Gaming Digital Combat Simulator Thread

[Marijn]

NEver mind, found the mistake myself. 320° not 220° Thats what you get for doing simple math in your head while doing other stuff.

I knew you meant 320 deg. And you are right. My statement was false. If the wind is 0.1, then the ship needs to cruise slightly faster than 30.

I should be doing other stuff..

 

[martins]

OK, I found another solution (forgot that asin has two solutions in [0,2pi]):

Course ship: 11.46 deg
Speed ship: 25.89

This one is a bit closer to Urwumpe's and doesn't have the tailwind problem, so is probably more sensible.

@Urwumpe: not sure about all acronyms, so I don't know if I interpret your solution correctly, but when I plug in your solution, I get a wind direction relative to the carrier direction of -7 deg instead of -9 deg. Did I make a mistake?

 

[Urwumpe]

@Urwumpe: not sure about all acronyms, so I don't know if I interpret your solution correctly, but when I plug in your solution, I get a wind direction relative to the carrier direction of -7 deg instead of -9 deg. Did I make a mistake?

I am not sure there either, I made it quick and dirty with a slide rule after lunch break. :lol:

The acronyms are from sailing:

AWA = apparent wind angle
AWS = apparent wind speed

TWA = true wind angle (actually it is TWD, I got it wrong again)
TWS = true wind speed.

apparent wind is the wind including the motion of the ship, while true wind is the wind for a stationary observer.

apparent wind is simply the addition of true wind vector and ship velocity vector (including drift and current)

Thus the idea was simply to take a target wind vector (apparent wind over the flight deck = 30 knots AWS@351°AWA), subtract the true wind vector from it and get the velocity vector to complete the wind triangle.

 

[Marijn]

If question I has more solutions, then I think that the answer to question II is the solution of question I which has the smallest angle between the heading of the plane and the runway.

Perhaps the solution where the nose of the plane is pointing towards the port side is the best solution, because all carriers have their bridge on the starboard side. That would be safer when evasive action is taken.

 

[Urwumpe]

The trivial solution to II is:

Ship speed = 30 knots
Deck angle = 0°
true wind angle = 0°
true wind speed = 0 knots


If you increase the wind speed and change the wind direction, the apparent wind angle of ship and plane will diverge.

If you increase the deck angle, there could be function of wind direction and wind speed, in which relative motion of the ship to the aircraft line of sight could be null.

 

[martins]

Or another way to look at it: when landing on a fixed runway, you prefer a direct headwind, because you don't have to compensate your course for the crosswind component.

In this case however, the runway is moving and it is not moving along the runway axis, so you _want_ just enough crosswind to compensate for the side drift of the runway.

 

[fred18]

a bit of OT:

AWA = apparent wind angle
AWS = apparent wind speed

TWA = true wind angle (actually it is TWD, I got it wrong again)
TWS = true wind speed.

you mentioned all the acronyms I'm used to deal with (semi-pro sailor here), do you know sailing?

 

[Urwumpe]

a bit of OT:



you mentioned all the acronyms I'm used to deal with (semi-pro sailor here), do you know sailing?

Yeah, I plan to learn it for my pre-midlife-crisis (between 40 and 50). :lol:



Still learning the basics here for getting the initial German certificate - not that I would need one outside Germany.

 

[martins]

Sorry for flogging a dead horse, but this is my brainteaser for the day, so I want to get to the bottom of it :lol:

Thus the idea was simply to take a target wind vector (apparent wind over the flight deck = 30 knots AWS@351°AWA), subtract the true wind vector from it and get the velocity vector to complete the wind triangle.

This was essentially my idea as well, but given that TWA is provided in the global frame and AWA in the carrier-fixed frame, can you just subtract the vectors?
In my solution, we know two sides of the wind triangle (a = 6, b = 30) and the angle opposite a (alpha=9). So we get the angle opposite b with the law of sines: beta = asin(b/a * sin(alpha)) [2 solutions!] and the third angle from gamma = 180-alpha-beta.

Now we can map that triangle into the global frame from the given direction rho=140 of a to get the carrier heading

tau = rho + gamma - pi + alpha

and the carrier speed as the length of the third side of the wind triangle, again from the law of sines:

c = a*sin(gamma)/sin(alpha)

with the two solutions I posted above, unless I made a mistake.

To mix it up a bit, I'd like to pose a modified problem:

Given the same wind of 6 ktn towards 140 deg and the same flight deck rotation of -9 deg, and given a constant carrier speed of 30 ktn, and a constant aircraft approach TAS of 100 ktn, which direction should the carrier go so that the pilot can simply line up with the runway centreline as if she was landing on a fixed runway with no wind?

[Hint: if the carrier was going at 6 ktn, the trivial answer would be 140 deg, but since it is faster it must be rotated to produce a larger cross-wind component]

 

[Marijn]

Given the same wind of 6 ktn towards 140 deg and the same flight deck rotation of -9 deg, and given a constant carrier speed of 30 ktn, and a constant aircraft approach TAS of 100 ktn, which direction should the carrier go so that the pilot can simply line up with the runway centreline as if she was landing on a fixed runway with no wind?

I'll go with no solution for now. The runway will move towards the right. I think that situation just cannot be equal to "as if she was landing on a fixed runway with no wind". An airplane can't move sideways. So it has to do a right and a left hand turn to arrive at the runway.

Perhaps you mean that the groundtrack aligns with the runway?

Edit: Hold on. I read over "given a constant carrier speed of 30 ktn"..
Edit: No. I'll stick with the answer above for now.

 

[Topper]

An airplane can move sideway relativ to the ground by the drag of the wind, and this movement has to fit the sideway movement of the runway

I'm a bit happy now... First, I thought this can not be so complicate and I'm stupit because it looks so easy, but if you think about it it's not so easy as it looks like and thats strange... Now it's to late for my brain to understand and solve this, I will try again tomorrow...

 

[Topper]

Wind direction is usually measured from the direction its coming.

Yes you are right but it's strange that you have to set the "to" direction in DCS... Maybe a mistake...
But antoher point is, you can't set the same wind speed for 10m altitude as for 500m altitude... Maybe because it's unrealstic???
You also can not set a different heading for 10 and 500m...

[EDIT]
Oh I have checked the ATC an he is telling the good from direction :thumbup:

 

[Marijn]

But antoher point is, you can't set the same wind speed for 10m altitude as for 500m altitude... Maybe because it's unrealstic???
You also can not set a different heading for 10 and 500m...

Windspeed normally decreases with altitude due to the friction with the surface of the earth. Over land, this is in the lowest 1000 meters of the atmosphere. Over sea, this effect will be somewhat lower I'd say. Also the wind direction changes with altitude. On the nothern hemisphere, it will go clockwise with altitude.

I don't know DCS, but being unable to set identical winspeed and direction for various altitudes is very realistic I would say.

I shudder when thinking about the math now..

 

[Topper]

Ok for those lazy guys who wants to build carrier missions, you just can type the following into google search. In this example, I have changed the wind speed to 6m/s (it was 6 knots), because in the mission editor you set it in m/s..., wind direction (140) is in "to" direction as it has to be set in the mission editor. If you have a negative value for the course, just substract it from 360.

For course:
"140 deg - (pi rad - asin(30knots/(6m/s) * sin(9 deg))) = ? deg"

For speed:
"6m/s * sin(pi rad - 9 deg - (pi rad - asin(30knots/(6m/s) * sin(9 deg)))) / sin (9 deg) = ? km/h"

 

[Fabri91]

Well, this is a new one:

This weekend the F/A-18C Hornet, the Persian Gulf map and the Su-33 will be available for free (limited to the duration of the weekend).

So, a good chance to try these modules out!

 

[Quick_Nick]

Well, this is a new one:

DCS World Hornet, Persian Gulf Map, and Su-33 Free Weekend! - YouTube

This weekend the F/A-18C Hornet, the Persian Gulf map and the Su-33 will be available for free (limited to the duration of the weekend).

So, a good chance to try these modules out!

They're also going to be in a bundle from October 1 to October 21.
If you own part of the bundle already, the remaining modules will be discounted.

Sent from my SM-G930U using Tapatalk

 

[Fabri91]

And it's being extended a couple of days:

Free Weekend Extended!

Due to new DCS World users being unable to take part in the Free Weekend on Friday, we are extending the Free Weekend to 2 October 2018 at 0900 UTC (Note that this only applies to the e-Shop version).

We hope you enjoy this taste and that you will take advantage of the Bundle Pack starting tomorrow that includes the Hornet, Persian Gulf Map, and Su-33.

Have fun!

 

[Fabri91]

So it turns out that MANPADs are a bit more than firecrackers:

The landing was a bit sporty, with the missing rudders and the single elevator making it impossible to fly with extended flaps:

 

[Topper]

 

[MaverickSawyer]

DCS: F-14 - Pre-Order / Gameplay Reveal Trailer - PRE ORDER NOW! - YouTube

My God.... It's even more glorious than I had dared to hope.

Now, if only my computer could run DCS... And I could find someone to be my RIO... That could be fun.