The Spieker Limit

Classical Mechanics Level pending

What is the limit of the height of this Y Y wire frame centre of mass from the base in terms of length L L ?

Give your answer up to 4 significant digits.


The answer is 0.8759.

W Rose by W Rose , 63, USA

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1 solution

W Rose
Jun 21, 2016

L = 1

D = L / 2^0.5

A = L^2 / 2^0.5

Limit = [ .25 x 2 x D x A ] + [ .75 x 2 x D x 2^0.5 x A ] + [ .75 x 2 x D x 2^0.5 x A ] / [ A ] + [ 2^0.5 x A ] + [ 2^0.5 x A ]

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

Can you elaborate what L, D and A are? Why is the final expression the limit?

L is the unit length of the longer leg(s) of the Y wire frame. D is the distance (vertical height) of the shorter leg while the overall Y wire frame height and width is 2 x D as the wire thickness approaches zero. A is the area of the shorter leg of the Y wire frame. The height of the center of mass of the Y wire frame approaches its maximum value when the wire thickness converges to zero. The Y wire frame configuration is constrained within a square box which has a fixed side dimension of D x 2.

W Rose - 4 years, 11 months ago

Maximum unclear!!!

And what about the lengths of the arms of the "Y"? How are they determined if they are longer than l? The drawing above doesn't deliver a unique measure!

In all: Confusing!

Andreas Wendler - 4 years, 10 months ago

The drawing is not to scale. The unit length of the upper arms are set to one. The maximum centroid location limit is relative to this unit length. The centroid is not positioned on the wire frame perimeter.

W Rose - 4 years, 10 months ago

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