a math question

[*brammie*]

i'm trying to find the pitch/roll of a terrain at a certain point.
now i have a little problem, as showed in this diagram:


i have a function to retrieve the height of a particular point in my terrain.
so. how can i achieve my goal?

oh, and by the way, this is what i got 

Code

GeSHi (cpp):
D3DXVECTOR3 Terrain::GetSlope(float x,float y,float size,float zrot,float roll_distance,float pitch_distance)
{
	float roll,pitch;
	float heights[4]={GetHeight(x+cos(RAD(zrot))*pitch_distance,y-sin(RAD(zrot))*pitch_distance,size),
					  GetHeight(x-cos(RAD(zrot))*roll_distance,y-sin(RAD(zrot))*roll_distance,size),
					  GetHeight(x-cos(RAD(zrot))*pitch_distance,y+sin(RAD(zrot))*pitch_distance,size),
					  GetHeight(x+cos(RAD(zrot))*roll_distance,y+sin(RAD(zrot))*roll_distance,size)};
	pitch=atan2(heights[2]-heights[0],pitch_distance*2);
	roll=atan2(heights[3]-heights[1],roll_distance*2);
	return D3DXVECTOR3(roll,pitch,zrot);
}
 
Created by GeSHI 1.0.7.18

GetHeight() is just a member of my terrain class return the height at a particular point.
and yes, i am using z as up...

[draco]

p = atan(a / c) ? You know c right? And A is just the height of the terrain at that point, which you can find, right?

[*brammie*]

the problem is,i dont know where a is as i don't know how long b is...
but i think i can do it like this:


and then p=atan(a/c) is right 

[*brammie*]

hmm.. actually that doesnt work.
it does work for the traingle, but no matter what pitch, a is always the same, but i you calculate b, you are abled to find out the pitch.

so that comes out to this:

Code

GeSHi (cpp):
D3DXVECTOR3 Terrain::GetSlope(float x,float y,float size,float zrot,float roll_distance,float pitch_distance)
{
    float roll,pitch;
    float middle=GetHeight(x,y,size);
    pitch_distance*=((pitch_distance)/(sqrt(pow(pitch_distance*2,2)-pow(middle,2))));
    roll_distance*=((roll_distance)/(sqrt(pow(roll_distance*2,2)-pow(middle,2))));
    float heights[4]={GetHeight(x+cos(RAD(zrot))*pitch_distance,y-sin(RAD(zrot))*pitch_distance,size),
                            GetHeight(x-cos(RAD(zrot))*roll_distance,y-sin(RAD(zrot))*roll_distance,size),
                            GetHeight(x-cos(RAD(zrot))*pitch_distance,y+sin(RAD(zrot))*pitch_distance,size),
                            GetHeight(x+cos(RAD(zrot))*roll_distance,y+sin(RAD(zrot))*roll_distance,size)};
    pitch=atan2(heights[2]-heights[0],pitch_distance*2);
    roll=atan2(heights[3]-heights[1],roll_distance*2);
    return D3DXVECTOR3(roll,pitch,zrot);
}
 
Created by GeSHI 1.0.7.18

i'm still debugging it so don't think it's working atm

(i'm just posting this in case someone needs it)
actually to get 150 posts but that doesnt care =3

[KTC]

(This is from your mathematical description, not try to understand your code.)

From your initial post, you just don't have enough information.

• b = c cos (A)
• a = c sin (A)
• A = tan (a/b)
• sin(A)/a = sin(B)/b = sin(C)/c
• a^2 = b^2 + c^2 - 2*b*c*cos(A)

In 4 of the above case, you're missing one piece of information. In the remaining case, you're missing two.

p = atan(a / c) ?

Um, no! *point to the formulas above*

the problem is,i dont know where a is as i don't know how long b is...
but i think i can do it like this:


and then p=atan(a/c) is right 

Huh? Then you do know "a"! *Slap* Instead of working with triangle ΔABC, just work with ΔAB'C' (i.e. the smaller triangle).

Similar triangle. The angle A ('p' in your original picture) doesn't change.

[FrozenKnight]

To solve a triangle you need at least pieces of information.
 since you know the length of d
 and we know the triangle is a right angle
 along with the length of side c we have enough to solve the triangle.
since the triangle is cba not cbd we need to multiply d*2 to get a.
now there is a nice formula for solving for a missing length of a triangle.
 A^2 + B^2 = C^2 //where C is the hypotenuse.
so to find side b you can use the following formula.
b = sqrt(c*c - 4*d*d);

to get the angle you can use
arcsin((2*d)/c);

You really need to learn Trig, it's not hard but is very rewarding for this kind of stuff.

[edit] sorry didn't see how old the thread was. I am a little tired right now.

[Seismosaur]

Actually, shouldn't he just have to find where one point on each end of the vehicle is, find the difference in height between the two to get eh slope of the line created, then use that information to build a right triangle on the two points and then find the angle?

Sort of like:

[*brammie*]

that's the point..
the vehicle is rotated because of the slope you are going to calculate..

[Seismosaur]

So then what's the problem?

All three sides of the triangle are known.

[FrozenKnight]

LOL, it's not as if all the sides are known, however we do have enough to solve the triangle.

As long as you have 3 values (and the 3 values are not all angles) you can find all values of a triangle. First you need to figure out what type of tangle you have. in this case it's a right triangle so sin cos and tan all work here. now for the tricky trig stuff.

sin p = <opposite side (in this case height or the y value)>/<hypotenuse (in this case the length your cart is traveling)>
cos p = <adjacent side>/<hypotenuse>
tan p = <opposite side>/<adjacent side>

from here it's just simple algebra.
to find the height. we can take sin p and work from there
sin p = <opposite side>/<hypotenuse>
so
<hypotenuse> * sin p = <opposite side>

so in his case
a = sin p * c
b = cos p * c

now that I have explained the basics of trig those of you out there who are geniuses should be able to find all the trig identities. (i know it's possible because with this simple information i was able to do just that.) of course this was after taking pre-calc and i did it just to answer a few questions.
ad for figuring out your carts rotation i suggest you do that before your translation. makes things easier.