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How to calculate the height of a lunar object from its shadow length

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Published: 13 January 2025

This is something that I have always wanted to do and I think this task is a very worthwhile exercise as apprentice Lunar observer.

On the 2025-01-09, we had a reasonably clear night and my favourite object, the Imbrium Basin was nicely illuminated. The moon was almost 80% full.

This image of the Northern part of Mare Imbrium was used as a basis for this example. The image was taken at 16:55 UT.

On the left hand (west) side, at the Northern end of the 'bay', Sinus Iridum, there is a distinctive triangular shadow that is due to the feature known as Promontorium Laplace. My task is to establish the height of this feature.

The image was cropped and measured as follows:

The large crater to the right (east) of the image is Plato, aged about 3.84MA. It is quite circular with an average diameter of 101km. We need to use this crater as a scale. Simply using a rule, the diameter of the crater was measured at 29mm across the peaks of the rim. Simple arithmetic (Crater Diameter/measured length) gives a scale of 3.48 km per mm.

Next thing to do is to measure the length of the shadow. This is measured from the highlight on the edge of the peak to the end of the shadow and I measured this at 11.5mm. Multiplying the 11.5mm by the scale value (3.48km/mm) gave me a shadow length of 40km.

 

We now have the length of the shadow. To calculate the height, we use the trigonometry formula Tan(sun angle) x Shadow Length.

We now need the Sun Angle. This can be read from almanacs, or, more simply in this world of computer systems there are other and probably more accurate ways of doing this.

This Sun Angle Calculator is an excel spreadsheet that will help you do just do that: https://the-moon.us/images/c/cc/Sun_Angle_Calculator.xls It's very simple to use, just enter the Selenographic co-ordinates of the Lunar Feature and co-ordinates of the Sub Solar Point

This website: https://www.internetsv.info/MoonCalc.html  will calculate the the Sub Solar point Longitude (Lo) and Latitude (Bo), in this case 56.3 E and -1.5 S

 

Note the unusual format for inputting the date and time. the date and time entries are separated by a . <full stop character> without spaces.

We enter the sub solar point co-ordinates onto the spreadsheet, exactly as displayed.

Next we need the selenographic co-ordinates of the feature, Promontorium Laplace. I use Wikipedia for this, I am  great fan and financial supporter of Wikipedia. Its not perfect, but it is certainly not bad. https://en.wikipedia.org/wiki/Promontorium_Laplace

OK, I can see that this also gives me the height as well, but that's not the point..

Enter the co-ordinates into the Lunar Feature box and the Sun Angle is displayed below: Note that you are entering Longitude as a negative number because it is west of the Central Meridian.

 

Sun Angle is 4.50 degrees.

From above, we can now calculate the height..

Height = Tan(sun angle) x Shadow Length.

Height = Tan(4.5) x Shadow Length

Height = 0.0787 x 40km

Height = 3.148km

Which is about 20% higher than the published figure of 2.6km.

Where did the error come from?

This is a very rough and ready method of calculating the height but there are some significant sources of error,

1. The measurements. It is very difficult to accurately measure the distances using a simple metric rule. The diameter of Plato is not easily defined, in this example, I measured across the rim which is apparently the correct way of measuring a crater diameter but the width of the rim and the sun angle makes the measurement points ambiguous. Additionally, the diameter quoted is the average diameter and several measurements across different aspects of the crater should be made and the result averaged. However, from the Earth viewpoint, the crater is oval shaped and this is due to foreshortening. You need to perform the measurement across the widest visible direction, i.e. East to West.

2. More measurements. I am trying to measure something that is approximately 40km long scaled from an object a few mm wide. It would have been much better if two separate images had been used (at the same magnification), a full frame higher resolution image of Plato and also an identically scaled image of the shade and the Promontorium.

3. Measuring directly with a rule on a screen is not ideal. From experience, draughtsmen (used to) use a pair of dividers to make a measurement and then transfer the spacing to a high precision rule. A more modern method would be to use a high end graphics program like 'Photoshop' or 'The Gimp' and measure the distances in pixels. 

4. I did not make any compensation in Longitude for any foreshortening, however, this would have made the apparent shadow length shorter.

I think this was still a very worthwhile exercise. I'll repeat it on some other objects where the depth or height has not been published.

Footnote:

https://the-moon.us/wiki/Promontorium_Laplace offers two different heights for Promontorium Laplace:

• Depth data from Kurt Fisher database
  • Viscardy, 1985: 2.6 km
  • Cherrington, 1969: 3.01 km

Clearly, pre Apollo era, others were having problems calculating the height as well.

A few other thoughts

To minimise errors, the shadow length needs to be as long as possible and across a flat surface (like a Mare).

In the top photograph, the crater to the upper left is La Condamine with a published diameter of 37km. Using the system described above, this gave a scaling factor of 1mm = 3.7km, and a shadow length of 42.55km..  This calculates the object height to nearly 3.5km which supports the thought that the image scales are too small.

Due to foreshortening, this system will only work for objects between 45W and 45E with errors increasing away from the central meridian. 

Conclusion

A worthwhile exercise. Providing that you accept that results may have a 20% error then this will give a very good indication of the height and depths of lunar features. The margin of error can be reduced by using a much larger image.