Re: Determining Curcuit Angle of View

[Joe McCary <mccary@xxxxxxxxx>]

Ok its time for me to step up to the plate and maybe make a big fool of
myself.

My HS geometry class taught me that the circumference of a circle is equal
to 2*Pi*Radius.  If I use my Noblex as an example; focal length is 50mm, the
full circle circumference would be 2*50* 3.14 (pi) for a total of 314mm.
That would be 360°.  If the  Noblex film is 120mm long (5x12) divide the
120mm into the 314mm and you get 2.61 parts of 360°.  Divide the 360 by the
2.61 and you get 137°.  That is ABOUT the horizontal angle of the Noblex
150.

So, what was the focal length of the lens used to create this 70" panorama?

Joe McCary
(hopping my logic is not too far off)

----- Original Message -----
From: "ADavidhazy" <ANDPPH@xxxxxxxxxxxxxxxxxx>
To: <panorama-l@xxxxxxxxxxxxxxxxx>
Cc: <ANDPPH@xxxxxxxxxxxxxxxxxx>
Sent: Thursday, August 24, 2000 11:43 PM
Subject: Re: Determining Curcuit Angle of View


> It seems to me this would be fairly difficult to do without any kind of
> "benchmark". One way to do it would be to go to the origial locations and
> determine the angle right there ... but, of course, this may not be
possible.
> Or get a map of the area and deal with it in a similar way by identifying
> landmarks.
>
> Otherwise ... I myself can not think of a way other than very rough
> approximation. On the other hand, if they are more than 360 degrees then
the
> situation becomes easier! Or if they are almost 360 degrees then it might
be
> possible to "fill in the gap" with imagination and deduce an approximate
angle
> that way.
>
> adios,
>
> Andrew  o o  0 0 o . o  Davidhazy, Imaging and Photo Tech
>          \/\/\/\/\/\/          http://www.rit.edu/~andpph
> __________|        |_____________________________________
>
>
> > Fellow Panoramic Photographers,
>
> > I have been given two, recently reprinted,
> > circuit panoramic contact prints 10" x 70".
>
> > I would like to determine the angle of view
> > if possible.
>
> > Can anyone help me out or point me in the
> > right direction?
>
> > Randy
>
>
>