Rotated strings height

Notice from the figure that the distance between the two shown points is: $2 = \sqrt{1^2+1^2+h^2}$ Hence $h=\sqrt{2}\approx\boxed{1.4142}$

We can solve this problem easily by taking into consideration only one of the string (taking all string at once won't make the difference). Now, what's left to do is to map out the movement of that string.

Let's look at both circles on top view:

• The Red X would be the tip of the string at the lower circle and the Red dot would be the tip of the string at the upper circle.

Before Rotation:

After Rotation: It would seem to appear then that it moved along the X direction by a distance of √2.

Using the information that the string length was 2 and it moved along X by √2. We can calculate the height.

$H^{2}$ = $2^{2}$ - $√2^{2}$

$H^{2}$ = 4 - 2

$H$ = $\boxed{√2}$ = $\boxed{1.414213562}$

The string length remains 2.

The horizontal distance between its ends being on a unit circle $90^\circ$ apart will be $\sqrt2$ .

So the vertical distance between the loops will be $\sqrt{2^2-(\sqrt2)^2}=\sqrt2\approx\boxed{1.414}$

We can define the lower unit circle to be $c_1$ and the upper unit circle to be $c_2$ . Because they are both unit circles where $c_2$ is rotated over 90 degrees (which equals $\frac{\pi}{2}$ rad.) we can say that for a point P on $c_1$ at time $t$ the coordinates of P would be $(x,y,z)=(cos(t),sin(t),0)$ and that for a point Q on $c_2$ at time $t$ (same variable) the coordinates of Q would be $(x,y,z)=(cos(t+\frac{\pi}{2}),sin(t+\frac{\pi}{2}),h)$ where $h$ is a constant (between 0 and 2). Our objective is to find the value for $h$ . We know that $|PQ|=2$ . This means that $(x_P-x_Q)²+(y_P-y_Q)²+(z_P-z_Q)² = 2² = 4$ and thus that $(cos(t)-cos(t+\frac{\pi}{2}))²+(sin(t)-sin(t+\frac{\pi}{2}))²+h²=4$ . Writing the squares out with some terms switching places gives us $(sin²(t) + cos²(t))+(sin²(t+\frac{\pi}{2})+cos²(t+\frac{\pi}{2})) - 2(cos(t)cos(t+\frac{\pi}{2}) + sin(t)sin(t+\frac{\pi}{2}))+h²=4.$ Furthermore, the term $cos(t)cos(t+\frac{\pi}{2}) + sin(t)sin(t+\frac{\pi}{2})$ equals $cos(t-(t+\frac{\pi}{2})) = cos(-\frac{\pi}{2})=0$ because $cos(u)cos(v)+sin(u)sin(v)=cos(u-v)$ . So we get $1+1-2\cdot 0 + h²=4$ , $h² = 2$ and this gives us the final answer: $h=\sqrt{2}=\boxed{1.414...}$ The method I used was more complicated than others I have seen for this problem, nevertheless I wanted to share my strategy for solving it.

I also made a figure of what I mean by all this in the picture below (not on scale (some lines are not correct)). The coordinates of point P and Q vary on the circles where they are at (component t involved).

The top view makes it clear that there are three mutually perpendicular directions with lengths 1, 1 and distance, d. The space diagonal is 2 because it is the unchanging string length.

The 3-d Pythagorean theorem for this is $1^{2}+1^{2}+d^{2}=2^{2}$ and solving gives $d=\sqrt{2}\approx1.41421356$

When looking from the top onto the cylinder the original point of each string and the new point create a right angled triangle.

Using Pythagoras the length of the long site can be calculated to be √(2).

When looking from the site onto the cylinder the strings form a right angled triangle again.

The length of the strings (2) is the longest site, while the √(2) calculated above is the base of the triangle. The height of the cylinder is the third site of the triangle.

Using Pythagoras again we can calculate, that

h² = 2² - (√(2))² Root and square cancel each other. h² = 4 - 2 h² = 2 h = √2

I just kinda got lucky, I realized it was between 0 and 2 but i thought it was exponential and I answered 1.41 or sqrt(2)