Misleadingly Simple Math Problem

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DigitalFury

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Posted: 18th Nov 2010 21:37 Edited at: 18th Nov 2010 21:50

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@Visigoth - lol. Yeah. I do have a good handle on trig. Few examples: Orbit Object, ATanFull to get X, Y, Z angles for Hands Point at Point example (Gun pointing at 3d position), and a few others.

I stated that:

Quote: "This function is based on a function lib I found for matrixes."

I found this function and wanted to learn how it works and re-write it using matrixes. I didn't completely understand it though.

Quote: "angC# = acos((a#^2 + b#^2 - c#^2) / (2*a#*b#))
angA# = asin((a# * sin(angC#)) / c#)
angB# = 180.00 - (angA# + angC#)"

Wouldn't ATanFull be easier?

+ Code Snippet

Ang1# = ATanFull(a, c) ` Angle For Point 1
Ang2# = ATanFull(a, b) ` Angle For Point 2
Ang3# = ATanFull(c, b) ` Angle For Point 3

Sorry. Still haven't looked at it yet. Working on my main computer. lolz

Mind explaining how this works:

+ Code Snippet

get_coordinates

Also, Does it output the point: (X, Y) (Or at least have global vars X and Y?

Thanks,

DigitalFury

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DigitalFury

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Posted: 18th Nov 2010 21:42 Edited at: 18th Nov 2010 21:42

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@Green Gandalf - I Forgot this rule:

Quote: "(a - b)^2 = a^2 - 2*a*b + b^2"

could be used to expand #2 like this:

Quote: "(x1 - x)^2 = x1^2 - 2*x1*x + x^2"

This just makes it and even longer linear equation. Fun!

lol

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Green Gandalf

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Posted: 18th Nov 2010 23:12

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Quote: "Fun!"

Indeed.

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Visigoth

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Posted: 19th Nov 2010 03:18

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@DigitalFury
yeah, I guess atanfull() would be easier, but atanfull() isn't a trig function by itself, its a DBPro function, so, wouldn't work with my source info. Good idea, this makes it EVEN easier.

second function determines the coordinates. Vars for sides and angles were declared as global. Its right at top of full code.

its just a quick example, just to show the method. Get the angles, then plot the coordinates.

all I did was follow the info from this source:
http://www.teacherschoice.com.au/Maths_Library/Trigonometry/solve_trig_SSS.htm

Honestly, I think you know exactly what you are doing.
Good day.

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DigitalFury

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Posted: 19th Nov 2010 03:39

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@Visigoth -

Quote: "all I did was follow the info from this source:
http://www.teacherschoice.com.au/Maths_Library/Trigonometry/solve_trig_SSS.htm"

I don't know how this problem solves for point (X, Y).

Quote: "Honestly, I think you know exactly what you are doing."

^^ Sometimes.

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Visigoth

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Posted: 19th Nov 2010 03:59

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@DigitalFury
because in your original post, there are no known angles.
First solve the triangle, to get the angles. That is first function.
Once you know the angles, you can plot the coords using sin() and cos() as a ratio of each length.

I don't know how to explain it any simpler, really. I thought the sample code would illustrate it.

See ya.

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DigitalFury

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Posted: 19th Nov 2010 04:09

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@Visigoth - I should be able to figure it out now. I couldn't figure it out just based on the website you gave me. Thanks.

DigitalFury

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DigitalFury

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Posted: 25th Nov 2010 21:02 Edited at: 25th Nov 2010 21:02

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I am going to put a pause on this problem until next year. I am working on a small video game demo.

P.S. Sorry for the double post.

DigitalFury

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DigitalFury

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Posted: 1st Dec 2010 00:07 Edited at: 1st Dec 2010 00:09

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@Visigoth - Finally got around to looking at your example. My main computer was down until now. Your example is awesome. That should be what I am looking for.

@Green Gandalf -
I am busy working on my current project. As soon as I get to the part of the project where I need this function I will try working on the problem again. Thanks for your help so far!

Check out my current project!

DigitalFury

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Visigoth

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Posted: 1st Dec 2010 03:44 Edited at: 1st Dec 2010 03:50

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@DigitalFury
I tried to tell ya!
just kidding. Hope it works out for you. I'm still looking for all the triangle solving code I wrote, but, sadly, I think it got wiped on an old harddrive. Come to think of it, I might have put them on the old code snippets board...anyway, one of these days I'll write them over. They're just translated from the math source link I posted earlier.

edit:
I noticed on your projects "other inputs" is next in line. Check out Todd Riggin's XBOX 360 dll....works great. I've tested it with 4 controllers, mix of wireless and even a guitar. Works.

http://forum.thegamecreators.com/?m=forum_view&t=155264&b=8

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DigitalFury

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Posted: 1st Dec 2010 04:10

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@Visigoth - Nice. The controller plugin looks a lot better then another one I am using. Input is a big feature in my game. Thanks for the link! Check out all my code snippets. Might be something of use.

DigitalFury

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Green Gandalf

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Posted: 1st Dec 2010 12:22

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Quote: "As soon as I get to the part of the project where I need this function I will try working on the problem again. Thanks for your help so far!"

Don't bother with my solution if you're happy with Visigoth's method - unless you just want to understand how mine works.

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DigitalFury

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Posted: 3rd Dec 2010 23:53 Edited at: 3rd Dec 2010 23:53

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@Visigoth - I spent a lot of time looking though your source following all the variables.

I have angles Angle1#, Angle2#, Angle3# corresponding to points 1, 2, 3. I have Dist1#, Dist2#, Dist3# corresponding to a, b, c.

The Type:

+ Code Snippet

Type Point2D
    X
    Y
EndType

The function is:

+ Code Snippet

FindPointInTriangle(Dist1#, Dist2#, Dist3#, Angle1#, Angle2#, Angle3#)
    Point As Point2D
    ` Point.X = 
    ` Point.Y =
EndFunction Point

To make things easier for me could you use your get_coordinates function and rewrite it using my function?

Quote: "eaA#,eaB#,eaC#,esa#,esb#,esc#"

I had a lot of problems following you with these variables.

Thanks,

DigitalFury

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Visigoth

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Posted: 6th Dec 2010 02:26 Edited at: 6th Dec 2010 02:30

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@DF
Only if Dist1#, Dist2#, and Dist3# include the lengths of the sides d,e,f in your first drawing.

Because, you said:

Quote: "The Known:
1, 2, 3 = The Points (X1, Y1), (X2, Y2), (X3, Y3)
a, b, c, d, e, f = Lengths of sides"

I could write it like this:

+ Code Snippet

FindPointInTriangle(Dist1#, Dist2#, Dist3#)
//where Dist1# = side a,b, or c
//where Dist2# = side d, e, or f
//where Dist3# = side d,e,or f
//so, the triangle being solved is one of the internal tris.
//3 choices
//sides b,f,e
//sides c,d,f
//sides a,e,d
    Point As Point2D
    ` Point.X = //my code
    ` Point.Y = //my code
EndFunction

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DigitalFury

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Posted: 6th Dec 2010 02:37 Edited at: 6th Dec 2010 03:06

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@Visigoth - Yep. I forgot that.

This is what I got:

+ Code Snippet

Dist1_3# = Dist3D(X1#, Y1#, Z1#, X3#, Y3#, Z3#)
Dist1_2# = Dist3D(X1#, Y1#, Z1#, X2#, Y2#, Z2#)
Dist2_3# = Dist3D(X2#, Y2#, Z2#, X3#, Y3#, Z3#)
   	
Angle1# = ATanFull(Dist1_2#, Dist1_3#)
Angle2# = ATanFull(Dist1_2#, Dist2_3#)
Angle3# = ATanFull(Dist1_3#, Dist2_3#)

The internal distances are:

+ Code Snippet

Int_Dist1_3#
Int_Dist1_2#
Int_Dist2_3#

I know these internal distances from some other algorithms.

Quote: "get_coordinates(aA#,aB#,aC#,sa#,sb#,sc#,eaA#,eaB#,eaC#,esa#,esb#,esc#)"

This function is what I confusing me a bit.

This is what I have so far:

+ Code Snippet

//internal tri
   	xA# = 0.0
	yA# = 0.0
	   	
   	//this would be the "random" point in the middle of the problem
	xC# = Dist1_3# * (Cos(Angle1#))
	yC# = Dist1_3# * (Sin(Angle1#))

	xB# = Dist1_2#
	yB# = 0.0
	//external
	e_xA# = 0.0
	e_yA# = 0.0
	
	e_xC# = esb# * (cos(eaA#))
	e_yC# = esb# * (sin(eaA#))

	e_xB# = esc#
	e_yB# = 0.0

I revised this:

I have angles Angle1#, Angle2#, Angle3# corresponding to points 1, 2, 3. I have A#, B#, C#, E#, F# corresponding to a, b, c, d, e, f.

The Type:

+ Code Snippet

Type Point2D
    X
    Y
EndType

The function is:

+ Code Snippet

FindPointInTriangle(A#, B#, C#, D#, E#, F#, Angle1#, Angle2#, Angle3#)
    Point As Point2D
    ` Point.X = 
    ` Point.Y =
EndFunction Point

You stated: xA,yA is the coords. I think that is (X, Y). The point I am trying to find.

I am just a bit lost is all with your variables and forgot about those internal distances.

Thanks,

DigitalFury

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Visigoth

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Posted: 6th Dec 2010 03:33

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I was going to write a big long explanation, but I forgot I put a diagram in the code:

+ Code Snippet

//where aA#, aB#,aC# are the angles
//where sa#,sb#,sc# are the sides
//angs are opposite sides
//largest side on bottom
//xA,yA, etc are the coordinates
//this diagram will be upside down on the display, but the data is correct
//                       xC,yC
//                        aC#
//                        /\
//                       /  \                                                y0
//                      /    \                                               | 
//                  sb#/      \ sa#                                          |
//                    /        \                                             |
//                   /          \                                            | 
//                aA# /____________\ aB#                                    x0 |---------- x screen width()
//                xA,yA      sc#     xB,yB                                       y screen height

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DigitalFury

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Posted: 6th Dec 2010 03:52

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@Visigoth - Already saw that diagram. I just didn't understand what this was:

+ Code Snippet

eaA#,eaB#,eaC#,esa#,esb#,esc#

. I just figured out it was the external and the

+ Code Snippet

aA#,aB#,aC#,sa#,sb#,sc#

are the internals.

What I didn't understand was this part:

+ Code Snippet

	//external
	e_xA# = 0.0
	e_yA# = 0.0
	
	e_xC# = esb# * (cos(eaA#))
	e_yC# = esb# * (sin(eaA#))

	e_xB# = esc#
	e_yB# = 0.0

I don't know where some variables are coming from. I don't really understand how it works.

Sorry I am still a bit confused.

Thanks for your help!

DigitalFury

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Visigoth

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Posted: 6th Dec 2010 04:02 Edited at: 6th Dec 2010 04:11

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@DF
ok, there are only 2 functions to do this. All the rest are just there to draw the results to the screen.
I'll try to explain the method.

You have to look back at your drawing in your first post. There are 4 triangles. One big one, and three littler ones inside it.
The function solves for one of the littler ones, and, the big one.

First, solving the triangle.
We need to solve it to get the angles.
We need the angles to figure the coordinates, later.
We solve the triangle using the SSS theorum, translated from the math page.
http://www.teacherschoice.com.au/Maths_Library/Trigonometry/solve_trig_SSS.htm

+ Code Snippet

function solve_triangle_sss(a#,b#,c#,ea#,eb#,ec#)
    //where c# is the longest side
    //find largest angle first
    //internal triangle
    angC# = acos((a#^2 + b#^2 - c#^2) / (2*a#*b#))
    //find next angle
    angA# = asin((a# * sin(angC#)) / c#)
    //find last angle
    angB# =  180.00 - (angA# + angC#)

    //THIS IS SOLVING THE BIG TRIANGLE
    //FORGET ABOUT THIS PART! JUST FOR DISPLAY!
    //e_ means EXTERNAL ANGLE
    //external triangle
    e_angC# = acos((ea#^2 + eb#^2 - ec#^2) / (2*ea#*eb#))
    //find next angle
    e_angA# = asin((ea# * sin(e_angC#)) / ec#)
    //find last angle
    e_angB# =  180.00 - (e_angA# + e_angC#)

endfunction

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Visigoth

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Posted: 6th Dec 2010 04:09 Edited at: 6th Dec 2010 04:15

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just input three valid side lengths of the internal triangle. Those would be params a#,b#,c#, with c# being the longest one. It will be the same number as ec#. (look at your drawing).

anything with an e in front of it represents the larger, external triangle. I had to do this, because if you change the length of some internal lines, it will "push" the external coordinates. If those external coordinates were fixed, and the internal lines were too long, then the internal "random point" would be outside the triangle.

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Visigoth

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Posted: 6th Dec 2010 04:15

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If your internal side lengths are correct, then the coordinate will be correct for the other two triangles. Like I said, we only need to solve for one.
That is the other function, to determine the coordinate.

function get_coordinates(aA#,aB#,aC#,sa#,sb#,sc#,eaA#,eaB#,eaC#,esa#,esb#,esc#)
//where aA#, aB#,aC# are the angles
//where sa#,sb#,sc# are the sides
//angs are opposite sides
//largest side on bottom
//xA,yA, etc are the coordinates
//this diagram will be upside down on the display, but the data is correct
//                       xC,yC
//                        aC#
//                        /\
//                       /  \                                                y0
//                      /    \                                               | 
//                  sb#/      \ sa#                                          |
//                    /        \                                             |
//                   /          \                                            | 
//                aA# /____________\ aB#                                    x0 |---------- x screen width()
//                xA,yA      sc#     xB,yB                                       y screen height
//                                                                                       
//                                                                           
//                                                                           
// xA,yA at 0,0
//
//internal tri
    xA# = 0.0
    yA# = 0.0
    //this would be the "random" point in the middle of the problem
    xC# = sb# * (cos(aA#))
    yC# = sb# * (sin(aA#))

xB# = sc#
    yB# = 0.0
//external
    e_xA# = 0.0
    e_yA# = 0.0
    
    e_xC# = esb# * (cos(eaA#))
    e_yC# = esb# * (sin(eaA#))

e_xB# = esc#
    e_yB# = 0.0

//now you can reposition points by offsetting ,rotating, or scaling them 
//see next few functions                
endfunction

+ Code Snippet

function get_coordinates(aA#,aB#,aC#,sa#,sb#,sc#,eaA#,eaB#,eaC#,esa#,esb#,esc#)
//where aA#, aB#,aC# are the angles
//where sa#,sb#,sc# are the sides
//angs are opposite sides
//largest side on bottom
//xA,yA, etc are the coordinates
//this diagram will be upside down on the display, but the data is correct
//                       xC,yC
//                        aC#
//                        /\
//                       /  \                                                y0
//                      /    \                                               | 
//                  sb#/      \ sa#                                          |
//                    /        \                                             |
//                   /          \                                            | 
//                aA# /____________\ aB#                                    x0 |---------- x screen width()
//                xA,yA      sc#     xB,yB                                       y screen height
//                                                                                       
//                                                                           
//                                                                           
// xA,yA at 0,0
//
//internal tri
    xA# = 0.0
    yA# = 0.0
    //this would be the "random" point in the middle of the problem
    xC# = sb# * (cos(aA#))
    yC# = sb# * (sin(aA#))

    xB# = sc#
    yB# = 0.0
//external
    e_xA# = 0.0
    e_yA# = 0.0
    
    e_xC# = esb# * (cos(eaA#))
    e_yC# = esb# * (sin(eaA#))

    e_xB# = esc#
    e_yB# = 0.0

//now you can reposition points by offsetting ,rotating, or scaling them 
//see next few functions                
endfunction

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DigitalFury

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Posted: 6th Dec 2010 04:15

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@Visigoth - Thanks Again. I'll give this another shot tomorrow.

Did ya check out all the code snippets yet. I updated the input code snippet to include multi-threading cuz I think you said you needed input functions.

DigitalFury

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Visigoth

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Posted: 6th Dec 2010 04:20

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Oh, do I. I'll rewrite the solve triangle function and the coordinate function to be more generic. I know its a little confusing. But you know how it is sometimes, you just get going, and it makes sense at the time.
Thanks man.

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DigitalFury

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Posted: 6th Dec 2010 04:39

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@Visigoth - What makes complete sense to one programmer could be complete gibberish to another. I changed my code habits years ago so I write everything so it is as easy as possible to understand. My code use to be completely unreadable for anyone reading it...even unreadable for me years later. lolz

Thanks for the re-write. It will help tremendously.

DigitalFury

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Visigoth

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Posted: 6th Dec 2010 06:12

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ok, as simple as I think I can make it:

Rem Project: Dark Basic Pro Project
Rem Created: Sunday, December 05, 2010

Rem ***** Main Source File *****
REM more fun with triangles by Visigoth

type triangle
    pointA_X as float
    pointA_Y as float
    pointB_X as float
    pointB_Y as float
    pointC_X as float
    pointC_Y as float
    angleA as float
    angleB as float
    angleC as float
    sideA as float
    sideB as float
    sideC as float
endtype

my_Tri as triangle

//init triangle side lengths
my_Tri.sideA = 80.0
my_Tri.sideB = 70.0
my_Tri.sideC = 80.0

///////////////////////////////////////////////////////////////////
do
    cls
    solve_triangle_sss(my_Tri.sideA,my_Tri.sideB,my_Tri.sideC)
    plot_triangle()
    display()     
LOOP
////////////////////////////////////////////////////////////////////

function solve_triangle_sss(sideA#,sideB#,sideC#)
    //where c# is the longest side
    my_Tri.angleC = acos((sideA#^2 + sideB#^2 - sideC#^2) / (2*sideA#*sideB#))
    //find next angle
    my_Tri.angleA = asin((sideA# * sin(my_Tri.angleC)) / sideC#)
    //find last angle
    my_Tri.angleB =  180.00 - (my_Tri.angleA + my_Tri.angleC) 
    get_coordinates(my_Tri.angleA, my_Tri.sideB, my_Tri.sideC)
ENDFUNCTION

function get_coordinates(angleA#, sideB#, sideC#)
    //first point in triangle is at 0,0
    my_Tri.pointA_X = 0.0  
    my_Tri.pointA_Y = 0.0
    
    //this would be the "random" point in the middle of the problem
    my_Tri.pointC_X = sideB# * (cos(angleA#))
    my_Tri.pointC_Y = sideB# * (sin(angleA#))
    
    //this point is based on the longest side, positioned along the x axis
    my_Tri.pointB_X = sideC#
    my_Tri.pointB_y = 0.0
    
//now you can reposition points by offsetting ,rotating, or scaling them 
//this assumes right handed triangle winding.
//so, point A is at 0,0
//point B is at (length side b, 0)
//point C is at the sin() and cos() functions based on angles from point A               
endfunction

//for display only
function plot_triangle()
    line my_tri.pointA_X,my_tri.pointA_Y,my_tri.pointB_X,my_tri.pointB_Y
    line my_tri.pointB_X,my_tri.pointB_Y,my_tri.pointC_X,my_tri.pointC_Y    
    line my_tri.pointC_X,my_tri.pointC_Y,my_tri.pointA_X,my_tri.pointA_Y    
ENDFUNCTION

function display()
    print my_tri.pointA_X, "," , my_tri.pointA_Y
    print my_tri.pointB_X, "," , my_tri.pointB_Y
    print "Random point: ,", my_tri.pointC_X, "," , my_tri.pointC_Y
ENDFUNCTION

+ Code Snippet

Rem Project: Dark Basic Pro Project
Rem Created: Sunday, December 05, 2010

Rem ***** Main Source File *****
REM more fun with triangles by Visigoth


type triangle
    pointA_X as float
    pointA_Y as float
    pointB_X as float
    pointB_Y as float
    pointC_X as float
    pointC_Y as float
    angleA as float
    angleB as float
    angleC as float
    sideA as float
    sideB as float
    sideC as float
endtype

my_Tri as triangle

//init triangle side lengths
my_Tri.sideA = 80.0
my_Tri.sideB = 70.0
my_Tri.sideC = 80.0

///////////////////////////////////////////////////////////////////
do
    cls
    solve_triangle_sss(my_Tri.sideA,my_Tri.sideB,my_Tri.sideC)
    plot_triangle()
    display()     
LOOP
////////////////////////////////////////////////////////////////////

function solve_triangle_sss(sideA#,sideB#,sideC#)
    //where c# is the longest side
    my_Tri.angleC = acos((sideA#^2 + sideB#^2 - sideC#^2) / (2*sideA#*sideB#))
    //find next angle
    my_Tri.angleA = asin((sideA# * sin(my_Tri.angleC)) / sideC#)
    //find last angle
    my_Tri.angleB =  180.00 - (my_Tri.angleA + my_Tri.angleC) 
    get_coordinates(my_Tri.angleA, my_Tri.sideB, my_Tri.sideC)
ENDFUNCTION

function get_coordinates(angleA#, sideB#, sideC#)
    //first point in triangle is at 0,0
    my_Tri.pointA_X = 0.0  
    my_Tri.pointA_Y = 0.0
    
    //this would be the "random" point in the middle of the problem
    my_Tri.pointC_X = sideB# * (cos(angleA#))
    my_Tri.pointC_Y = sideB# * (sin(angleA#))
    
    //this point is based on the longest side, positioned along the x axis
    my_Tri.pointB_X = sideC#
    my_Tri.pointB_y = 0.0
    
//now you can reposition points by offsetting ,rotating, or scaling them 
//this assumes right handed triangle winding.
//so, point A is at 0,0
//point B is at (length side b, 0)
//point C is at the sin() and cos() functions based on angles from point A               
endfunction

//for display only
function plot_triangle()
    line my_tri.pointA_X,my_tri.pointA_Y,my_tri.pointB_X,my_tri.pointB_Y
    line my_tri.pointB_X,my_tri.pointB_Y,my_tri.pointC_X,my_tri.pointC_Y    
    line my_tri.pointC_X,my_tri.pointC_Y,my_tri.pointA_X,my_tri.pointA_Y    
ENDFUNCTION

function display()
    print my_tri.pointA_X, "," , my_tri.pointA_Y
    print my_tri.pointB_X, "," , my_tri.pointB_Y
    print "Random point: ,", my_tri.pointC_X, "," , my_tri.pointC_Y
ENDFUNCTION

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DigitalFury

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Joined: 30th Jul 2010

Location: United States

Posted: 6th Dec 2010 06:13

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@Visigoth - Thanks a lot. I think it is ready for the code snippet forums. Your demo was great!

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DarkBASIC Professional Discussion / Misleadingly Simple Math Problem