Triangulation chart


D

Duncs

Hope I can do the explanation justice, and that someone cna help.

I'm looking for a chart that will show three values, as follows:

The chart will take the form of a triangle. In each of the corners
are the values 'A', 'B' & 'C'. Each of the sides represent a value
from 0 to 100. The idea is to take the data below and plot it into
the graph as follows:

A - 70
B - 60
C - 55

A line will go from corner A onto side BC, and stop at the indicator
for 70. A line will go from corner B onto side CA, and stop at the
indicator for 60. Finally, a line will go from corner C onto the side
AB and stop at the indicator for 55. The aim is then to identify
where the three lines merge, if at all.

Is this possible, in Excel 2003, and if so how?

TIA

Duncs
 
Ad

Advertisements

X

Xt

Hope I can do the explanation justice, and that someone cna help.

I'm looking for a chart that will show three values, as follows:

The chart will take the form of a triangle.  In each of the corners
are the values 'A', 'B' & 'C'.  Each of the sides represent a value
from 0 to 100.  The idea is to take the data below and plot it into
the graph as follows:

A - 70
B - 60
C - 55

A line will go from corner A onto side BC, and stop at the indicator
for 70.  A line will go from corner B onto side CA, and stop at the
indicator for 60.  Finally, a line will go from corner C onto the side
AB and stop at the indicator for 55.  The aim is then to identify
where the three lines merge, if at all.

Is this possible, in Excel 2003, and if so how?

TIA

Duncs

Is ABC equilateral?
By "identify" do you mean by eye or by calculation?
Is this a standard chart used in some analysis and usually drawn by
hand?
 
D

Duncs

Is ABC equilateral?
By "identify" do you mean by eye or by calculation?
Is this a standard chart used in some analysis and usually drawn by
hand?- Hide quoted text -

- Show quoted text -

The triangle is equilateral, with all sides showing a uniform range of
0 to 100.

Preferably, I'd like the graph to plot the three lines and see
visually where the lines converge. However, to get a number
representing the intersection point of the three lines would be good.

The graph is currently charted by hand, but there are approximately 12
of them at the moment all representing different areas of a project,
and then copied over multiple projects--with different values.

The ability to generate the graph automaticall from the data would be
a great help.

Can it be done?

Duncs
 
X

Xt

The triangle is equilateral, with all sides showing a uniform range of
0 to 100.

Preferably, I'd like the graph to plot the three lines and see
visually where the lines converge.  However, to get a number
representing the intersection point of the three lines would be good.

The graph is currently charted by hand, but there are approximately 12
of them at the moment all representing different areas of a project,
and then copied over multiple projects--with different values.

The ability to generate the graph automaticall from the data would be
a great help.

Can it be done?

Duncs- Hide quoted text -

- Show quoted text -

I'm sure it can be done. There is no standard chart. You would have
to construct your own or get an expert to do it.

In principle there are no real problems. The vertices are, say, A
(0,0), B (50, 86.6) - [that's 50*SQRT(3)] and C (100,0) so you have
the starting points of your three cutting lines. Opposite each vertex
we have a point a certain percentage along the opposite line. An
example. To find the point X which is 70% of the way along BC
starting from B is to combine 70% of C and 30% of B. Note that the
percentage is applied to the "opposite" end. You are closer to C so
you take more of C.

Think of it as 70%*(100,0)+30%*(50, 86.6) although that doesn't quit
make sense. The point X is (70%*100+30%*50,70%*0+30%*86.6) = (85,
26.0) So now you have a line from A to X. Repeat with the other two
lines. You now have start and end of each of three sides and three
"%" lines. Plot away.

If you need scales, make separate a separate data series for each side
and label the points using the chart labeller addin.

Here is something related which isn't what you want but the idea is
probably worth looking at.
http://www.lboro.ac.uk/research/phys-geog/tri-plot/index.html

A mathematical measure of the closeness of the intersection is
possible too but is more complicated and can wait until you have this
going.

Derek
 
D

Duncs

The triangle is equilateral, with all sides showing a uniform range of
0 to 100.
Preferably, I'd like the graph to plot the three lines and see
visually where the lines converge.  However, to get a number
representing the intersection point of the three lines would be good.
The graph is currently charted by hand, but there are approximately 12
of them at the moment all representing different areas of a project,
and then copied over multiple projects--with different values.
The ability to generate the graph automaticall from the data would be
a great help.
Can it be done?
Duncs- Hide quoted text -
- Show quoted text -

I'm sure it can be done.  There is no standard chart.  You would have
to construct your own or get an expert to do it.

In principle there are no real problems.  The vertices are, say, A
(0,0), B (50, 86.6) - [that's 50*SQRT(3)] and C (100,0) so you have
the starting points of your three cutting lines.  Opposite each vertex
we have a point a certain percentage along the opposite line.  An
example.  To find the point X which is 70% of the way along BC
starting from B is to combine 70% of C and 30% of B. Note that the
percentage is applied to the "opposite" end.  You are closer to C so
you take more of C.

Think of it as 70%*(100,0)+30%*(50, 86.6) although that doesn't quit
make sense. The point X is (70%*100+30%*50,70%*0+30%*86.6) = (85,
26.0)  So now you have a line from A to X.  Repeat with the other two
lines.  You now have start and end of each of three sides and three
"%" lines.  Plot away.

If you need scales, make separate a separate data series for each side
and label the points  using the chart labeller addin.

Here is something related which isn't what you want but the idea is
probably worth looking at.http://www.lboro.ac.uk/research/phys-geog/tri-plot/index.html

A mathematical measure of the closeness of the intersection is
possible too but is more complicated and can wait until you have this
going.

Derek- Hide quoted text -

- Show quoted text -

Derek,

Sorry, but that's not going in and staying in!

If I have the following values:

A = 66
B = 48
C = 75

In a clockwise direction, and 'A' is the bottom left corner, the
corners are A, B & C. Points to plot woould be:

Value A, line BC, (82,31)
X calc = ((64*100)+(36*50))/100
Y calc = ((64*0)+(36*86.6))/100

Value B, line CA, (76,45)
X calc = (((100-C7)*100)+(C7*50))/100
Y calc = ((C7*0)+((100-C7)*86.6))/100

Value C, line AB, (88,22)
X calc = ((C9*100)+((100-C9)*50))/100
Y calc = ((C9*0)+((100-C9)*86.6))/100

This doesn't seem to be plotting correctly. All my plot points are
appearing on the right of the chart, near the line BC. As you can
see, Value C doesn't have a Y value of 0 as I would expect.

Am I totally missing the point and just not understanding your whole
explanation?

Please help!!

TIA

Duncs
 
X

Xt

I'm sure it can be done.  There is no standard chart.  You would have
to construct your own or get an expert to do it.
In principle there are no real problems.  The vertices are, say, A
(0,0), B (50, 86.6) - [that's 50*SQRT(3)] and C (100,0) so you have
the starting points of your three cutting lines.  Opposite each vertex
we have a point a certain percentage along the opposite line.  An
example.  To find the point X which is 70% of the way along BC
starting from B is to combine 70% of C and 30% of B. Note that the
percentage is applied to the "opposite" end.  You are closer to C so
you take more of C.
Think of it as 70%*(100,0)+30%*(50, 86.6) although that doesn't quit
make sense. The point X is (70%*100+30%*50,70%*0+30%*86.6) = (85,
26.0)  So now you have a line from A to X.  Repeat with the other two
lines.  You now have start and end of each of three sides and three
"%" lines.  Plot away.
If you need scales, make separate a separate data series for each side
and label the points  using the chart labeller addin.
Here is something related which isn't what you want but the idea is
probably worth looking at.http://www.lboro.ac.uk/research/phys-geog/tri-plot/index.html
A mathematical measure of the closeness of the intersection is
possible too but is more complicated and can wait until you have this
going.
Derek- Hide quoted text -
- Show quoted text -

Derek,

Sorry, but that's not going in and staying in!

If I have the following values:

A = 66
B = 48
C = 75

In a clockwise direction, and 'A' is the bottom left corner, the
corners are A, B & C.  Points to plot woould be:

Value A, line BC, (82,31)
     X calc = ((64*100)+(36*50))/100
     Y calc = ((64*0)+(36*86.6))/100

Value B, line CA, (76,45)
     X calc = (((100-C7)*100)+(C7*50))/100
     Y calc = ((C7*0)+((100-C7)*86.6))/100

Value C, line AB, (88,22)
     X calc = ((C9*100)+((100-C9)*50))/100
     Y calc = ((C9*0)+((100-C9)*86.6))/100

This doesn't seem to be plotting correctly.  All my plot points are
appearing on the right of the chart, near the line BC.  As you can
see, Value C doesn't have a Y value of 0 as I would expect.

Am I totally missing the point and just not understanding your whole
explanation?

Please help!!

TIA

Duncs- Hide quoted text -

- Show quoted text -

Unfortunately I'm not sure what the C9 etc hold but

I'm assuming that A, B and C go clockwise from A(0,0) and the various
percentages go from 0 to 100 along each line clockwise as well.

The pattern for the first one is correct (except that 66% has changed
to 64% on the way down the page)

For the 48% from B onto CA, C(100,0) and A(0,0)
(X,Y) = 48%*A + 52%*C = ([48*0+52*0]/100,[48*100+52*0]/100)=(52,0)

For the 75% from C onto AB, A(0,0) and B(50,86.6)
(X,Y) = 75%*B + 25%*A = ([75*50+25*0]/100,[75*86.6+525*0]/
100)=(37.5,65)

% x y
A to (B to C) 64% 82 31.2
B to (C to A) 48% 52 0
C to (A to B) 75% 37.5 65.0

I've coded this into Excel and drawn the graph. It all seems to
work. Send me your email if you want a copy.

On the other hand it may be me that has got the wrong end of things.
On my graph if I put 30%, 40% and 78% the lines pretty well cross. Is
that what should be?

Cheers

Derek
 
Ad

Advertisements

D

Duncs

Hope I can do the explanation justice, and that someone cna help.
I'm looking for a chart that will show three values, as follows:
The chart will take the form of a triangle.  In each of the corners
are the values 'A', 'B' & 'C'.  Each of the sides represent avalue
from 0 to 100.  The idea is to take the data below and plot it into
the graph as follows:
A - 70
B - 60
C - 55
A line will go from corner A onto side BC, and stop at the indicator
for 70.  A line will go from corner B onto side CA, and stop at the
indicator for 60.  Finally, a line will go from corner C ontothe side
AB and stop at the indicator for 55.  The aim is then to identify
where the three lines merge, if at all.
Is this possible, in Excel 2003, and if so how?
TIA
Duncs
Is ABC equilateral?
By "identify" do you mean by eye or by calculation?
Is this a standard chart used in some analysis and usually drawn by
hand?- Hide quoted text -
- Show quoted text -
The triangle is equilateral, with all sides showing a uniform rangeof
0 to 100.
Preferably, I'd like the graph to plot the three lines and see
visually where the lines converge.  However, to get a number
representing the intersection point of the three lines would be good.
The graph is currently charted by hand, but there are approximately12
of them at the moment all representing different areas of a project,
and then copied over multiple projects--with different values.
The ability to generate the graph automaticall from the data would be
a great help.
Can it be done?
Duncs- Hide quoted text -
- Show quoted text -
I'm sure it can be done.  There is no standard chart.  You would have
to construct your own or get an expert to do it.
In principle there are no real problems.  The vertices are, say, A
(0,0), B (50, 86.6) - [that's 50*SQRT(3)] and C (100,0) so you have
the starting points of your three cutting lines.  Opposite each vertex
we have a point a certain percentage along the opposite line.  An
example.  To find the point X which is 70% of the way along BC
starting from B is to combine 70% of C and 30% of B. Note that the
percentage is applied to the "opposite" end.  You are closer to C so
you take more of C.
Think of it as 70%*(100,0)+30%*(50, 86.6) although that doesn't quit
make sense. The point X is (70%*100+30%*50,70%*0+30%*86.6) = (85,
26.0)  So now you have a line from A to X.  Repeat with the othertwo
lines.  You now have start and end of each of three sides and three
"%" lines.  Plot away.
If you need scales, make separate a separate data series for each side
and label the points  using the chart labeller addin.
Here is something related which isn't what you want but the idea is
probably worth looking at.http://www.lboro.ac.uk/research/phys-geog/tri-plot/index.html
A mathematical measure of the closeness of the intersection is
possible too but is more complicated and can wait until you have this
going.
Derek- Hide quoted text -
- Show quoted text -

Sorry, but that's not going in and staying in!
If I have the following values:
A = 66
B = 48
C = 75
In a clockwise direction, and 'A' is the bottom left corner, the
corners are A, B & C.  Points to plot woould be:
Value A, line BC, (82,31)
     X calc = ((64*100)+(36*50))/100
     Y calc = ((64*0)+(36*86.6))/100
Value B, line CA, (76,45)
     X calc = (((100-C7)*100)+(C7*50))/100
     Y calc = ((C7*0)+((100-C7)*86.6))/100
Value C, line AB, (88,22)
     X calc = ((C9*100)+((100-C9)*50))/100
     Y calc = ((C9*0)+((100-C9)*86.6))/100
This doesn't seem to be plotting correctly.  All my plot points are
appearing on the right of the chart, near the line BC.  As you can
see, Value C doesn't have a Y value of 0 as I would expect.
Am I totally missing the point and just not understanding your whole
explanation?
Please help!!

Duncs- Hide quoted text -
- Show quoted text -

Unfortunately I'm not sure what the C9 etc hold but

I'm assuming that A, B and C go clockwise from A(0,0) and the various
percentages go from 0 to 100 along each line clockwise as well.

The pattern for the first one is correct (except that 66% has changed
to 64% on the way down the page)

For the 48% from B onto CA, C(100,0) and A(0,0)
(X,Y) = 48%*A + 52%*C = ([48*0+52*0]/100,[48*100+52*0]/100)=(52,0)

For the 75% from C onto AB, A(0,0) and B(50,86.6)
(X,Y) = 75%*B + 25%*A = ([75*50+25*0]/100,[75*86.6+525*0]/
100)=(37.5,65)

                        %       x       y
A to (B to C)   64%     82      31.2
B to (C to A)   48%     52      0
C to (A to B)   75%     37.5    65.0

I've coded this into Excel and drawn the graph.  It all seems to
work.  Send me your email if you want a copy.

On the other hand it may be me that has got the wrong end of things.
On my graph if I put 30%, 40% and 78% the lines pretty well cross.  Is
that what should be?

Cheers

Derek- Hide quoted text -

- Show quoted text -

Derek,

You've got my possibly rather poor explanation of it perfectly.
However, I'd appreciate it if you could send me a copy of the file.
My e-mail address is kilted_scot at hotmail dot com.

Many, many thanks for your help.

Duncs
 
Ad

Advertisements

D

Duncs

Hope I can do the explanation justice, and that someone cna help.
I'm looking for a chart that will show three values, as follows:
The chart will take the form of a triangle.  In each of the corners
are the values 'A', 'B' & 'C'.  Each of the sides represent avalue
from 0 to 100.  The idea is to take the data below and plot it into
the graph as follows:
A - 70
B - 60
C - 55
A line will go from corner A onto side BC, and stop at the indicator
for 70.  A line will go from corner B onto side CA, and stop at the
indicator for 60.  Finally, a line will go from corner C ontothe side
AB and stop at the indicator for 55.  The aim is then to identify
where the three lines merge, if at all.
Is this possible, in Excel 2003, and if so how?
TIA
Duncs
Is ABC equilateral?
By "identify" do you mean by eye or by calculation?
Is this a standard chart used in some analysis and usually drawn by
hand?- Hide quoted text -
- Show quoted text -
The triangle is equilateral, with all sides showing a uniform rangeof
0 to 100.
Preferably, I'd like the graph to plot the three lines and see
visually where the lines converge.  However, to get a number
representing the intersection point of the three lines would be good.
The graph is currently charted by hand, but there are approximately12
of them at the moment all representing different areas of a project,
and then copied over multiple projects--with different values.
The ability to generate the graph automaticall from the data would be
a great help.
Can it be done?
Duncs- Hide quoted text -
- Show quoted text -
I'm sure it can be done.  There is no standard chart.  You would have
to construct your own or get an expert to do it.
In principle there are no real problems.  The vertices are, say, A
(0,0), B (50, 86.6) - [that's 50*SQRT(3)] and C (100,0) so you have
the starting points of your three cutting lines.  Opposite each vertex
we have a point a certain percentage along the opposite line.  An
example.  To find the point X which is 70% of the way along BC
starting from B is to combine 70% of C and 30% of B. Note that the
percentage is applied to the "opposite" end.  You are closer to C so
you take more of C.
Think of it as 70%*(100,0)+30%*(50, 86.6) although that doesn't quit
make sense. The point X is (70%*100+30%*50,70%*0+30%*86.6) = (85,
26.0)  So now you have a line from A to X.  Repeat with the othertwo
lines.  You now have start and end of each of three sides and three
"%" lines.  Plot away.
If you need scales, make separate a separate data series for each side
and label the points  using the chart labeller addin.
Here is something related which isn't what you want but the idea is
probably worth looking at.http://www.lboro.ac.uk/research/phys-geog/tri-plot/index.html
A mathematical measure of the closeness of the intersection is
possible too but is more complicated and can wait until you have this
going.
Derek- Hide quoted text -
- Show quoted text -

Sorry, but that's not going in and staying in!
If I have the following values:
A = 66
B = 48
C = 75
In a clockwise direction, and 'A' is the bottom left corner, the
corners are A, B & C.  Points to plot woould be:
Value A, line BC, (82,31)
     X calc = ((64*100)+(36*50))/100
     Y calc = ((64*0)+(36*86.6))/100
Value B, line CA, (76,45)
     X calc = (((100-C7)*100)+(C7*50))/100
     Y calc = ((C7*0)+((100-C7)*86.6))/100
Value C, line AB, (88,22)
     X calc = ((C9*100)+((100-C9)*50))/100
     Y calc = ((C9*0)+((100-C9)*86.6))/100
This doesn't seem to be plotting correctly.  All my plot points are
appearing on the right of the chart, near the line BC.  As you can
see, Value C doesn't have a Y value of 0 as I would expect.
Am I totally missing the point and just not understanding your whole
explanation?
Please help!!

Duncs- Hide quoted text -
- Show quoted text -

Unfortunately I'm not sure what the C9 etc hold but

I'm assuming that A, B and C go clockwise from A(0,0) and the various
percentages go from 0 to 100 along each line clockwise as well.

The pattern for the first one is correct (except that 66% has changed
to 64% on the way down the page)

For the 48% from B onto CA, C(100,0) and A(0,0)
(X,Y) = 48%*A + 52%*C = ([48*0+52*0]/100,[48*100+52*0]/100)=(52,0)

For the 75% from C onto AB, A(0,0) and B(50,86.6)
(X,Y) = 75%*B + 25%*A = ([75*50+25*0]/100,[75*86.6+525*0]/
100)=(37.5,65)

                        %       x       y
A to (B to C)   64%     82      31.2
B to (C to A)   48%     52      0
C to (A to B)   75%     37.5    65.0

I've coded this into Excel and drawn the graph.  It all seems to
work.  Send me your email if you want a copy.

On the other hand it may be me that has got the wrong end of things.
On my graph if I put 30%, 40% and 78% the lines pretty well cross.  Is
that what should be?

Cheers

Derek- Hide quoted text -

- Show quoted text -

Derek,

I've replied to your e-mail.

Many thanks for the spreadsheet. As I mentioned, it works a treat.

Many, many, many thanks for your help with this. It has been much
appreciated.

Duncs
 

Ask a Question

Want to reply to this thread or ask your own question?

You'll need to choose a username for the site, which only take a couple of moments. After that, you can post your question and our members will help you out.

Ask a Question

Top