Measuring angular diameters
We have inspiration from the observations of Samanta Chandrasekhar of Orissa, who had observed (perhaps also predicted) the 1874 transit of Venus, with handmade instruments, and made such accurate determination of the relative angular sizes of the Sun and Venus.
We need to measure, independently, the angular diameters of the Sun and Venus. Angular diameter of the Sun - this is a measurement that we could do at any time of the year, surely? Yes, just keep in mind though, that the angular diameter of the Sun does very during the year (Earth's orbit being ever so slightly elliptical), but, the variation is rather small - about a minute of the arc during a year's time.
In any case, close enough to the transit, we can make this measurement at any time. One simple method to measure the angular diameter of the Sun would be to project its image using a telescope, binoculars, a simple pinhole camera or a heliostat and allow the image to drift across the projection screen, due to the rotation of the Earth.
We only need to time, as accurately as possible, the time the projected image takes, to drift across one whole diameter of the image. It will help if we set up our projection apparatus, fix the distance where the screen is placed and thereby fix the physical dimensions of the projected image of the Sun. By placing a graph sheet on the projection screen, one could mark quickly (before the image drifts out) a few points on the circumference of the projected image of the Sun and thereby determine the diameter of the projected image. Draw a circle of this diameter on a white paper, remove the graph sheet and insert this paper on the projection screen.
Adjust the projection so that the image could be poised to enter this circle. Wait for the image of the Sun to exactly drift into this circle. Start timing at this instant and time the event to the point when Sun just drifts out of this circle. This gives the time that is taken for a whole diameter of the Sun to move, on account of the rotation of the Earth.
It seems a straightforward calculation after this - given that the Earth rotates through 360 degrees in 24 hours - if the Sun takes T minutes to drift out of the circle - its angular diameter should be (T * 360 degrees) /(24 * 60) ?
It would be that, when we are close enough to the equinoxes - when the Sun shines directly overhead at the Equator.
At other times we need to make some additional geometric corrections to this. This is clear from the figure below -
a is the angle that the celestial sphere turns through, in a given time T. If an object is on the celestial Equator, it will appear to turn through the same angle a in time T.
But, an object not on the celestial equator, will turn through a smaller angle than a in time T - the angle A'B' being - r * a. Here, r is smaller than R the radius of the assumed celestial sphere.
r is then R Cos d where, d is the declination of the object in the sky. Declination, for an object in the sky, is the equivalent of lattitude on the Earth. It is the angular distance, measured in degrees, north or south of celestial equator (projection of Earth's equator onto the sky). Northerly declinations are positive and southerly ones negative.
The angular diameter of the object in the sky is then
angular diameter = A'B'/R
= a R Cos d /R
= a Cos d
a can be determined observationally as (T * 360 degrees) /(24 * 60), T being the time taken by the object to drift across its projected diameter.
We need to know the declination of the object, of course and that is another story :-)