Distance from node to node

[admgamer]

Hi,

I have a SceneNode that is in a static location, and i basically want to check if the FPS camera that you control is within a certain distance from this SceneNode.

I thought you would need to do something such as get the node position with getPosition, and then check if the player controlled camera is within a certain radial distance but i'm having trouble figuring out how to easily do this.

does anyone know any functionality that can do this? i think i'm right in thinking that just using XYZ of scene nodes isnt enough as this doesn't help one bit with the player being somewhere within the circumference?

[MasterGod]

You are in the right direction:

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n1->getAbsolutePosition().getDistanceFrom(n2->getAbsolutePosition())

While n1 is the node and n2 is the camera.

[admgamer]

thanks for the reply, i JUSt saw this getDistanceFrom and realised how much of a moron i am lol

cheers

[Masterhawk]

do you just want to have a distance between two points?

d = point1 - point2

distance = sqrt((d.x)^2 + (d.y)^2 + (d.z)^2)


i think that should do the trick...

EDIT: too late and too complicated....

[Acki]

distance = sqrt((d.x)^2 + (d.y)^2 + (d.z)^2)

but this wouldn't work like you expected...
in C/C++ the ^ doesn't mean "power of" it means xor !!!

so it should be:

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distance = sqrt(pow(d.x, 2) + pow(d.y, 2) + pow(d.z, 2));

or:

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distance = sqrt((d.x * d.x) + (d.y * d.y) + (d.z * d.z));

[Halifax]

Yeah, I would go with the latter, Acki. I don't know if I am correct, but the pow() function would present some overhead, especially since I believe it uses a genric algorithm so that you could also calculate square roots with pow() etc. Either way, it is still fast, but I just wanted to put that out there.

Correct me if I am wrong, because this has always puzzled me.

[Acki]

I'm not sure about the overhead, too...
I for myself would also prefere the second line I posted...

and of course you can do a square root, or any other root, with pow(), you just need to calculate in the power of the reciprocal value...

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// get the square root 
f32 val = pow(4, 1.0 / 2.0); // val = 2.0
f32 val = pow(9, 1.0 / 2.0); // val = 3.0

or the cubic root:

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// get the cubic root 
f32 val = pow(8, 1.0 / 3.0); // val = 2.0
f32 val = pow(27, 1.0 / 3.0); // val = 3.0

[Masterhawk]

distance = sqrt((d.x)^2 + (d.y)^2 + (d.z)^2)

but this wouldn't work like you expected...
in C/C++ the ^ doesn't mean "power of" it means xor !!!

Of course you're right. This peace of "code" should only show the mathematical way and wasn't intended to be "correct c/c++"

[Halifax]

distance = sqrt((d.x)^2 + (d.y)^2 + (d.z)^2)

but this wouldn't work like you expected...
in C/C++ the ^ doesn't mean "power of" it means xor !!!

Of course you're right. This peace of "code" should only show the mathematical way and wasn't intended to be "correct c/c++"

I'm sure he was just correcting you, so that possibly you would show a correct form next time. It is assumed that everyone on this forum uses C/C++ so it is best to put it in that language, for the newbies.

[rogerborg]

Just as an aside, getDistanceFromSq() (and comparing it to Distance * Distance) is more efficient, as it avoids the sqrt().

[hybrid]

pow is equivalent to the direct multiplication once you disable IEEE precision requirements. That's e.g. possible by passing the -ffast-math to the gcc (and therefore made in the Makefile for the optimized compile). Otherwise it's usually a little slower, but more accurate.

[Halifax]

pow is equivalent to the direct multiplication once you disable IEEE precision requirements. That's e.g. possible by passing the -ffast-math to the gcc (and therefore made in the Makefile for the optimized compile). Otherwise it's usually a little slower, but more accurate.

Ah, okay, thanks for clearing that up hybrid. Now I know, because I used to be a heavy advocate of not using it for something that could be represented with straight multiplication.