Gaming Digital Combat Simulator Thread

Damn... I really would love to play this sim again, but my hardware is damaged and the replacement way down in the priority list now. :(
 
Heatblur just announced a new project:

[ame="http://www.youtube.com/watch?v=B30QigkUy-4"]Shoulder to Shoulder - A-6E AI for DCS World - YouTube[/ame]

Also, note the mention of 7 Oct. I suspect that may be the day they release their F-14A/B pack.
 
Give 'em time. They're still working out the last kinks in their F-14 pack... which includes a fairly comprehensive "GIB" AI called "Jester". This video demonstrates that AI in action:

DCS F-14 Valley Run - YouTube

At least we are getting a Forrestal class carrier, that is already pretty.

Now just a A-4E Skyhawk and a John McCain memorial paintjob....
 
Hello I know it's a misc topic between this and the math forum, but I want to create a dcs carrier mission and have some special questions so I decided to ask it here...


The angle between an aircraft carriers course and the flight deck is -9° (as far as i know). Let's say wind speed is 6 knots and winds direction is (to) 140 deg


I) How to calculate the course and speed of the ship to have a wind speed of 30knots down the flight deck?
(I read that this is the official procedure...?)

II) How to calculate the course and speed of the ship under the same conditions (30 knots relativ to the flight deck, don't care about exact direction relativ to the flight deck...) to have a wind speed so that an incoming airplane can aim the flight deck (nose is pointing to the flight deck and into the direction of flight deck) during it's approach? (So that the "wind side effect" fits the "flight deck site movement effect" resulting of the movement of the ship and the delta angle of the flight deck)?
 
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The first questions seems simple, but I am not sure what you are exactly after. My first thought is:
140+180=320+9=329 at 24knots. But you will experience some wind component from the side. I think that cannot be avoided when a flight deck has a fixed angle with respect to it's heading.

---------- Post added at 10:47 AM ---------- Previous post was at 09:35 AM ----------

Correction on question 1:

̶C̶o̶u̶r̶s̶e̶ ̶s̶h̶i̶p̶:̶ ̶1̶4̶0̶+̶1̶8̶0̶+̶9̶=̶3̶2̶1̶
̶S̶p̶e̶e̶d̶ ̶s̶h̶i̶p̶:̶ ̶C̶O̶S̶(̶-̶9̶)̶*̶3̶0̶=̶2̶9̶.̶6̶3̶

Course ship=140+180-(-9)=329
Speed ship=(30-6) / COS(-9) = 24.3

Spreadsheet: =(30-6)/(COS(-9*(PI()/180)))
 
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Interesting geometric problem.

I get:
Course ship: 88.54 deg
Speed ship: 33.39

Any other offers?

The wind is towards 140 degrees. So if the ship's course is 89 degrees, then that would be a tailwind situation. I don't think that can be right..
 
Wind direction is usually measured from the direction its coming.
So a wind from the North would have 0 degrees heading.

Water currents are usually specified in the direction they are flowing, so a current flowing South would carry you in a 180 degree direction.


all above from memory, so please check!

N.
 
Wind direction is usually measured from the direction its coming.

Correct, but Topper stated that the wind is towards 140.

To answer question II, I think we need the speed of the plane as well in order to solve it. Not sure though.
 
Wind direction is usually measured from the direction its coming.
So a wind from the North would have 0 degrees heading.

Water currents are usually specified in the direction they are flowing, so a current flowing South would carry you in a 180 degree direction.


all above from memory, so please check!

N.

You are right, but the original problem definition stated
Let's say wind speed is 6 knots and winds direction is (to) 140 deg
and I was assuming that the "(to)" indicated the direction the wind is going towards.

Edit: :ninja:
 
To answer question II, I think we need the speed of the plane as well in order to solve it. Not sure though.

You may be right there. My answer was for somebody standing on deck measuring the relative wind speed, but that is probably not relevant for the landing plane.
 
Well, using R² math, I get:

GOAL: 351° AWA and 30 knots AWS
SITUATION: 320° TWA (to 140°) and 6 knots TWS

[math]\begin{pmatrix}30\cos(-9)-6\cos(320)\\30\sin(-9)-6\sin(320)\end{pmatrix} = \vec{v}[/math]

This gets me an AWA of 358° and a AWS of 25 knots for the ship to experience 30 knots AWS in direction of the flight deck.
 
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Although it is not explicitly stated that the ship will launch and retrieve it's planes into a headwind, I think this is a safe assumption we could make.

That means that the speed of the ship will always be 30 or less.
 
Although it is not explicitly stated that the ship will launch and retrieve it's planes into a headwind, I think this is a safe assumption we could make.

That means that the speed of the ship will always be 30 or less.

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