Is a Heightfield a Scalar or a Vector and How Many Dimensions Does it Have?

  • Thread starter Thread starter hexa
  • Start date Start date
  • Tags Tags
    Scalar Vector
AI Thread Summary
A heightfield is classified as a scalar because it assigns a height value to each point defined by coordinates (x, y). This means that for every pair of x and y values, there is a corresponding height, making it a 2-dimensional representation in a 3-dimensional space. The confusion may arise from visualizing heightfields as vector fields, but they do not possess direction in this context. Therefore, heightfields are best understood as scalar functions describing a surface. Overall, they illustrate a 2D surface within a 3D framework.
hexa
Messages
32
Reaction score
0
Hello,

I've been wondering if a heightfield is a scalar or a vector, and how many dimensions in space it has.

Sure, if it's a scalar it only has a magnitude, and as a vector magnitude and direction but I cannot see either, or keep them apart in this case. Density or gravityacceleration are easier I guess.

Dimensions: 2 would work out: h=(x,y) but would 1d and 3d possible as well?
Hexa
 
Last edited:
Physics news on Phys.org
Hello Hexa,

in the case of h=h(x,y) you get for each set of values x and y a corresponding height h, which is a scalar. Or to put it another way: For every point on the x-y-surface you get a certain height h. So altogether the function describes a 2-Dimensional surface in a 3-Dimensional space.

I hope that was helpful for you.
 
Hi David,

that was also my intuition. I got confused by an old exam question about visualising such heightfield as vectorfield:confused: and could not find a direction.

Thanks a lot

Hexa

DavidBektas said:
Hello Hexa,

in the case of h=h(x,y) you get for each set of values x and y a corresponding height h, which is a scalar. Or to put it another way: For every point on the x-y-surface you get a certain height h. So altogether the function describes a 2-Dimensional surface in a 3-Dimensional space.

I hope that was helpful for you.
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...

Similar threads

Replies
6
Views
2K
Replies
44
Views
4K
Replies
1
Views
1K
Replies
4
Views
3K
Replies
2
Views
2K
Replies
35
Views
5K
Replies
4
Views
3K
Back
Top