Can radius be found?

[tmoria]

In the diagrams below there are 3 circles vertically seperated by distance 'h'. A straight line passes through the centre of each chord. ABC and DEF are straight lines, OP passes through the centre of each circle.

a, b, c and h are known.

The second diagram is to show that the circles can be moved in their respective planes whilst the knowns and unknowns remain the same.
I do know that if 'd' is increased and OP still passes through the centre of each circle them ABC and DEF are no longer straight lines.

Is it possible to obtain equations for R and d in terms of these known quantities?

Many thanks for any help.

 

[DrPhil]

In the diagrams below there are 3 circles vertically seperated by distance 'h'. A straight line passes through the centre of each chord. ABC and DEF are straight lines, OP passes through the centre of each circle.

a, b, c and h are known.

The second diagram is to show that the circles can be moved in their respective planes whilst the knowns and unknowns remain the same.
I do know that if 'd' is increased and OP still passes through the centre of each circle them ABC and DEF are no longer straight lines.

Is it possible to obtain equations for R and d in terms of these known quantities?

Many thanks for any help.

I am sure the sliding is proportional to the height of the circle: if the middle one shifts right by x, then the top one will shift right by 2x. In that case that ABC, OP, and DEF remain straight lines, as does the line through the centers of the chords. d does NOT change, and the three radii remain R, (R+d), and (R+2d). What we do NOT know after the shift is the angle of line OP - and the slope of the line through the chords was not known either before or after the shift.

Since there are two unknowns, you will need two independent relationships. I suggest looking at the distances of the three chords from the respective circle centers:
(y_1)^2 = R^2 - a^2
(y_2)^ = (R+d)^2 - b^2
(y_3)^2 = (R+2d)^2 - c^2
Those three have to fall on a straight line.
Those have to fall on a straight line.
Also has to be true after shifting.

See how far that gets you!

 

[tmoria]

Thanks for that DrPhil.

Working on what you suggest I think I require a fourth circle at distance \(\displaystyle h\) below the current bottom circle because I can't create two independant equations.
i.e.

The slope of the straight line you are suggesting would be

\(\displaystyle m=\dfrac{\sqrt{(R+d)^2-b^2}-\sqrt{R^2-a^2}}{h}=\dfrac{\sqrt{(R+2d)^2-c^2}-\sqrt{(R+d)^2-b^2}}{h}=\dfrac{\sqrt{(R+2d)^2-c^2}-\sqrt{R^2-a^2}}{2h}\)

Although it is no problem to get the fourth circle, for which the relevant line-length would be\(\displaystyle \ \ \ \sqrt{(R+3d)^2-e^2}\) (where \(\displaystyle e\) is known)

it looks as though it would be very difficult (if not impossible) to isolate \(\displaystyle R\) and \(\displaystyle d\).

 

[DrPhil]

Thanks for that DrPhil.

Working on what you suggest I think I require a fourth circle at distance \(\displaystyle h\) below the current bottom circle because I can't create two independant equations.
i.e.

The slope of the straight line you are suggesting would be

\(\displaystyle m=\dfrac{\sqrt{(R+d)^2-b^2}-\sqrt{R^2-a^2}}{h}=\dfrac{\sqrt{(R+2d)^2-c^2}-\sqrt{(R+d)^2-b^2}}{h}=\dfrac{\sqrt{(R+2d)^2-c^2}-\sqrt{R^2-a^2}}{2h}\)

Although it is no problem to get the fourth circle, for which the relevant line-length would be\(\displaystyle \ \ \ \sqrt{(R+3d)^2-e^2}\) (where \(\displaystyle e\) is known)

it looks as though it would be very difficult (if not impossible) to isolate \(\displaystyle R\) and \(\displaystyle d\).

But the original question was,
"Is it possible to obtain equations for R and d in terms of these known quantities?"
They didn't ask you actually to DO it - just determine is a solution exists.

After I posted before, I realized that there was only ONE relationship with R and d, for the same reason you found. In my notation that relation would be found by setting (y_2 - y_1) = (y_3 - y_2).

I wouldn't expect it to "fair" to draw another circle. Instead, I wonder if you can make something out of the fact that line DEF remains a straight line after the shift. Why did they emphasize those vertical green triangles on the right?

 

[tmoria]

But the original question was,
"Is it possible to obtain equations for R and d in terms of these known quantities?"

Hi again.

This isn't a homework question, it is something I am working on myself.
Adding another circle (or more if necessary) isn't a problem as I would know the line-lengths involved.
From the further work I've done since my previous post it appears I'll have two (using four circles) quartic equations in \(\displaystyle R\) so it might not be insurmountable.

I created the drawings myself. I emphasised the green triangles to show that all three remain identical whether the circles are shifted horizontally in their relative planes.

Also, the reason I stated that if \(\displaystyle d\) is changed then, if OP still runs through the centre of each circle, ABC and DEF are no longer straight lines, is that I believe this is the reason that \(\displaystyle R\) and \(\displaystyle d\) will be unique for a given \(\displaystyle a, b, c\) and \(\displaystyle h\).

btw, I have actually tried changing \(\displaystyle d\) so I know the last statement re: ABC and DEF to be true.

 

[tmoria]

OK, here's the current state of play...

I can get R in quadratic form...

I have that \(\displaystyle 2\sqrt{(R+d)^2-b^2}=\sqrt{R^2-a^2}+\sqrt{(R+2d)^2-c^2}\)

which can be squared and rearranged twice to give

\(\displaystyle 8(a^2-2b^2+c^2)R^2+8d(3a^2-4b^2+c^2)R+\left((a^2-4b^2)-2(a^2+4b^2)c^2+16a^2d^2+c^4\right)=0\)

to give

\(\displaystyle R=\dfrac{-2d(3a^2-4b^2+c^2)\pm\sqrt{2(a+2b+c)(a+2b-c)(a-2b+c)(a-2b-c)(2d^2-c^2+2b^2-a^2)}}{4(a^2-2b^2+c^2)}\)

I've realised that my earlier idea of adding extra circles (below or between) is no advantage as they will have exactly the same relationship as above.

\(\displaystyle h\) is known so I think I need to get this included to find another equation for \(\displaystyle R\) but have hit a wall again.

DrPhil's input was immensely helpful in pushing me in the right direction. Anyone any other ideas how I can get \(\displaystyle h\) included in the mix?

Many thanks again

 

[DrPhil]

I can get R in quadratic form...

I have that \(\displaystyle 2\sqrt{(R+d)^2-b^2}=\sqrt{R^2-a^2}+\sqrt{(R+2d)^2-c^2}\)

which can be squared and rearranged twice to give

\(\displaystyle 8(a^2-2b^2+c^2)R^2+8d(3a^2-4b^2+c^2)R+\left((a^2-4b^2)-2(a^2+4b^2)c^2+16a^2d^2+c^4\right)=0\)

to give

\(\displaystyle R=\dfrac{-2d(3a^2-4b^2+c^2)\pm\sqrt{2(a+2b+c)(a+2b-c)(a-2b+c)(a-2b-c)(2d^2-c^2+2b^2-a^2)}}{4(a^2-2b^2+c^2)}\)

\(\displaystyle h\) is known so I think I need to get this included to find another equation for \(\displaystyle R\) but have hit a wall again.

You have done a LOT of work to get this explicit form for \(\displaystyle R\) as a function of \(\displaystyle d\). Unfortunately I don't think \(\displaystyle h\) will help find another relationship, since it doesn't do anything except allow ratios to be used. You may have gone about as fer as you kin go.

Thus you have a continuum of solutions in \(\displaystyle R\! -d\) space. If you have some other criterion - a symmetry you would like, or some subjective factor, you could search for satisfactory solutions. One choice would be \(\displaystyle R = d\).

I don't know if the other tutors have been following this, but now is the time to chime in!

 

[tmoria]

Unfortunately I don't think \(\displaystyle h\) will help find another relationship, since it doesn't do anything except allow ratios to be used.

My reasoning for thinking \(\displaystyle h\) is an important known value is...
in the image below, the 6 red circles highlight 6 points on a hyperbola with a known equation. If \(\displaystyle h\) is changed then we have a different hyperbola with a different equation. It may turn out that you're right but I think I'll keep banging at it for a while

 

[tmoria]

In case it is of any use to anyone searching the internet here's the result I have obtained.

My original question was can the radius (R) and difference in radii (d) be found. I wanted to know this because I wanted to know if a given hyperbola is from a unique cone.
The answer to this is no.

From the previous work I have \(\displaystyle \sqrt{R^2-a^2}+\sqrt{(R+2d)^2-c^2}=\sqrt{(R+d)^2-b^2}\)

Squaring and rearranging a couple of times gives...

\(\displaystyle 8(a^2-2b^2+c^2)R^2+8d(3a^2-4b^2+c^2)R+(a^4+16b^4+c^4-8a^2 b^2-2a^2 c^2-8b^2 c^2+16a^2 d^2)=0\)

Taking the hyperbola \(\displaystyle x^2-y^2=1\) for positive x and y values we can take the following three points, \(\displaystyle (2,\ \sqrt{3}),\ (3,\ \sqrt{8}),\ (4,\ \sqrt{15})\)

giving values in the original post as \(\displaystyle a=\sqrt{3}\ \ \ \ b=\sqrt{8}\ \ \ \ c=\sqrt{15}\ \ \ \ h=1\)

Now, let \(\displaystyle R=4\) and substitute these values in the above equation and we get

\(\displaystyle d=\dfrac{8\pm\sqrt{13}}{3}\ \ \ \ \ \ \ i.e.\ \ \ d\approx 1.465\ \ \ \ or\ \ \ \ \ d\approx 3.869\)

These values of \(\displaystyle d\) will produce cones with different apex angles, therefore different cones. Different values can also be chosen for \(\displaystyle R\) with similar results.

\(\displaystyle \therefore\ \ \ \ \ \ \ \)a given hyperbola is not part of a unique cone.