TYPES OF PARALLEL PROJECTION (See Sec 6.1.2.)
Various special cases of parallel projections are given special names because of their frequent occurrence in practice (e.g. in engineering or architectural drawings). The book discusses these, but is somewhat confusing in several respects. For that reason I am expanding on the book with the following discussion. Again, this applies only to parallel projections.
Definitions:
Orthographic Projections:
DOP is perpendicular to the projection plane.
If Projection plane is one of xy, yz or zx these are called front, top or side elevation (not necessarily in that order - it depends on which waythe object is oriented).
Otherwise (i.e. if projection plane is not perpendicular to a coord axis) they are called axonometric orthographic projections.
A special case of axonometric orthographic is isometric projection where projection plane normal (and DOP) makes equal angles with all three axes. Since there are eight octants in 3D, there are 8 possible cases of isometric projections depending on which of the 8 octants the DOP points into.
Oblique Projections:
DOP is not perpendicular to the projection plane.
At page 235 the book says (sentence 1) that oblique projections have projection plane normal to a principal axis. This is not necessary at all and is clearly an error.
Special cases of Oblique Projections are defined as follows: Cavalier Projection: DOP is at 45 degrees to projection plane normal. Cabinet Projection: DOP is at 63.4 degrees to projection plane normal.
The purpose of the special cases is to choose views that show more than just a face, while at the same time preserving some of the length information in a view.
Cavalier:
In the cavalier case, lengths perpendicular to the projection plane are preserved, and as in all parallel projections, lengths parallel to the projection plane are preserved. Thus in this case lengths in all three directions are preserved. For example the projection of a unit cube using a cavalier projection will be an object whose sides are all of length 1, although angles will not be 90 of course.
There are an infinite number of different cavalier projections corresponding to different DOP directions that all make a 45 degree angle with the projection plane. You can think of the DOP as being allowed to rotate around on a cone at a 45 degree vertex angle at the origin.
Note that to achieve equal length between parallel and perpendicular lines, the DOP will need to treat these directions equally, or in other words the angle betwewen DOP and each must be equal. This can occur only if the angles are 45. Hence the above definition for cavalier.
Cabinet:
In the cabinet case, lines perpendicular to the projection plane project at 1/2 their length, while lines parallel to plane project at full length. Thus in this case their is a foreshortening which is more realistic to the eye, while at the same time allowing measurements to be made from a drawing, if one remembers the factor 2 scaling in the one direction. For example the projection of a unit cube using a cabinet projection will be an object 2 of whose sides are of length 1, with one side of length 1/2. Angles will not be 90 between these sides of course.
There are an infinite number of different cabinet projections corresponding to different DOP directions that all make a 63.4 degree angle with the projection plane. You can think of the DOP as being allowed to rotate around on a cone at a 63.4 degree vertex angle at the origin.
Note that to achieve factor 2 shortening in the perpendicular direction the ratio of parallel to perpendicular needs to be 2 - in other words the tangent of the angle between DOP and projection plane should be 2. The angle whose tangent is 2 is 63.4 degrees. Hence the above definition for cabinet.
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I would now like to explain the pictures on pages 235 and 236 (numbers depends on edition!) of cavalier and cabinet projections. It is essential to realize that the angle described as 45 in these pictures has NOTHING to do with the angle 45 between DOP and projection plane in the definition of cavalier.
To find the angle alpha in the pictures, you need to look at figure 6.12 whch is the only picture that shows the direction of the DOP. I will work out all of the data needed to fully understand the pictures. The descriptions in the book are far from complete to say the least.
Figure 6.10 shows two examples of cavalier, labeled (a) and (b).
In (a) the DOP = (1/sqrt(2),1/sqrt(2),-1) and proj plane is z=0, with normal N = (0,0,1). The angle between DOP and N can be found from the fact that
u . v = ||u|| ||v|| cos(a) where a is the angle between two vectors.
In this case we have chosen both DOP and N to be unit vectors. So it follows that the angle a between them is:
cos(a) = DOP . N / (||DOP|| ||N||) = (1/sqrt(2), 1/sqrt(2), -1) . (0,0,1) / ||DOP|| = (-1) / sqrt(2)
since ||N|| = 1 and ||DOP|| = sqrt(2).
Since cos(45) = 1/sqrt(2) it follows that DOP does make an angle of 45 with the plane so this is cavalier.
Next we compute angle alpha as in fig 6.12. The projection will take point P(0,0,1) into P' = P + DOP = (1/sqrt(2), 1/sqrt(2), 0). Thus the line from origin to P makes a 45 angle with x axis - this is the angle alpha in picture - since tan(alpha) = OP'y/OP'x = 1/sqrt(2) / 1/sqrt(2) = 1 and this is the tan of 45 degrees.
In picture (b) the DOP = (sqrt(3)/2,1/2, -1) and so here we get cos(a) = DOP . N / (||DOP|| ||N||) = (sqrt(3)/2, 1/2, -1) . (0,0,1) / ||DOP|| = -1 / sqrt(2)
since ||N|| = 1 and ||DOP|| = sqrt(2). Again it follows that DOP does make an angle 45 with the plane, so this is cavalier.
Next we compute angle alpha as in fig 6.12. The projection will take point P(0,0,1) into P' = P + DOP = (sqrt(3)/2, 1/2, 0). Thus the line from origin to P makes an 30 angle with x axis - this is the angle alpha in picture - since tan(alpha) = OP'y/OP'x = 1/2 / sqrt(3)/2 = 1/sqrt(3) and this is the tan of 30 degrees.
Figure 6.11 shows two examples of cabinet projections, labeled (a) and (b).
In a) we have DOP = (sqrt(2)/4, sqrt(2)/4, -1) so arguing as before we get: cos(a) = -1/sqrt(20/16) = -sqrt(4/5) It follows that sin(a) = sqrt(1 - cos(a)*cos(a)) = sqrt(1/5) and so: tan(a) = sin(a)/cos(a) = sqrt(1/5)/sqrt(4/5) = 1/2. Thus the angle between DOP and normal has tangent 1/2. The complementary angle bewtween DOP and plane is 90-a and therefore has tangent 1/tan(a) or 2. (because tan (90-a) = 1/tan(a)).
Next we compute angle alpha as in fig 6.12. The projection will take point P(0,0,1) into P' = P + DOP = (sqrt(2)/4, sqrt(2)/4, 0). Thus the line from origin to P makes a 45 angle with x axis - this is the angle alpha in picture - since tan(alpha) = OP'y/OP'x = sqrt(2)/4 / sqrt(2)/4 = 1 and this is the tan of 45 degrees.
In picture (b) we have DOP = (sqrt(3)/4, 1/4, -1) so arguing as before we get: cos(a) = -1/sqrt(20/16) = -sqrt(4/5) It follows that again tan(a) = 1/2. Thus the angle between DOP and normal has tangent 1/2. The complementary angle bewtween DOP and plane is 90-a and therefore has tangent 1/tan(a) or 2.
Next we compute angle alpha as in fig 6.12. The projection will take point P(0,0,1) into P' = P + DOP = (sqrt(3)/4, 1/4, 0). Thus the line from origin to P makes a 30 degree angle with x axis - this is the angle alpha in picture - because OP'y/OP'x = tan(alpha) = 1/4 / sqrt(3)/4 = 1/sqrt(3) and this is the tan of 30 degrees.