# NavList:

## A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding

**Re: Calculating Horizontal Parallax**

**From:**Frank Reed

**Date:**2014 Apr 2, 14:59 -0700

Yes. The parallax of any (reasonably distant) celestial body when it is seen on the horizon, and thus known as the "horizontal parallax", is identical to the apparent semi-diameter of the Earth as seen from that celestial body. Draw a picture, and you'll see that this makes sense. To calculate the angular size of any (reasonably distant) object, you divide the linear size measured across the line of sight by the distance to the object. That works for lighthouses, stratospheric aircraft, distant galaxies, and nearby planets, too. In this case, take the radius of the Earth and divide by the distance from the celestial body to the Earth (both in the same linear units, whether miles, kilometers, or whatever is most convenient). The result is the angle as a pure ratio (also known as an angle "in radians"). Note that the sine function is not really necessary here (for "reasonably distant" objects). To convert to degrees multiply by 180/pi. Or to convert to minutes of arc multiply by 60*180/pi, which is a very handy "magic number" nearly equal to 3438. THIS WORKS ANYWHERE. Whenever you want an angular size in minutes of arc, divide the linear dimension ACROSS the line of sight by the distance OUT to the object along the line of sight and then multiply by 3438. In a short equation:

M.O.A. = 3438*(distance across)/(distance out)

You asked: "Is a figure of 6,378 km for the radius of Earth sufficient for most calculations?"

Yes. If more accuracy than three significant figures were ever required, you would also need to worry about the Earth's oblateness (polar flattening). But normally you wouldn't worry about that, and you would not need more accuracy. In fact if you're only trying to get the HP of Venus or Mars to the nearest tenth of a minute of arc, then a radius of 6000 km is sufficient.

You also asked:

"Also, MICA gives distances to nine decimal places. Is it necessary to use all of them?"

Try it and see! If you round to TWO decimal places, what happens?? Calculation is cheap these days, so you can easily try modifications to the accuracy of your inputs. If you wiggle it up or down a little, does it significantly affect the accuracy of the result you're looking for? The key, by the way, is to look at significant figures, which usually do not correspond to "decimal places". Ask if that's not clear.

-FER

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