Flexible planar frame is made of 7 rigid rods, connected at their ends by bearings?

Question

http://alexandersemenov.tripod.com/ya/vrtr/index.htm

The frame can slide on the flat top of the desk.
Two red rods of length 30 can revolve around fixed point A and one green rod of lenght 2 can revolve around fixed pont B, |AB| = 34.

The lenght of four blue rods is 18 each.

What is the area of the loop made by joint C, after green rod completes one revolution?

Answer 1

This is a Peaucellier linkage, which has the property of inversion, which means C traces out another circle. What's the diameter? Well, we fiddle through some Pythagorean math. Let x = 1/2 of the distance between the ends of the 2 red rods. Then we find out what x is when the end of the linkage is at 32 and at 36 from A. We plug it back into some more equations to find the min/max distances from A to C, which comes out to 16 and 18. The difference is the diameter of the circle traced out by C, and the area is just π.

Addendum: Nice work, Dr D.

Answer 2

Let r = variable distance between C and A
θ = angle of elevation of AC from original position

After all the tedious work, it can be shown that
r^2 = (290 - 578sin^2 θ) +/- sqrt[(290 - 578sin^2 θ)^2 - 288^2]

One of the keys in obtaining this relation is recognizing from the geometry that A, C and F must always be colinear, where F is the other end of the 2 units long link connected to B.

The area traced by C can be computed by performing the double integral of
r*dr*dθ between the limits
r^2 = (290 - 578sin^2 θ) - sqrt[(290 - 578sin^2 θ)^2 - 288^2]
and
r^2 = (290 - 578sin^2 θ) + sqrt[(290 - 578sin^2 θ)^2 - 288^2]

And θ = -arcsin(1/17) to θ = +arcsin(1/17)

You end up having to integrate
34* integral sqrt(1 - 289sin^2 θ) *cosθ dθ
which is
arcsin(17sinθ) + 17*sinθ*sqrt(1 - 289sin^2 θ)

Apply the limits to get π

If you knew in advance that the locus of C would be a circle (as opposed to an ellipse), then it's not difficult to find that the diameter is 2. If you didn't know it would be a circle, then you'd have to do what I did.

Answer 3

Okay, now Scythian1950 you are right. Sorry, i made a mistake,the circles are not identical. you got it right. up-rating your answer...
I will update my answer slightly, but you deserve the credits.


The green rod will pull the 2 blue right rods . These blue rods will try to pull the red rods. However, since point A is fixed, the red rods will tend to close and rotate at the same time. And therefore will force the rhombus to rotate and squeeze vertically. This will force the point C to rotate but in the opposite direction.
More details:
-------------------
Let theta be the angle between the two red rods (the smaller angle).
Let i be the angle bisector of the two intersecting red rods.

Consider clockwise rotation.

Green Rod revolve:
-) 0 to 90, then i will rotate counter-clockwise, and theta will decrease due to the limitation of the rods length. This will cause the rhombus to squeeze vertically and hence point C will be moving in addition to the rotation caused by the i.

-) 90 to 180, i (clockwise), and theta will remain decreasing.
-) at 180 the theta reached its minmum and the difference between the initial C and the current C will be maximum which is the diameter of the newly fowmed circle.
The bisector i returned to its initial position
-) 180 to 270, i will remain turning clockwise, theta will increase. C will be returning back.
-) finally, 270 to 360, i will turn counter-clockwise, theta remain increasing.
-) at 360 C revolved back to its initial position and formed a circle with smaller diameter.