Is the height of sand as a function of time any different between a 2D hourglass and its 3D counterpart? I'm trying to (eventually) model a 3D hourglass by calculating the height of sand in both upper and lower sections as a function of (percent of total) time, and representing that in 2D.
So I suppose there will be further questions.
ESTABLISHED CONDITIONS:

*

*we have an hourglass (sand timer) that is derived from a mathematical shape such as two mirrored cones or something more complex like a lemniscate.

*the volume of sand is equal to exactly one half of the hourglass -- one of the two sections

*I am not modeling the physical properties of sand -- we can assume a level top surface

*sand falls at a constant rate

*we know the amount of time is takes for all the sand to fall

Since sand falls at a constant rate, we know that the volume of sand (as a percentage of total volume) in the lower section will always be directly proportional to the percent of time expired.
Therefore, the height of the sand in either upper or lower sections should always be some function of:

*

*The shape of the section

*The amount of the sand in it

But before I start trying to do this, my first question is, am I correct to assume that we will get the same height value whether we think/calculate in terms of the 2d version (using only areas) vs the 3d version (using only volumes) of the hourglass -- the "real" hourglass vs its own [perfect/unskewed/unaltered] shadow on the wall?
This would not be true for a shape whose 3d "version" could not be inferred from its 2d "version."
But, in this case, I think this is true because it makes intuitive sense, and the 2d version of an hourglass is simply a "subsection/cross-section" (and is the same at any angle of rotation) of the 3d version.
But on the other hand, I think it could be false because if we were to take the same logic and step down to a one-dimensional cross section, it would become nonsensical and obvious that we are losing information.
 A: For a specific example, let's consider an hourglass shaped like a finite double cone:

The shadow of this hourglass on the wall is just two isosceles triangles meeting at their apexes.
If you fill the bottom cone so that the height of the sand is half the height of the cone, you will have filled $7/8$ of the cone's volume. If the hourglass is a one-hour hourglass it will take $52.5$ minutes to fill it to that level.
But if you draw a horizontal line across the triangular shadow of the cone, exactly halfway between the bottom edge of the triangle and the top vertex of the triangle,
the area below that line will be just $3/4$ of the area of the triangle.
So if you were to use the area of the vertical cross-section (shadow) of the hourglass in order to estimate how long it takes the sand to reach a certain level, you would think that it takes $45$ minutes for the sand to reach the level halfway between the bottom and the top of the lower cone of the hourglass.
So the answer to this question is "no":

am I correct to assume that we will get the same height value whether we think/calculate in terms of the 2d version (using only areas) vs the 3d version (using only volumes) of the hourglass -- the "real" hourglass vs its own [perfect/unskewed/unaltered] shadow on the wall?

The "using only areas" approach will work if the bottom of the hourglass is a cylinder with a vertical axis. Otherwise not.
