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schematic view of the problem

The more I try to figure this one out, the more I realize I don't recall everything I need to attack this problem, despite doing well in geometry classes (about 15 years ago). It appears to be a problem in two parts.

A line, $\overline{bf}$, is tangent to arc segment $\overline{ab}$ with center $c$, and fixed at point $a$. Line $\overline{bf}$ is rotated by $6$ degrees to become line $\overline{df}$, causing the arc radius to decrease to $R$, but the arc remains fixed at point $a$ and tangent to the original arc. The center $c$ moves to point $e$ due to the decrease in radius, but remains coincident with line $\overline{ag}$. Line $\overline{fg}$ has fixed length $W$ and is equal to the starting radius of arc segment $\overline{ab}$, but not equal to the finishing radius of arc segment $\overline{ad}$.

I need to find (a) the new radius $R$ and (b) the values of $dx$ and $dy$ for the tangent point ($x$ and $y$ distances to the new point of tangency). An algebraic form solution would be appreciated. Thanks very much in advance.

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  • $\begingroup$ Just to clarify, $H'$ is meant to represent the length of the line segment closest to the "$6^{\circ}$," not the length of $\overline{bf}$, right? In other words, is it true that $H'\ne H$? $\endgroup$ – Robert Howard Jul 19 '18 at 17:54
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    $\begingroup$ H' is the length of _bf; H is the overall height of the opposite side (that is, H' plus the radius of the starting arc segment ab, which is equal to W, so H' = H - W. Thank you for asking. $\endgroup$ – Rory Buszka Jul 19 '18 at 18:03
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Your drawing is very... unusual. What's wrong with capital letters?

Note that $\angle dea=90^\circ-6^\circ=84^\circ$.

$$\overline{bc} = dx+R\sin \angle dea$$

$$\overline{ga}=\overline{df}\cos \angle bfd-R\cos \angle dea + R\tag{2} $$

This leads to:

$$\overline{df} \sin 6^\circ+R\sin 84^\circ=W\tag{3}$$

$$\overline{df}\cos 6^\circ-R\cos 84^\circ + R=W+H\tag{4} $$

This linear system of two equations, (3) and (4), has two unknows, $\overline{df}$ and $R$, and can be easily solved in terms of $W$ and $H$.

The rest is easy:

$$dx=\overline{df}\sin 6^\circ$$

$$dy=\overline{df}\cos 6^\circ-H$$

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  • $\begingroup$ Edit time limits are a POS. So let's see if I get this. Given Eqns (3) and (4) above, $\overline{df}\cos 6^\circ\ - R\cos 84^\circ + R = \overline{df}\sin 6^\circ + R\sin 84^\circ + H$ and some algebra later $R - R\cos 84^\circ - R\sin 84^\circ = \overline{df}\sin 6^\circ - \overline{df}\cos 6^\circ + H$, then distributing out $R$ and $\overline{df}$ I get $R(1 -\cos 84^\circ -\sin 84^\circ) = \overline{df}(\sin 6^\circ - \cos 6^\circ) + H$ $\endgroup$ – Rory Buszka Jul 19 '18 at 19:14
  • $\begingroup$ ...which gives the new radius $R$ as $R = \frac{\overline{df}(sin 6^\circ - \cos 6^\circ) + H}{1 - \cos 84^\circ - \sin 84^\circ}$ given $\overline{df}$ and $H$ - can someone please check? $\endgroup$ – Rory Buszka Jul 19 '18 at 19:18
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    $\begingroup$ @RoryBuszka I can hardly follow your calculations. My advice: multiply (3) with $\cos 6^\circ$ and (4) with $-\sin 6^\circ$, then add these two equations. $df$ will be eliminated and you will get a single equation with $R$ remaining as a single unknown. Also consider upvoting answers that you find useful. $\endgroup$ – Oldboy Jul 19 '18 at 20:03
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Calling the small circle $C$ and the tangent slanted line $L$

$$ C \to x^2+(y-y_c)^2=r^2\\ L \to y = f_y+m(x-f_x) $$

with $m = \tan\theta_0, f = (f_x,f_y)$

the tangency between $L$ and $C$ is obtained as follows

Substituting $y$ from $L$ into $C$ we have

$$ (m (x-f_x)+f_y-y_c)^2-r^2+x^2 = 0 $$

solving for $x$ we have

$$ x=\frac{\pm\sqrt{\left(m^2+1\right) r^2-(y_c+f_x m-f_y)^2}+m (y_c+f_x m-f_y)}{m^2+1} $$

but tangency imposes

$$ \sqrt{\left(m^2+1\right) r^2-(y_c+f_x m-f_y)^2}=0 $$

or solving for $y_c$

$$ y_c = f_y-m f_x\pm r\sqrt{m^2+1} $$

taking the condition that in $C$ we have $0^2+(R-y_c)^2 = r^2$ and solving

$$ y_c = f_y-m f_x\pm r\sqrt{m^2+1}\\ (R-y_c)^2 = r^2\\ $$

we obtain $r, y_c$. Now the tangency point $d$ is obtained knowing that

$$ t_0 = \sqrt{||f-e||^2-r^2} $$

and then

$$ p = f + t_0(\cos\theta_0,\sin\theta_0) $$

etc,

Attached a plot for the realization with

$$ f = (-1,2)\\ \theta_0 = 84^{\circ}\\ R = 1 $$

enter image description here

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  • $\begingroup$ IMHO, too complicated $\endgroup$ – Oldboy Jul 19 '18 at 20:36
  • $\begingroup$ @Oldboy Opinions are of the domain of Psychology, not Mathematics. $\endgroup$ – Cesareo Jul 19 '18 at 20:41
  • $\begingroup$ Ok, in that case I will only say: your solution is too complicated, the picture does not help to understand it at all and ithecsolution ends with “etc”. BTW, what is the value of the drawing? The origimal picture is much more usefull and you did not add anything to it. $\endgroup$ – Oldboy Jul 19 '18 at 21:23
  • $\begingroup$ And people do have opinions, even in mathematics. For example, solutions can be more or less elegant and there is no exact way to determine which one is the most beautiful. That judgement is opinion based. Is Rieman hypothesis true? Nobody knows and mathematicians have different opinions and talk freely about them. $\endgroup$ – Oldboy Jul 19 '18 at 21:37
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    $\begingroup$ @Oldboy Give it up. You’re not going to get through to him. math.stackexchange.com/a/2847571/265466 $\endgroup$ – amd Jul 19 '18 at 22:00
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You can compute all of these values directly. I’m going to rotate the diagram so that $c$ is at the origin, $a$ is at $(W,0)$, $b$ is at $(0,W)$ and $f$ is at $(-H,W)$.

The center of a circle lies on one of the angle bisectors of a pair of its tangents. This can be used to find the new center $e$ as follows: Using homogeneous coordinates, the tangent line at $a$ is $\mathbf l = [1:0:-W]$ and the rotated line through $f$ is $\mathbf m = [\sin\theta:\cos\theta:H\sin\theta-W\cos\theta]$. Both of these representations have unit normals, so the angle bisectors of the two lines are simply the sum and difference of these vectors. For this problem, the appropriate bisector to use is the difference. Its $x$-intercept is the new center $e = (\mathbf l-\mathbf m)\times[0:1:0]$, which is easily computed to be $$x_e = {W(1-\cos\theta)+H\sin\theta \over 1-\sin\theta}.$$ This is the amount by which the original circle center is offset toward $a$. The new radius is then simply $r = W-x_e$ and the point $d = e+r(\sin\theta,\cos\theta)$. Finally, its offset from $b$ is $e-b+r(\sin\theta,\cos\theta)$. I’ll leave expanding this to you.

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