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Let's use transverse Mercator with a custom zone, with central meridian 30, the zone is 3 degrees wide at the equator, with a scale factor k=1, it means that the cylinder touching the spheroid along the central meridian.

However, with scale factor k=0.9996, the cylinder is secant of the spheroid at two meridians, one at right of Central Meridian and the other at left. How to find these meridians for a given scale factor k at equator?

mkennedy
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user9322960
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    OP, I removed the reference to a conical projection (Lambert conformal?) because it makes the question too broad. Please feel free to undo my changes if you disagree. – mkennedy Dec 04 '18 at 20:25
  • Related: https://gis.stackexchange.com/questions/122703/influence-of-the-scale-factor-on-the-projection/122751#122751 Please check it out. – mkennedy Dec 04 '18 at 20:27
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    Would a spherical approximation be sufficient for you? Also, secant lines in transverse mercator do not follow meridians, so just to clarify, are you looking for the meridian (longitude) of the secant line at the equator only? Or at any given parallel? – FSimardGIS Dec 04 '18 at 21:40
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    @RalphTee -- Spheroid = Ellipsoid – Martin F Dec 05 '18 at 20:28
  • @MartinF You are right! I misread the definition (i.e., 2 axes of equal length vs 3 axes). I had deleted my previous comment to avoid confusion. – Ralph Tee Dec 06 '18 at 02:03

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According to wikipedia.../Universal_Transverse_Mercator_coordinate_system

In each zone the scale factor of the central meridian reduces the diameter of the transverse cylinder to produce a secant projection with two standard lines, or lines of true scale, about 180 km on each side of, and about parallel to, the central meridian (Arc cos 0.9996 = 1.62° at the Equator). The scale is less than 1 inside the standard lines and greater than 1 outside them, but the overall distortion is minimized.

You should interpret that 1.62° at the Equator as the angular distance of the standard lines (not meridians) either side of the central meridian. A more precise value for arc cos 0.9996 is 1.62062° or 1°37'14.2" but note that all the above is for the spherical case; the ellipsoidal case is more complicated.

Martin F
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  • I think, It would be sufficient to me with this spherical case bcz the ellipsoidal case is more complicated as you say, but software like QGIS and co. handles the +proj=tmerc ... as spherical case also? – user9322960 Dec 05 '18 at 08:58
  • do QGIS and ArcGIS handle the projection from a sphere if I understand good from you (at least with a transverse cylindrical projection)? – user9322960 Dec 05 '18 at 22:37
  • I suggest you ask a different, specific question. I'm not knowledgeable enough of those softwares. – Martin F Dec 06 '18 at 04:27
  • Ok, I will do ;) – user9322960 Dec 06 '18 at 08:41