Tag Archives: Combined Angle

Derivation of combined angle- Version 2

The simple analysis of combined angle (or compound angle) was described in this article.There the derivation was based on two deflection angles only. And hence the solution was approximate. For more accurate analysis, same approach can be used. The additional parameter included is the slope of incoming line.

However, a different approach which uses only coordinates can be used to calculate the actual deflection angle of two lines in 3D space. This approach will not only give the deflection angle but also provide accurate length of those lines.


Let us suppose three coordinates in space and let X be the angle between these two lines

The vector AB is given by


Similarly BC is given by



The dot product of vector is given by AB.BC =AB*BC*CosX

Here AB and BC=length of segment AB and BC respectively and is given by


Thus CosX=AB.BC/(L1*L2)

Total Lent of segment L=L1+L2

A worksheet using this approach can be downloaded here . The sheet also contains VBA codes to visualize the alignment.


Derivation of Combined Angle

Abstract: Derivation of combined angle may become useful when you want to know the magnitude and direction of force due to bend in both XY and XZ plane. For eg in design of bend in pipeline.

Let us consider a straight line which deflects at O along OA as shown in the figure. The deflection angles are also shown in the figure.

Let R be the position vector of A. OB is projection of OA on XY plane and AB is projection of OA on a plane parallel to YZ plane. Similarly, OC is projection of line OB in X axis and BC is projection of OB on axis parallel to Y axis.

From figure it can inferred that

OB=R cosα

AB=R sinα

OC=OB cosβ = R cosα cosβ

BC= OB sinβ =R cosα sinβ

Therefore, the position vector of A is OA which is given by

R=|R| (cosα cosβ i + cosα sinβ j + sinα k )

And the unit vector along R is given by

r= (cosα cosβ i + cosα sinβ j + sinα k )

Now the combined deflection angle is given by

r . i =|r| |i| cosδ

Or, r . i =1*1 cosδ

Or, cosδ = cosα cosβ+0+0

Or, δ =cos-1(cosα cosβ)

The angle δ lies in the plane OAC. This plane is shown in the figure in hatch line.


If F be the magnitude of force due to the deflection δ (e.g. deflection in pipeline) then the vertical and horizontal component of this force is given by,

Fx1 =F r . i =F cosα cosβ

Fx2 =F r . j =F cosα sinβ

Fx =F

=F cosα

Fy =F r . j

=F sinα