It needs a wee bit of validation from a pilot who knows how it's supposed to work. Screenshots and theory are one thing. I have to say, I don't really care that much if it's accurate or not. It does the job, if I was being paid to write milspec simulations I'd go the extra nautical but there's a game to finish.
Lets take a closer look at the slip-ball and turn rate elements.
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| In the "doghouse" Turn rate and slip-ball |
The turn rate indicator is the square sitting in middle of a row of 'doghouses'. It shows the direction of turn and rate. This has a scale of +- 5 degrees per second, when turning left at 5 dps it will hit the left stop. Consequently it often slams against the sides whenever I'm flying.
A couple of hours every night and we have a decent representation of the Apache FLT page. Building it showed some gaps in our simulation variables.
Flight path indicator
I ran over a conversation I had with Fred Naar about the flight-path indicator and adjusting it for different fields of view (PC users tend to have an annoying variety of display sizes, virtualising symbology required melding aircraft velocity with variable fields of view).
The MPD page has a fixed resolution and non-virtual which makes the math is considerably easier (and faster). We can skip a lot of math and resort to velocity divided by the total velocity as a simple offset from the centre.
Vtot is Sqr(Vtot.x * Vtot.x + Vtot.y * Vtot.y + Vtot.z * Vtot.z); // TOTAL VELOCITY
fpx = AirState.Velocity.x / Vtot
fpy = AirState.VelocityGlobal.y / Vtot
This results in an offset we can plot from the centre of the MPD. The only two things we need to consider are total velocities less than 1 (we don't want divide by zeros so we clamp Vtot) and the x/y ratio of the MPD to suit a range pilots will be happy with. I don't have any empirical data on the range of motion consequently it's set it to something I find useful.
To finish off, we'll position the slip and turn co-ordinator.
* edit *
Turn co-ordinator and Slip-Ball math
This is proving to be quite interesting. To calculate the sideways G load for the turn co-ordinator first find the radius of the turn.
radius = ground_speed ^2 / (9.8 * Tan(bank_angle))
9.8 is a constant (gravity in meters per second)
For a speed of 25 meters per second and bank angle of 30 degrees that results in a turn radius of 110.46 meters. Double the speed to 50 meters per second and our radius is 441.8 (which is 110.46 * 4). Double the speed the radius is squared. We can verify these results using a handy javascript turn-rate calculator.
To calculate Centripetal force, the sideways force towards the centre of a turn we use:
centripetal acceleration = velocity ^ 2 / radius
If we use the result of 384 from our above example of 50 meters per second at 30 degrees of bank, we get a force of 5.65 meters per second.
I need to verify I'm not mixing my units, Fred will have a better idea about the g-force loading so I'll pass that on to him to have a look when he has a free moment.
really interesting work, thanks for the details
ReplyDeleteHmm.. Your G force calculations seems to be off.
ReplyDelete1. Mass don't affect. Only acceleration
2. In your current example you need to think triangle. In this case vertical G force is 1 (9.8 m/s^2=1 G) and vertical G force would be 0.58 ( 5.65/9.8)
For total G force you need to use Pythagorean theorem a^2+b^2=c^2 which in thise case would be horizontal G force^2+vertical G force^2=total G force^2.
In your current example that would be
0.58^2+1^2=total G force^2 which would result total G force of 1.15 (to get total G force of 2 while doing coordinated turn you would have to use bank angle of 60 degrees)
Check this picture http://selair.selkirk.ca/Training/Aerodynamics/graphics/g-force_definition2.gif to get better understanding about what I mean.
ps.. accidentaly deleted my 1st comment but is was same except couple typos
Thank you, I knew someone would correct me before I got around to checking it. It didn't make sense at the time I was writing the entry. At 60 degree of bank you get up to around 2G in most aircraft. I'll go back and fix the entry later (and I removed the empty post).
ReplyDeleteThanks for your help in clearing that up.
Plugged in, that's happy now. And I'm happy too :)
ReplyDelete