--- In

PanoToolsNG@yahoogroups.com, Fabio Bustamante <contato@...> wrote:

>

> Now I wondered if this relation between the equirectangular resolution

> and the "true" resolution would be the same between the

> equirectangular's area and the surface of the sphere.

>

> If so...

>

> In a 6000x3000 pano we could assume this sphere has an equator

> circunference of 6000 pixels.

>

> If pi x diameter = 6000 then the diameter of this sphere would be of

> about 1910 pixels and the radius 955 pixels.

>

> Now, the surface's area equation is A = 4 x pi x R^2. So this sphere

has

> a surface of around 11460844 pixels, or roughly 11.4mp.

>

> Could it be that simple?

>

>

Yes, but as Erik mentioned in this thread already, it depends on the

viewing projection used by the user.

One way to simplify the computation above is to use Lambert Equal Area

projection: the number of pixels per surface is always constant

regardless of the pitch. So you only have to multiply its horizontal

and vertical dimensions to get a feeling of its "true resolution".

dmg