I don’t think I get accurate results, so I must be doing something wrong along the way, as I’m by no means an expert at maths or physics. I’m not looking at lift just yet, but obviously I will at some point.
I’m sure I’m making plenty of silly mistakes, so please correct me where I go wrong. I’ve been attempting to get my head around this for a couple of days but I don’t seem to be making much progress.

For example, let’s take a Tornado GR4 with the following specs:
(http://www.fas.org/man/dod-101/sys/ac/row/tornado.htm)
Weight: 13890kg
Thrust: 38.7kN
Wing area: 26.6m^2

Am I right in thinking I should be looking at rolling resistance rather than static friction in order to get the object (aircraft) moving along the ground?

Rolling resistance: F = C * Nf
Where F is the rolling resistance force
C is the rolling resistance coefficient
Nf is the normal force

Let’s take C to be 0.01 (ordinary car tyres on concrete).

Would the mass of the object be 13890 kg or 13890 / 9.81 = 1415.9 kg?
I’m a bit confused because force in Newtons has the units of kg ms^-2 but mass has the units of kg, although the units given for the weight (force), is in kg? If I were to take 1415.90 kg to be the mass, the normal force (Nf) would be 13890 N (1415.9 * 9.81)? I’m not sure that sounds right though :/

So the force required to get the object moving would have to be greater than 0.01 * 13890 = 138.9N?


For drag: D = Cd * A * 0.5 * r * V^2
Where D is the drag
Cd is the drag coefficient. Say 0.02
A is the reference area (for an aircraft, it’s the wing viewed from above which makes it easier/comparable to calculate the lift?). 26.6 m^2
r is the density of the air. 1.225 kg m^-3 (at sea level)
V is the velocity. Say 44 m^s-1
D = 0.02 * 26.6 * 0.5 * 1.225 * 44^2 = 630.8456 N

So how would I then calculate how much to slow the object down by? This is quite challenging for me as you can probably tell, but I hope to eventually grasp it. I’d appreciate any help that anyone is able to offer :)

Many thanks.
 
Physics news on Phys.org
Big problems when the specs list the weight of the aircraft in kilograms. 13890 kg is actually the mass. Weight = mg = 13890(9.8) Newtons. Max takeoff weight is 28000(9.8) N. Adjust rolling friction force accordingly (yes, it's rolling friction, not static or kinetic friction). Your drag formula is based on a certain speed...that should not be max takeoff speed, as drag is dependent on speed, which varies during takeoff...you'd have to use a weighted average speed, since calcs are a bit complex... I'm not sure what you mean by slowing down...F_net = ma, where F_net is the thrust less the friction less the drag.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top