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This comment, got posted and I did not understand it.

for those interested, one rod per moonhour is about a quarter furlong per fortnight.

If I googled correctly and jumped to the right conclusions...

The day night line on the Moon moves at about 9.3 miles per hour (14.9KPH)

Which means you could easily drive around the moon at the equator and always stay in day or night (your choice). At lunar noon, drive 20MPH for 4 hours, stop for 4 hours, and it would be lunar noon again.

How fast would you need to average at the lunar equator to stay in eternal day light?

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    $\begingroup$ Using the diameter 3476 km and the moon day of 29.53 Earth days I calculated a speed of 15.4 km/h for day night line at the Moon equator. $\endgroup$ – Uwe Nov 12 '19 at 14:03
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    $\begingroup$ @Uwe while short, your comment is the answer. $\endgroup$ – Carl Witthoft Nov 12 '19 at 14:09
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    $\begingroup$ See the short story "A Walk in the Sun" by Geoffrey Landis: en.wikipedia.org/wiki/A_Walk_in_the_Sun_(short_story) $\endgroup$ – Mark Foskey Nov 12 '19 at 19:59
  • $\begingroup$ @MarkFoskey walking on the Moon in a space suit with 15.4 km/h for 30 days continously is impossible anyway. Far away from the lunar equator the walking speed may be possible for some hours but not for many days. $\endgroup$ – Uwe Nov 12 '19 at 21:27
  • $\begingroup$ A single circumnavigation beginning at local sunrise and arriving at local sunset could go slower. A lot of rough terrain - better have a course well mapped out. $\endgroup$ – Ken Fabian Nov 12 '19 at 22:21
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Using the diameter 3476 km and the Moon solar day of 29.53 Earth days I calculated a speed of 15.4 km/h for the day night line at the Moon equator.

$$\frac{3476 km*π} {29.53*24 h} = 15.4 km/h = 4.28 m/s$$

That is the necessary average speed at the lunar equator to stay in eternal day light.

  • A rod is 5.0292 m, so a rod per moon-hour is 0.0473 mm/s
  • A furlong is 201.168 m, so a furlong per fortnight is 0.598 m/h or 0.1663 mm/s
  • The ratio is then 0.0473/0.1663 or about 0.284
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  • $\begingroup$ @JamesJenkins I face-palmed on that as well. $\endgroup$ – uhoh Nov 12 '19 at 16:17
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    $\begingroup$ @uhoh Many thanks for editing. $\endgroup$ – Uwe Nov 12 '19 at 18:10
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A more generic answer could be:

Angular speed (constant in any point on surface) is 2*pi radians every 29.53*24 hours, which gives 0.00443 rad/h.

Linear speed depend on local distance from rotation axis, which is r * cos(latitude).

Hence linear speed in any point on Moon surface is:

v = 0.00443 * r * cos(lat)

[rad/h]*[km]

Of course for lat=0° it becomes just 0.0443 * r = 0.00443 * 3476 = 15.4 km/h

The "standard walk speed" is around 3 km/h:

3 = 0.00443 * 3476 * cos(lat)

--> lat = arccos(3/(0.00443*3476)) = 78.8°

But sun would be vety low over the horizon, hence maybe covered by craters rims.

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