Volume displacement.  You can also calculate density using the Archimedes
principle.

Kevin


> Not even close.  Even using two orthogonal projections directly, rather
> than just the summary statistic of area or perimeter, doesn't uniquely
> determine volume.  If you had three orthogonal projections you could begin
> to produce a rough approximation of the volume, for example by
> discretization using the intersections of the pixels to determine the
> voxels occupied.  This still wouldn't account for any concavities or
> hollows in the object, which could be significant depending upon the
> application (e.g. in estimating the volume of a bell this way you could be
> off by orders of magnitude), or fine scale structure of the object.
> Three projections is hardly the magic number either.  Imagine your object
> was a sphere, if you took three orthographic projections of the sphere
> they would all be circles.  If you pasted these circles on rectangular
> solid (e.g. block of wood) and cut them out in turn (e.g. with a band
> saw), what you would get would hardly be a polished sphere.  Also, photos
> are not orthographic projections, so even after accounting for lens
> distortion your pixels define pyramidal sections in space, not rectangles,
> so their intersection defines complicated 3d polygons rather than simple
> rectagular voxels.
>
> Best of luck,
>
>  - Mike
>
> On Sun, 22 Jul 2007, A. Rabatsky wrote:
>
>> I'd like to know how to calculate volume using 2 measures of either area
>> or perimeter. I have 2 digital photographs, one provides a top view and
>> one provides a side view, of the object that I want to calculate the
>> volume of. I need to know if there's a mathematical formula that I can
>> plug in area from the top view & area from the side view to get volume.
>> I
>> can add photos later if need be.
>>
>> Thanks.
>>
>
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