NavList:

Alexander Duytschaever you wrote:
The number at the bottom center seems to be a Right Ascension reference.

Looking at slide with RA19.0 labeled at the bottom, it seems that the delta time from Altair (~0:17) to Sirius/Procyon (~0:58) is 0:41. On a star map, that distance covers (along a great circle, the center lines look like great circles) ~170 degrees, which makes 360 degrees cover in 1h26m. 

Now that Ed has produced the more photographs, I can see that what I thought was the smudge of some far off galaxy does in fact say RA.  I don’t think the right ascension is measured along the strips.  I think it’s a way of referencing the strips.  How about this for a thesis?

The object has a period of approximately 88 minutes, which means it must be in quite a low orbit.  The orbit is tilted at approximately 31 degrees to the celestial equator.  It either crosses the celestial equator at a different point on each pass, or the strips are to cover the possibility of the orbit not being in quite the orbit planned.  The RA figure shown for a particular strip is the hour angle that the orbit for that strip crosses the celestial equator moving from south to north.
   
Test: On the RA=20.0 orbit strip, the line should cross the celestial equator at                 SHA=360-20x15=060. 

Looking at my diagram here: http://fer3.com/arc/m2.aspx/Curious-star-charts-DavidPike-nov-2020-g49154 , it does.

The time down the left-hand edge of the strip is the time to reach that position after the body crosses the celestial equator from south to north.  This should help us get an orbital period.  If we can find two points 180degrees SHA (12 hours RA) apart, we can double the orbit time between the two points and get the period.  E.g. on my diagram above, the body passes Alpheratz after 29 minutes and the lower tip of Corvus after 73 minutes, so the half period is 73-29=44 minutes and the full period is 88minutes or 1 hour 28 minutes.
Because the height corresponding to a particular period changes so quickly close to the Earth’s surface, I don’t think I can suggest a height bearing in mind the approximateness of my working.  DaveP