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From: Beho nashed on 19 Jun 2010 12:05
"Roger Stafford" <ellieandrogerxyzzy(a)mindspring.com.invalid> wrote in message <hvgt8p$ss7$1(a)fred.mathworks.com>...
> Beho <newparadize2002(a)gmail.com> wrote in message <e615968b-18d6-489f-bbfc-343a546f8869(a)g19g2000yqc.googlegroups.com>...
> > can I just use this (2 points)
> >

Hello Roger,
so how to change the code just to measure the angle, between the two lines.
should I use two points.

I used subspace, I selected two points (giving an approximate 45 deg), but it gave me 4 degree and sometimes 14 deg, I am not gonna use this one.

I am not sure Roger how to calculate the angle? is the y-axis messing up the angle?
Thanks
From: Roger Stafford on 19 Jun 2010 13:40
"Beho nashed" <beho86(a)yahoo.com> wrote in message <hvips2$nuh$1(a)fred.mathworks.com>...
> Hello Roger,
> so how to change the code just to measure the angle, between the two lines.
> should I use two points.
> I used subspace, I selected two points (giving an approximate 45 deg), but it gave me 4 degree and sometimes 14 deg, I am not gonna use this one.
> I am not sure Roger how to calculate the angle? is the y-axis messing up the angle?
> Thanks
- - - - - - - - - -
When speaking of the angle between two lines it is necessary to distinguish between two possible meanings. If the lines are "undirected", meaning that there is no preferred direction along either line, then there are two angles between the lines to be considered, one being the supplement of the other, and their sum is 180 degrees (or pi radians.) Usually the smaller of the two is chosen as being the defined angle between the lines, and that is the one given by matlab's 'subspace' function for two lines through the origin. Such an angle cannot exceed 90 degrees by definition.

On the other hand there is the concept of the angle between two directed lines, meaning that there is a preferred "positive" direction along each line. This is equivalent to the notion of the angle between two vectors which have a common base point. Such an angle can range from 0 to 180 degrees. It is this latter kind of angle that I have been giving the formulas for. Matlab's 'atan2' function is actually capable of producing angles anywhere between minus pi and plus pi radians, but in all these formulas, the first argument was always non-negative, so that restricts the angle to the first two quadrants, and 'atan2' can then only produce results for these that range from zero to pi radians (0 to 180 in degrees.)

In neither kind of situation can one find the angle with only two points! You will always need either two points on each line to establish a direction for each one, or else the point of intersection of the two lines and one other point on each line. If both the two lines pass through the coordinate origin, then that point of intersection is the origin and only two other points on the lines are required.

If the lines are considered as directed, then on each line the coordinates of the point less far out in the positive direction are subtracted from those of the other point farther out, and this establishes the components of the two vectors needed to determine the angle between the directed lines.

I hope this finally winds up this thread about finding the "angle between two vectors" which has now gone on for forty-eight articles and stretched over a span of almost three years. That is quite enough in my opinion for such an elementary concept.

Roger Stafford
From: Bard Romstad on 17 Jul 2010 14:50
"Roger Stafford" <ellieandrogerxyzzy(a)mindspring.com.invalid> wrote in message <fl3moi$pvc$1(a)fred.mathworks.com>...
> "baris kazar" <mbkazar.nospam(a)gmail.com> wrote in message <fl377q$4ip
> $1(a)fred.mathworks.com>...
> > Hi,-
> > how can we generalize this to 3-D vectors? Think of a
> > plane on 3-D space and you have vectors on this plane. I
> > wanna know the angle between 2 vectors in the range 0-2pi.
> > or at least -pi to pi. I have 3 vectors. One (First)
> > vector is the same all the time. I wanna know the relative
> > positions of the other two vector wrt the firsy one. Thus,
> > i need angles in the range 0 to 2pi or -pi to pi.
> > Thanks in advance
> ---------
> As Bruno has pointed out, the angle between two three-dimensional vectors
> depends on which sense one gives to a vector orthogonal to their plane. It
> isn't clear what you meant by, "the relative positions of the other two vector
> wrt the firsy one." The vector cross product of the second two vectors will be
> a vector orthogonal to their plane. Perhaps you mean that the angle between
> them is to be considered positive if this cross product lies on the same side
> of the plane as this first vector, and negative otherwise. If that is the case,
> then let your first, second, and third vectors be designated as x, y, and z,
> respectively. A matlab formula for calculating the angle between y and z will
> then be:
>
> c = cross(y,z);
> angleyz = sign(dot(x,c))*atan2(norm(c),dot(y,z));
>
> The value of 'angleyz' will range from -pi to +pi. If you want it to range from
> 0 to 2*pi, then apply the 'mod' function as I did on Dec. 11 in this thread.
>
> Roger Stafford
>

This will fail if all three vectors x, y and z lie on the same plane because then dot(x,c)=0 (and sign(0)=0 -> angleyz = 0).
Example:
x = [1 0 1];
y = [0 1 1];
z = [0.5 0.5 1];
x=x/norm(x);y=y/norm(y);z=z/norm(z);
c = cross(y,z);
angleyz = sign(dot(x,c))*atan2(norm(c),dot(y,z))

Any suggestions?

Bård Romstad
From: Roger Stafford on 17 Jul 2010 17:31
"Bard Romstad" <this.is.not.a.real.address(a)gmail.com> wrote in message <i1su0s$deu$1(a)fred.mathworks.com>...
> "Roger Stafford" <ellieandrogerxyzzy(a)mindspring.com.invalid> wrote in message <fl3moi$pvc$1(a)fred.mathworks.com>...
> > "baris kazar" <mbkazar.nospam(a)gmail.com> wrote in message <fl377q$4ip
> > $1(a)fred.mathworks.com>...
> > > Hi,-
> > > how can we generalize this to 3-D vectors? Think of a
> > > plane on 3-D space and you have vectors on this plane. I
> > > wanna know the angle between 2 vectors in the range 0-2pi.
> > > or at least -pi to pi. I have 3 vectors. One (First)
> > > vector is the same all the time. I wanna know the relative
> > > positions of the other two vector wrt the firsy one. Thus,
> > > i need angles in the range 0 to 2pi or -pi to pi.
> > > Thanks in advance
> > ---------
> > As Bruno has pointed out, the angle between two three-dimensional vectors
> > depends on which sense one gives to a vector orthogonal to their plane. It
> > isn't clear what you meant by, "the relative positions of the other two vector
> > wrt the firsy one." The vector cross product of the second two vectors will be
> > a vector orthogonal to their plane. Perhaps you mean that the angle between
> > them is to be considered positive if this cross product lies on the same side
> > of the plane as this first vector, and negative otherwise. If that is the case,
> > then let your first, second, and third vectors be designated as x, y, and z,
> > respectively. A matlab formula for calculating the angle between y and z will
> > then be:
> >
> > c = cross(y,z);
> > angleyz = sign(dot(x,c))*atan2(norm(c),dot(y,z));
> >
> > The value of 'angleyz' will range from -pi to +pi. If you want it to range from
> > 0 to 2*pi, then apply the 'mod' function as I did on Dec. 11 in this thread.
> >
> > Roger Stafford
> >
>
> This will fail if all three vectors x, y and z lie on the same plane because then dot(x,c)=0 (and sign(0)=0 -> angleyz = 0).
> Example:
> x = [1 0 1];
> y = [0 1 1];
> z = [0.5 0.5 1];
> x=x/norm(x);y=y/norm(y);z=z/norm(z);
> c = cross(y,z);
> angleyz = sign(dot(x,c))*atan2(norm(c),dot(y,z))
>
> Any suggestions?
>
> Bård Romstad
- - - - - - - - - -
Well of course it fails, based as it is on a desperate guess as to what Baris Kazar meant when he said (back on December 28, 2007) "the relative positions of the other two vector wrt the firsy one"! This was an attempt on my part to interpret his statement as indicating that the first vector establishes which side of the plane containing the second two vectors is to be considered as its "positive" side. If that first vector lies in this plane, then of course it fails to do so.

With three 3D vectors all lying in a plane through their origin, even if one of them is regarded as some sort of reference, there is no inherent "clockwise" or "counterclockwise" sense to angles in that plane. Only some reference direction outside the plane or else a pair of "standard" vectors within the plane can do that. When we say "clockwise" it is said from the point of view of a person seeing a clock from outside the plane of the clock hands from the usual side of the clock. If a transparent clock is viewed from its rear, what was the clockwise direction now becomes the counterclockwise direction (if the backward numerals are ignored.) The famous "right-hand rule" depends on one's right-hand thumb lying outside the plane of one's curled finger directions.

Bard, if you want a formula for the angle between Baris Kazar's second two vectors ranging from 0 to 2*pi, please explain to me your idea of what you think he meant by the above quotation given that the first vector lies in their plane.

Roger Stafford
From: Bard Romstad on 18 Jul 2010 06:55
"Roger Stafford" <ellieandrogerxyzzy(a)mindspring.com.invalid> wrote in message <i1t7en$gvq$1(a)fred.mathworks.com>...
> "Bard Romstad" <this.is.not.a.real.address(a)gmail.com> wrote in message <i1su0s$deu$1(a)fred.mathworks.com>...
> > "Roger Stafford" <ellieandrogerxyzzy(a)mindspring.com.invalid> wrote in message <fl3moi$pvc$1(a)fred.mathworks.com>...
> > > "baris kazar" <mbkazar.nospam(a)gmail.com> wrote in message <fl377q$4ip
> > > $1(a)fred.mathworks.com>...
> > > > Hi,-
> > > > how can we generalize this to 3-D vectors? Think of a
> > > > plane on 3-D space and you have vectors on this plane. I
> > > > wanna know the angle between 2 vectors in the range 0-2pi.
> > > > or at least -pi to pi. I have 3 vectors. One (First)
> > > > vector is the same all the time. I wanna know the relative
> > > > positions of the other two vector wrt the firsy one. Thus,
> > > > i need angles in the range 0 to 2pi or -pi to pi.
> > > > Thanks in advance
> > > ---------
> > > As Bruno has pointed out, the angle between two three-dimensional vectors
> > > depends on which sense one gives to a vector orthogonal to their plane. It
> > > isn't clear what you meant by, "the relative positions of the other two vector
> > > wrt the firsy one." The vector cross product of the second two vectors will be
> > > a vector orthogonal to their plane. Perhaps you mean that the angle between
> > > them is to be considered positive if this cross product lies on the same side
> > > of the plane as this first vector, and negative otherwise. If that is the case,
> > > then let your first, second, and third vectors be designated as x, y, and z,
> > > respectively. A matlab formula for calculating the angle between y and z will
> > > then be:
> > >
> > > c = cross(y,z);
> > > angleyz = sign(dot(x,c))*atan2(norm(c),dot(y,z));
> > >
> > > The value of 'angleyz' will range from -pi to +pi. If you want it to range from
> > > 0 to 2*pi, then apply the 'mod' function as I did on Dec. 11 in this thread.
> > >
> > > Roger Stafford
> > >
> >
> > This will fail if all three vectors x, y and z lie on the same plane because then dot(x,c)=0 (and sign(0)=0 -> angleyz = 0).
> > Example:
> > x = [1 0 1];
> > y = [0 1 1];
> > z = [0.5 0.5 1];
> > x=x/norm(x);y=y/norm(y);z=z/norm(z);
> > c = cross(y,z);
> > angleyz = sign(dot(x,c))*atan2(norm(c),dot(y,z))
> >
> > Any suggestions?
> >
> > Bård Romstad
> - - - - - - - - - -
> Well of course it fails, based as it is on a desperate guess as to what Baris Kazar meant when he said (back on December 28, 2007) "the relative positions of the other two vector wrt the firsy one"! This was an attempt on my part to interpret his statement as indicating that the first vector establishes which side of the plane containing the second two vectors is to be considered as its "positive" side. If that first vector lies in this plane, then of course it fails to do so.
>
> With three 3D vectors all lying in a plane through their origin, even if one of them is regarded as some sort of reference, there is no inherent "clockwise" or "counterclockwise" sense to angles in that plane. Only some reference direction outside the plane or else a pair of "standard" vectors within the plane can do that. When we say "clockwise" it is said from the point of view of a person seeing a clock from outside the plane of the clock hands from the usual side of the clock. If a transparent clock is viewed from its rear, what was the clockwise direction now becomes the counterclockwise direction (if the backward numerals are ignored.) The famous "right-hand rule" depends on one's right-hand thumb lying outside the plane of one's curled finger directions.
>
> Bard, if you want a formula for the angle between Baris Kazar's second two vectors ranging from 0 to 2*pi, please explain to me your idea of what you think he meant by the above quotation given that the first vector lies in their plane.
>
> Roger Stafford

Hi Roger,
Giving it a second look I'm not quite certain what Baris ment after all, but what I am trying to find is how to calculate the signed angle (-pi to pi) between two vectors, A and B, projected onto the vertical plane passing through vector C, viewed from the direction defined by cross(C,[0 0 1]).
To clarify (or confuse):
In my specific problem A and B are surface normals of two neighbouring patches on a 2.5D topographic surface (no caves/overhangs -> normals are always pointing upwards). I want to investigate the convexity of the topographic surface along a given direction (defined by the horizontal component of C). My idea was to do this by finding the angle between A and B (relative to C).
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