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I would like to know how to calculate the grid convergence angle between True North and Grid North for UTM maps.

I am using ArcGIS 10.0 (ArcView licence). and have used the Calculate Grid Convergence Angle tool from the Cartography toolbox.

However, I would like to know how to calculate this angle myself (using Excel) to compare against the ArcGIS result?

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    Excel will only be able to compute the grid convergence for projections whose formulas you have code for. A general solution, applicable on any platform that will project geographic coordinates, appears in my reply at gis.stackexchange.com/a/5075/664. – whuber Sep 29 '14 at 15:26
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The formula for calculating grid convergence (sometimes called meridian convergence) for spherical UTM projections was given (very incorrectly until just now) at How to Calculate North?

In case that is not clear

γ = arctan [tan (λ - λ0) × sin φ]

where

γ is grid convergence,
λ0 is longitude of UTM zone's central meridian,
φ, λ are latitude, longitude of point in question

See Transverse_Mercator_projection#Convergence where it also gives γ as a function of grid coordinates.

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  • Where do I find λ0 for a given UTM zone? Do you have a link to a list? – evolved Feb 22 '18 at 15:24
  • @evolved: Try this jaworski.ca/utmzones.htm or this apsalin.com/utm_central_meridian_lookup.aspx – Martin F Feb 23 '18 at 23:56
  • Just curious, what is the difference between this answer and @V. Kelly Bellis answer? The results come out close. – Rex Aug 10 '18 at 14:03
  • @Rex: The other seems to be an approximation. – Martin F Aug 10 '18 at 17:48
  • I'm surprised I haven't found a formula yet for grid convergence for ellipsoidal UTM projections. AFAIK UTM doesn't typically use a spherical projection. Spherical transverse mercator is common in web maps, however. – Andy Nov 16 '19 at 7:10
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I've always used this formula:

CA = (λ - λCM) × sin φ

Given: CA is the grid Convergence Angle for any transverse Mercator projection; λCM is the longitude of the zone's Central Meridian; φ, λ are the latitude and longitude of the point in question.

Pay attention to the resulting algebraic sign!

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    That is true for small angles: arctan [tan (λ - λ0) × sin φ] ~ (λ - λ0) × sin φ – Hans Erren Dec 5 '20 at 14:45
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The book "The Mercator Projection" by Peter Osborne (https://zenodo.org/record/35392#.XvGjn5MzY3h) goes a bit in depth on this topic, but it seems to be no trivial topic for the general transverse mercator projection of the ellipsoid, however it agrees with Martin and Bellis for the spherical case.

It should still be the same angle regardless of wheter it is calculated in the projection or on the ellipsoid/tangential plane by the conformal property of the projection.

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Complementing the answers by @Martin F and @V. Kelly Bellis, an article by Ontario land surveyor Tim Hartley (see p.4) gives the formula:

CA = 32.39 arc seconds * (distance from central meridian in km) * tan(latitude)

The other answers require longitude of each point where being calculated, while this formula can be used with the UTM/MTM X coordinate, yielding:

CA = (8.997E-06) degrees * (delta X from central meridian in m) * tan(latitude)

You still have to calculate tan(latitude), but in tropical and temperate regions tan(latitude) changes much, much less per delta Y than the linear dependency on delta X, (and the convergence error of true longitude vs delta X itself) so you can do this once for your whole map for a survey area.

I expect this is equivalent to @V. Kelly Bellis's formula, with the tan() representing the 1/cos() dependency of (spherical) ground scale by latitude combined with the sin() in that formula, with units converted.

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Complementing the answers by @Martin F

Grid Convergence = atan {tan [longitude - (6 × Grid Zone Designation Number - 183)] × sin (latitude)}

The longitude of the zone's Central Meridian can be calculated using this formula: (6 × Grid Zone Designation Number - 183).

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