Maps in the Lambert equivalent azimuthal projection are often used in normal aspect in atlases for maps of the polar regions, i.e., the northern and southern hemispheres. The transverse aspect is usually used for atlas maps of the eastern and western hemispheres. The oblique aspect is used for atlas maps of continents and oceans. The projection is recommended for maps of approximately circular areas that need to preserve the area [8].

In the simple normal aspect of this projection, the meridian images are straight lines that intersect in the image of the pole. The angles between the meridian images are equal to the angles between the meridians on the sphere. The parallel images are concentric circles. The distances between the parallel images gradually decrease with distance from the pole. In this projection, the entire sphere of the Earth can be shown (Fig. 1). The distortion is moderate in the image of a hemisphere but becomes very large on a map of the whole Earth (Fig. 2).

Let us derive the equations of the simple normal Lambert equivalent azimuthal projection. The equivalence condition in normal azimuthal projections of the sphere is expressed by the equation (10)p=hk=1,where h and k are local linear scale factors along the meridian and parallel [8]. Inserting the expression for the local linear scale factors (7) into (10) we obtain (11)−dρdφρcos⁡φ=1,from where (12)ρdρ=−cos⁡φdφ.

By integrating, we get (13)ρ22=−sin⁡φ+K.

If we set the condition that the North Pole is mapped as a point (for φ=π/2 it should be ρ=0), we get (14)K=1.

By inserting this value into (13) we get (15)ρ2=2(1−sin⁡φ).

This expression can also be written in the form (16)ρ2=4sin2⁡(π4−φ2),so it is finally (17)ρ=2sin⁡(π4−φ2).

We chose the sign in (17) so that ρ≥0.

If we insert the obtained value for ρ into the expression for the local linear scale factor k (9), we obtain (18)k=2sin⁡(π4−φ2)cos⁡φ=1cos⁡(π4−φ2),and then (19)h=1k=cos⁡(π4−φ2).

Therefore, the mapping equations and expressions for the local linear scale factors and distortion in simple normal equivalent azimuthal projections are: (20)ρ=2sin⁡(π4−φ2),x=ρsin⁡λ,y=−ρcos⁡λ (21)h=cos⁡(π4−φ2),k=1h (22)p=1,sin⁡ω2=|h−k|h+k.

A map of the world in the simple normal equivalent azimuthal projection is shown in Fig. 1. From expression (20) we see that the images of all parallels are circles. If φ=π/2, then ρ=0, the north pole is mapped to a point. If φ=0, then ρ=2. The Equator is mapped into a circle of radius 2. If φ=−π/2, then ρ=2. The South Pole is mapped into a circle of radius 2 (Fig. 1).

For a point to be standard for a certain normal azimuthal projection, it is necessary and sufficient that at that point [7] (23)h=k=1.

For the normal Lambert equivalent azimuthal projection (20), condition (23) reads: (24)cos⁡(π4−φ2)=1cos⁡(π4−φ2)=1,from which immediately follows the unique φ=π/2 from the interval [−π/2,π/2] and then ρ=0. So, in that projection, we have one standard point with the coordinates (x,y)=(0,0). The distribution of linear distortions is shown in Fig. 2.

If we set the condition that the North Pole is mapped not only as a point, but as a circle (Fig. 3), it follows from (13) (See also Section 3.3)(25)ρ=2(K−sin⁡φ),and in that case it should be (26)K≥1,so that the expression under the root is non-negative. From (25) it is easy to see that the North Pole will be mapped into a circle of radius 2(K−1), the Equator into a circle of radius 2K, and the South Pole into a circle of radius 2(K+1).

For the radius of the parallel images defined in (25), condition (23) gives (27)2(K−sin⁡φ)=cos⁡φ.

After squaring, expression (27) can be written in the form (28)(sin⁡φ−1)2=2(1−K).

Since the left side of (28) is always non-negative, and by assumption (26) and (28) will be fulfilled only for K=1. Therefore, the only simple normal azimuthal equivalent projection that has a standard point is the one for which K=1, i.e., the one for which Eq. (20) hold, which is the simple Lambert equivalent azimuthal projection.

In the cartographic literature, tangent and secant projections are often mentioned. I am committed that the terms tangent and secant projection should be avoided in general, because they only make sense in perspective projections [9]. First, the derivation or definition of the Lambert equivalent azimuthal projection makes no mention of tangency or contact. The position of the projection plane in relation to the sphere being mapped is completely irrelevant. Furthermore, conclusions about an undistorted circle if a secant plane is used are a complete failure, because in the previous section we showed that an undistorted circle exists only for K=1, when the pole is mapped to a point.

The radius of the parallel corresponding to the latitude φ on the unit sphere is equal to cos⁡φ. In order for that parallel to be “secant” for the Lambert equivalent azimuthal projection, ρ(φ)=cos⁡φ should be valid, and this is the relation (27) that leads to the conclusion stated after Formula (28).

If K≥1, we have the most general case of the normal Lambert equivalent azimuthal projection (Eq. (25)). The North Pole will be mapped into a circle of radius 2(K−1) (Fig. 3). The local linear scale factor along meridians h will always be less than 1, while the local scale factor of linear along parallels k will be greater than 1 (Fig. 4). The blue and red lines in Fig. 4 nowhere reach the value 1. Thus, for K>1, there is no parallel for which h=1 or k=1 would hold, and especially h=k=1. In other words, for K>1 there is no parallel whose image in the Lambert equivalent azimuthal projection would be undistorted.

If K>1, the entire world is depicted in a circular wreath. This general case of a normal Lambert equivalent azimuthal projection has not been investigated in the theory of map projections so far and seems to have no special practical value.

To conclude, it makes no sense to talk about tangent and secant projections in general, because the position of the projection plane is in no way connected to the sphere being mapped except for perspective projections. In addition, the introduction of the names tangent and secant can lead to wrong conclusions about the distribution of distortions.

It is commonly assumed that standard parallels and parallels that appear as the intersections of a developable surface and a sphere or ellipsoid coincide. In a previous paper [9], it has been shown that this is not true for azimuthal projections which are equidistant along meridians, equidistant along parallels (orthographic), or equivalent, because there is no standard parallel at all in such projections. Only some azimuthal conformal (stereographic) projections have a standard parallel, in which case it coincides with the secant parallel. Furthermore, there are azimuthal projections with two secant parallels and with infinitely many secant parallels. One can read more about this in Lapaine’s articles [9,10].

In the transverse aspect, the central meridian is shown as a straight line. Meridians that are 90° from the central meridian form a circle. The images of the other meridians are complex curves that intersect at both poles. The distance between the images of the meridians decreases with distance from the central meridian. The image of the Equator is a straight line. The images of the other parallels are complex curves concave towards the nearer Pole. The distance between the images of the parallels decreases with distance from the Equator. Along the image of the meridian that is 90° from the central meridian, the parallels are placed at equal intervals (Fig. 5).

Let us recall how the equations of the transverse Lambert projection are formally derived from the equations of the normal Lambert projection. First, Eq. (20) are written as follows: (29)ρ=2sin⁡(π4−φ′2),x′=ρsin⁡λ′,y′=−ρcos⁡λ′,and then a substitution is made: cos⁡φ′cos⁡λ′=cos⁡φsin⁡λ cos⁡φ′sin⁡λ′=sin⁡φ (30)sin⁡φ′=cos⁡φcos⁡λ.

Substitution (30) is a special case of the general transition from the normal aspect of the projection to the transverse one: cos⁡φ′cos⁡λ′=cos⁡φcos⁡(λ−λP) cos⁡φ′sin⁡λ′=sin⁡φPcos⁡φsin⁡(λ−λP)+cos⁡φPsin⁡φ sin⁡φ′=−cos⁡φPcos⁡φsin⁡(λ−λP)+sin⁡φPsin⁡φ,which is Formula (10) in the article by Lapaine and Frančula [11]. When the values φP=0, λP=π/2 are taken for the coordinates of the new pole, expressions (30) are obtained. Let us first calculate ρ=2sin⁡(π4−φ′2)=2(1−sin⁡φ′)=2(1−cos⁡φcos⁡λ), sin⁡λ′=sin⁡φcos⁡φ′=sin⁡φ1−cos2⁡φcos2⁡λ, (31)cos⁡λ′=cos⁡φsin⁡λcos⁡φ′=cos⁡φsin⁡λ1−cos2⁡φcos2⁡λ,and then according to (29) (32)x′=2sin⁡φ1+cos⁡φcos⁡λ,y′=−2cos⁡φsin⁡λ1+cos⁡φcos⁡λ.

To get a map with the cardinal points arranged as we are used to, i.e., North is up and South is down, it is necessary to rotate the image by 90°, i.e., apply the expressions (33)x=−y′,y=x′.

In this way we get the equations of the simple transverse Lambert projection (34)x=2cos⁡φsin⁡λ1+cos⁡φcos⁡λ,y=2sin⁡φ1+cos⁡φcos⁡λ.

Let us notice that the expression in the denominator in (34) can be equal to zero. Since φ and λ are geographical coordinates, this will happen if φ=0 and at the same time λ=π or λ=−π. We cannot map a point on the sphere whose coordinates are (φ,λ)=(0,π) or (φ,λ)=(0,−π) in the projection plane using Eq. (34). We will omit this point from further considerations.

With (35)cos⁡z=cos⁡φcos⁡λ

Eq. (34) look a little simpler (36)x=21+cos⁡zcos⁡φsin⁡λ,y=21+cos⁡zsin⁡φ.

Let us calculate x2+y2=21−cos2⁡z1+cos⁡z=2(1−cos⁡z)≤4.

This means that a map of the entire world made in the projection to which Eq. (34) are applied will be placed in a circle of radius 2 (Fig. 5).

For φ=±π/2, cos⁡z=0, x2+y2=2, x=0, y=±2. So, the Poles are mapped to points.

For λ=±π/2, cos⁡z=0, x2+y2=2, we have the equation of a circle with its center at the origin and radius 2. The images of two meridians λ=±π/2 form a circle.

The considerations on the transverse Lambert equivalent azimuthal projection carried out in Section 4.1 are sufficient to achieve my goal, i.e., to apply it in further research on the Hammer projection. However, for the sake of completeness, let us briefly present the general case. If instead of the equation ρ=2(1−sin⁡φ)=2sin⁡(π4−φ2) when deriving the equations of the transverse Lambert equivalent azimuthal projection we start from the more general expression ρ=2(K−sin⁡φ) with K≥1, and apply the same procedure as in Section 4.1, we will obtain equations that can be written in the form (37)x=2(K−cos⁡z)1−cos2⁡zcos⁡φsin⁡λ,y=2(K−cos⁡z)1−cos2⁡zsin⁡φ.

In the special case for K=1, Eq. (37) reduce to the already known (36). For K>1 we will obtain an image of a pseudopol in the form of a circle of radius 2(K−1) (Fig. 6). The entire world is depicted in a circular wreath. This general case of a transverse Lambert equivalent azimuthal projection has not been researched so far in the theory of map projections.

Partial derivatives are needed to estimate projection distortions. We will focus on the special case of the transverse Lambert equivalent azimuthal projection and Eq. (36), because we have seen the general case is not promising, while for further investigation of the Hammer projection only a simple form is needed. Let us do the math first: ∂∂φcos⁡z=−sin⁡φcos⁡λ,∂∂λcos⁡z=−cos⁡φsin⁡λ (38)∂∂φ1+cos⁡z=−sin⁡φcos⁡λ21+cos⁡z,∂∂λ1+cos⁡z=−cos⁡φsin⁡λ21+cos⁡z.

After that, we can calculate the partial derivatives: (39)∂x∂λ=cos⁡φ2(1+cos⁡z)3[cos⁡λ(2+cos⁡z)+cos⁡φ] (40)∂x∂φ=−2sin⁡φsin⁡λ(1+cos⁡z)3(2+cos⁡z) (41)∂y∂φ=12(1+cos⁡z)3[cos⁡λ+cos⁡φ(2+cos⁡z)] (42)∂y∂λ=sin⁡φcos⁡φsin⁡λ2(1+cos⁡z)3.

Let us look for a different way to write the Lambert projection equations that would be more convenient for calculating partial derivatives. The following relation is known from trigonometry (43)1+cos⁡z=2cos2⁡z2.

Using (43), the equations of the transverse Lambert equivalent azimuthal projection (36) can be written in the form (44)x=cos⁡φsin⁡λcos⁡z2,y=sin⁡φcos⁡z2,cos⁡z=cos⁡φcos⁡λ.

The partial derivatives obtained based on expression (44) after rearrangement are: (45)∂x∂λ=cos⁡φcos⁡z2(cos⁡λ+cos⁡φsin2⁡λ4cos2⁡z2) (46)∂x∂φ=−sin⁡φsin⁡λ2cos⁡z2(1+12cos2⁡z2) (47)∂y∂φ=1cos⁡z2(cos⁡φ+sin2⁡φcos⁡λ4cos2⁡z2) (48)∂y∂λ=sin⁡φcos⁡φsin⁡λ4cos3⁡z2.

Following Solov’ev [12]: (49)x=2cos⁡φsin⁡λv,y=2sin⁡φv,v=1+cos⁡φcos⁡λ.

The partial derivatives will be: (50)∂x∂λ=cos⁡φ2v(cos⁡λ+cos⁡φ+cos⁡λv2) (51)∂x∂φ=−sin⁡φsin⁡λ2v(1+1v2) (52)∂y∂φ=12v(cos⁡φ+cos⁡φ+cos⁡λv2) (53)∂y∂λ=sin⁡φcos⁡φsin⁡λ2v3.

Thus, we have three equivalent but different partial derivative notations for the simple Lambert equivalent azimuthal projection. Partial derivatives are needed to determine projection distortions. We will deal with this in the next section.

Since the equations of the Lambert transverse equivalent azimuthal projection can be written in the form (21), which is formally equal to the form (18), then for the Lambert transverse equivalent azimuthal projection we can write the local linear scale factors according to the model of the normal projection. (62)h=cos⁡(π4−φ′2),k=1cos⁡(π4−φ′2)=1hand conclude that (19) will hold for the transverse Lambert equivalent azimuthal projection if (63)cos⁡(π4−φ′2)=1cos⁡(π4−φ′2)holds. From there, φ′=π/2 and then ρ=0 immediately follows. So, in that projection, we have one standard point with the coordinates (x,y)=(0,0). The distribution of linear distortions is identical to the distribution of distortions of the normal simple Lambert equivalent azimuthal projection (Fig. 2).

A third, even simpler method without additional calculations would be to apply the property of map projections according to which all aspects of a projection (normal, oblique and transverse) have the same distribution of distortions.

Transverse and oblique projections are created by reparametrization from normal aspect projections. Since the first differential form is invariant to the change of parameters [13,14], the distortion distribution will not change with the transition from normal to transverse or oblique projection. Since the normal simple Lambert equivalent azimuthal projection (20) has only one standard point located at the origin of the coordinate system, then this is also true for the transverse aspect of this projection.