Creating a triangle - Part 1

[jgbexcel]

I have 2 connected triangles so I will post each separately to avoid confusion (mainly on my part!)
They are part of a design for an articulated arm. Excel will define the dimensions that I need to create my CAD drawing with.
My trig skills are non-existant, but I am fairly well versed in Excel and Lotus 123 before it (and its predecessor on my Radio-Shack TSR-80. Yeah, I'm that old.)

So, 1st triangle.
I have 3 sides known.
3 vertices are F C & B.
F & B are connected elsewhere to known XY co-ordinates.
Side FC and side BC are defined fixed lengths.
Side FB is calculated elsewhere, as point F will move.
I need to calculate XY of C
Included angles are irrelevant.
(Note: C & B are also part of the next triangle problem which I will post tomorrow)

jgb

 

[Perpa]

jgb,
This requires a 'coordinate geometry' solution. Surveyors do this calculation quite often.
I retired from survey work about 16 years ago. The results you are looking for can be had using online
surveyor's calculation sheet typically called 'COGO'. (See Q-Cogo)

Given the coordinates of your 2 known points, F and B, and the radii from each point, there will
be 2 solution points.

I ran a sample to check out how the program works and with very little effort, came up with the
the coordinates of both solutions. You need only enter the coordinates as North for the Y values, and
East for the X values and viola! The coordinates of both solutions are shown after you run an Intersect
of your known Horizontal Distances (HD).
You need not re-invent the wheel with an Excel spread sheet and numerous trigonometric formulae.
Perpa

 

[jgbexcel]

jgb,
This requires a 'coordinate geometry' solution. Surveyors do this calculation quite often.
I retired from survey work about 16 years ago. The results you are looking for can be had using online
surveyor's calculation sheet typically called 'COGO'. (See Q-Cogo)

Given the coordinates of your 2 known points, F and B, and the radii from each point, there will
be 2 solution points.

I ran a sample to check out how the program works and with very little effort, came up with the
the coordinates of both solutions. You need only enter the coordinates as North for the Y values, and
East for the X values and viola! The coordinates of both solutions are shown after you run an Intersect
of your known Horizontal Distances (HD).
You need not re-invent the wheel with an Excel spread sheet and numerous trigonometric formulae.
Perpa
View attachment 91413

Thanks for that. But I am looking for the actual trig formula I can put into a pair of cells to determine the X and Y position of point C.
jgb

 

[Tetra201]

Let's assume your xF, yF, xB, yB, FC, BC are in range A2:A7. Then, xC and yC formulas for solution 1 are:

=A2+(A7^2-A6^2+(A4-A2)^2+(A5-A3)^2)*(A4-A2)/(2*((A4-A2)^2+(A5-A3)^2))-SQRT(A7^2-((A7^2-A6^2+(A4-A2)^2+(A5-A3)^2)/(2*SQRT((A4-A2)^2+(A5-A3)^2)))^2)*(A5-A3)/SQRT((A4-A2)^2+(A5-A3)^2)
=A3+(A7^2-A6^2+(A4-A2)^2+(A5-A3)^2)*(A5-A3)/(2*((A4-A2)^2+(A5-A3)^2))+SQRT(A7^2-((A7^2-A6^2+(A4-A2)^2+(A5-A3)^2)/(2*SQRT((A4-A2)^2+(A5-A3)^2)))^2)*(A4-A2)/SQRT((A4-A2)^2+(A5-A3)^2)

and for solution 2:

=A2+(A7^2-A6^2+(A4-A2)^2+(A5-A3)^2)*(A4-A2)/(2*((A4-A2)^2+(A5-A3)^2))+SQRT(A7^2-((A7^2-A6^2+(A4-A2)^2+(A5-A3)^2)/(2*SQRT((A4-A2)^2+(A5-A3)^2)))^2)*(A5-A3)/SQRT((A4-A2)^2+(A5-A3)^2)
=A3+(A7^2-A6^2+(A4-A2)^2+(A5-A3)^2)*(A5-A3)/(2*((A4-A2)^2+(A5-A3)^2))-SQRT(A7^2-((A7^2-A6^2+(A4-A2)^2+(A5-A3)^2)/(2*SQRT((A4-A2)^2+(A5-A3)^2)))^2)*(A4-A2)/SQRT((A4-A2)^2+(A5-A3)^2)

 

[Perpa]

Trebor201,
Heads up...
I set up a spreadsheet using your formulae and ran a comparison between those results and the results
I obtained from the cogo surveying program i had suggested. There was a difference in the solutions per your formulae.
In looking at the cell data it appears the lengths of BC and FC should be switched, ie. BC should be in A6
and FC should be in A7.

I checked the lengths using the coordinates, and confirmed that the switch gives the same results as the cogo program.
Perpa

 

[Tetra201]

Thank you Perpa for catching this. Yes, the correct order of values in range A2:A7 is xF, yF, xB, yB, BC, FC.

I used different variable names in deriving my formulas (f for the side opposite to F, b for the side opposite to B) and did not transcribe them correctly in the input list.

 

[jgbexcel]

Thanks. I will try this over the weekend. It seems like a "brute force" solution. I was hoping for a "simpler" formula.
jgb

 

[jgbexcel]

OK, here is the description of both problems. 2 connected triangles.

The bottom one (part 1) is F B C. I need to find the XY of C
Side FC is input defined Side BC is input defined. Side FB is defined by calculation elsewhere.
Internal angles are irrelevant.

The top (Part 2) is B C D. I need to find the XY of D
Side BD is input defined Side BC is input defined. Side CD is irrelevant.
The angle BD:BC is input defined in degrees All the other angles are irrelevant.

This is a robot arm for a special need.
The whole pair of triangles pivot on B relative to point F which moves in both X and Y within a set range.
So FB changes length, but that should not matter.
I need to know where point D will be (XY) as point F moves in XY.
I will play in Excel with the lengths of FC, BC and BD, as well as angle DBC till I find a set that fits my constraints.
Those dimensions will be incorporated in my 3D CAD drawing of the mechanism.

 

[Perpa]

jgb,
Here is one solution for Point D coordinates with the assumed known values.
I hope this is useful.
Perpa

 

[jgbexcel]

View attachment 91926View attachment 91927
jgb,
Here is one solution for Point D coordinates with the assumed known values.
I hope this is useful.
Perpa

Thank you. Looks like the best solution I've ever had on this problem. Let me try it out and I'll get back to you.
jgb