[TEC] SH5.exe patches to fix bugs and add functionality

[TheDarkWraith]

Originally Posted by Gammel

so add 80.000 would be 00409c47

Yes

[Gammel]

cool, lets have children...

haha your're even in range sometimes. I saw in your posts you went to Osterode/Harz some time? thats not very far. O. is not the best place in DE imho.
Harz mountains is great...

The Converter you recomment is clean afaik, checked with virustotal. Very handy tool
http://www.mediafire.com/download/jg...oatConvert.zip
You never know

[TheDarkWraith]

Originally Posted by Gammel

haha your're even in range sometimes i saw in your posts you went to Osterode/Harz some time? thats not very far. O. is not the best place in DE imho.

Yes I usually make it out to Osterode once or twice a year. Walking the streets of Osterode is like stepping back into the 18th century. It's so peaceful and relaxing compared to the US. Are you familiar with the big Piller plant/factory in Osterode or Bilshausen?

[Gammel]

Originally Posted by TheDarkWraith

Yes I usually make it out to Osterode once or twice a year. Walking the streets of Osterode is like stepping back into the 18th century. It's so peaceful and relaxing compared to the US. Are you familiar with the big Piller plant/factory in Osterode or Bilshausen?

A very good friend of mine did his job trainig as a toolmaker (that CNC stuff etc.) at Piller.
So yes i know it exists but not from a personal point of view.

No luck with the render patch btw activating mid campaign - CTD
Single mission works with no noticable loss of performace afaik
will have to wait when back in port for campaign testing. This will happen this evenig.

[Gammel]

to get fully OT

some of the last and very fierce WW2 battles on the western front were fought out
in that Area beginning with Osterode -> Herberg -> Bad Lauterberg -> Braunlage and finally Blankenburg were the "mighty" 12. Army and General Lucht surrendered to american forces.

American (and German) losses were high in that area in that last day of WW2 in April 1945.

Sad stories but also fascinating i pretend you're interested in that stuff as well.

I was born and grew up in Bad Lauterberg.

Maybe you could do some research for that events in those places when you come again some day.

"Festung" Harz 1945:
http://www.alpha64.de/1945.htm only german

[TheDarkWraith]

Today the SH4 forums gave me a great idea for a new patch: have the flags blow in the wind's direction

I fired up OllyDebug and figured out where the object's matrix data is and where it's combined matrix data is (it's matrix data * parent's matrix data). After locating those two key bits of info I then had to figure out the kernel functions for reading angles from the matrix data and storing new angles to the matrix data. Luckily for me, the kernel just happened to have those two functions After a bit of playing around to figure out which value of the 3 angles returned was the yaw value (of the pitch, yaw, and roll returned) I was able to modify the yaw value returned for the flag. I modified the yaw value to have the flag yaw (rotate around it's Y axis) +45 degrees. I then called the kernel function to set the new angles in the matrix and it worked first try

What I'm trying to derive currently is how to figure out how much the wind influences the flag's yaw. For instance, if the unit's speed is 8 knots @ 90 degrees, the wind's speed is 5 knots @ 315 degrees, how much should the flag yaw

[Tonci87]

Originally Posted by TheDarkWraith

Today the SH4 forums gave me a great idea for a new patch: have the flags blow in the wind's direction

I fired up OllyDebug and figured out where the object's matrix data is and where it's combined matrix data is (it's matrix data * parent's matrix data). After locating those two key bits of info I then had to figure out the kernel functions for reading angles from the matrix data and storing new angles to the matrix data. Luckily for me, the kernel just happened to have those two functions After a bit of playing around to figure out which value of the 3 angles returned was the yaw value (of the pitch, yaw, and roll returned) I was able to modify the yaw value returned for the flag. I modified the yaw value to have the flag yaw (rotate around it's Y axis) +45 degrees. I then called the kernel function to set the new angles in the matrix and it worked first try

What I'm trying to derive currently is how to figure out how much the wind influences the flag's yaw. For instance, if the unit's speed is 8 knots @ 90 degrees, the wind's speed is 5 knots @ 315 degrees, how much should the flag yaw

I have drawn something to illustrate how it works IRL. [/URL] Using tangens you should be able to get the correct angle.

[TheDarkWraith]

Actually I think the law of cosines applies here:

ship's heading (90) - winds heading (from 315 or 45 degrees) = 45 degrees
square root of (8^2 + 5^2 + (2*8*5*COS(45))) = 12.065

12.065 is the length of the 3rd side of the triangle.

flag's angle in relation to ship = COS-1((8^2 + 12.065^2 - 5^2) / (2*8*12.065)) = 17.04 degrees

flag's angle = 17.04 + 90 = 107.04 degrees

[TheDarkWraith]

flags affected by wind patch coming along nicely

[Sjizzle]

Originally Posted by TheDarkWraith

Yes I usually make it out to Osterode once or twice a year. Walking the streets of Osterode is like stepping back into the 18th century. It's so peaceful and relaxing compared to the US. Are you familiar with the big Piller plant/factory in Osterode or Bilshausen?

i live in germany and Osterode is a pretty nice place for a relaxing vacation but
if u wish to go back in time i recomand u a vacation in Sighisoara ( my home town )
sorry for offtopic



Dracula's ( Vlad the impaler ) birth place

http://upload.wikimedia.org/wikipedi...birthplace.jpg

[Sartoris]

Originally Posted by TheDarkWraith

flags affected by wind patch coming along nicely

A great idea for a patch, really looking forward to it!

[Tonci87]

Originally Posted by TheDarkWraith

Actually I think the law of cosines applies here:

ship's heading (90) - winds heading (from 315 or 45 degrees) = 45 degrees
square root of (8^2 + 5^2 + (2*8*5*COS(45))) = 12.065

12.065 is the length of the 3rd side of the triangle.

flag's angle in relation to ship = COS-1((8^2 + 12.065^2 - 5^2) / (2*8*12.065)) = 17.04 degrees

flag's angle = 17.04 + 90 = 107.04 degrees

Maybe I´m wrong (just got up) but why do you need the lenght of the 3rd side of the triangle? It is not needed unless you want to determine how high the flag should wave. I you know the lenghts of the opposite and adjacent sides then you can calculate the angle by using tangent. Right?

[TheDarkWraith]

Originally Posted by Tonci87

Maybe I´m wrong (just got up) but why do you need the lenght of the 3rd side of the triangle? It is not needed unless you want to determine how high the flag should wave. I you know the lenghts of the opposite and adjacent sides then you can calculate the angle by using tangent. Right?

I don't see how you can use TAN. There are no right triangles. Unless I'm forgetting some form of math the only way I can see to solve this problem is with law of cosines . If you can show me an eaiser/better/another way I'm all ears

[Tonci87]

Originally Posted by TheDarkWraith

I don't see how you can use TAN. There are no right triangles. Unless I'm forgetting some form of math the only way I can see to solve this problem is with law of cosines . If you can show me an eaiser/better/another way I'm all ears

Yeah, now that I´m more awake I can see why tangens won´t work.

[Viktor_Prien]

flags affected by wind patch coming along nicely

Wow!Every day another piece of the puzzle comes up!Great great work TDW and thank you very much for all your efforts to render the game more realistic and playable!Very well done!
Salute Herr Kaleun!