Can the Law of Cosines Solve the Distance Difference Problem?

[daviddoria]

See the image in the attached document. I am looking for a function which will make
$$f(d1-p)=f(d2-d1)=f(d3-d2)$$ (see the very last part of the document)

I thought it would be as simple as dividing by the angle between the lines, but that doesn't seem to work. Is it reasonable to do this?

Thanks,

David

 

[daviddoria]

I guess even better would be some transform of each distance, so that:

$f(d1)-f(p)=f(d2)-f(d1)$

of course f() may not be exactly the same function, it may depending on the position (i.e. it could be

$f(d1)-f(p)=g(d2)-g(d1)$

or something like that).

 

[daviddoria]

I guess another way to say it is:

"I need a difference function which will produce 'x' for f(p,d1) and also produce 'x' for f(d2,d1)"

 

[Redbelly98]

Use the Pythagorean theorem:

f(d_i) = sqrt(d_i² - p²) = i·x
​

This gives f(d1) = x, f(d2) = 2x, f(d3) = 3x, etc. I.e., it is the distance from where line p meets the plane to where line d_i meets the plane, which you have set up to be simply i·x in your figure. This satisfies your condition:

f(d1) - f(p) = f(d2) - f(d1) = f(d3) - f(d2) = constant = x​

 

[daviddoria]

Gah, you are right. However, this requires I have p. What if I don't have p?

 

[Redbelly98]

I'm not sure you can do this without knowing p.

What information do you have?
How do you get the values of the d_i's to begin with?
How do you plan to evaluate f(p) without knowing p?

 

[daviddoria]

I think it works if you just use the law of cosines. See attached.