<p>So, I want to find the distance to the Moon/any body after performing the correction for diurnal parallax. Unfortunately, the resources I have do not cover how to do this. Is there an easy algorithm to adjust the distance to an object after switching to topocentric coordinates, assuming an ellipsoidal Earth?</p>

CC BY-SA 4.0
2020-05-06T21:32:27.547Z
Sure it's possible. It would be an exercise in trigonometry.  I'm curious about the reason that you need this.  Do you want the distance to a particular point on the moon, or to the moon's centre, or just the closest point on the moon to your position?
James K
CC BY-SA 4.0
2020-05-06T22:21:23.190Z
@James K The moon's center. I'm writing an eclipse predictor and simulator, and I need the distance from the surface to the moon's center to calculate it's semidiameter to accurately simulate what eclipses look like. Currently, I'm just using the distance from Earth's center, but that's not ideal.
John Dumancic
<distances><coordinate><parallax>
Determine distance to Moon after diurnal parallax?
John Dumancic
# Determine distance to Moon after diurnal parallax?

(Asked by: username_0 on 2020-05-06)

So, I want to find the distance to the Moon/any body after performing the correction for diurnal parallax. Unfortunately, the resources I have do not cover how to do this. Is there an easy algorithm to adjust the distance to an object after switching to topocentric coordinates, assuming an ellipsoidal Earth?

**Tags:** distances, coordinate, parallax
> **Comments:**
> **username_1**: Sure it's possible. It would be an exercise in trigonometry.  I'm curious about the reason that you need this.  Do you want the distance to a particular point on the moon, or to the moon's centre, or just the closest point on the moon to your position?  
> **username_0**: @James K The moon's center. I'm writing an eclipse predictor and simulator, and I need the distance from the surface to the moon's center to calculate it's semidiameter to accurately simulate what eclipses look like. Currently, I'm just using the distance from Earth's center, but that's not ideal.  

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