Symphony # 5

[batman8100]

Does anybody know, how many times the mofit at the beggining of the symphony 5(tha, tha, that tham), repeats itself alone the whole symphony? I try to count them and I get different numbers.

[robert newman]

It is said of Anton Bruckner that he used to count the leaves on trees, the number of bricks in any wall he saw, and even the number of roses on bushes. Let us assume these distrubing reports are true. Let us further assume that, over a period of time, the great Bruckner stumbled across that mathematical marvel that is the Fibonnaci series - so that the time taken to finish his calculations was greatly reduced.

(Well, that's where my theory collapses - since Bruckner would simply have used the same time by finding other trees).

I estimate that Beethoven used this motif 147 times in the 1st movement alone - this based on the assumption that the square of the longest side of a parallel rhomboid is 0.453 of the circumference of a sphere that can be drawn within it muliplied by the tangent of its shortest side. This, plus the date of Beethoven's birth gives us, roughly, 147 or (as Bruckner would say) about 9 seconds of tree leaf counting.

[sjwenger]

Leonard Slatkin (he's no Felix Slatkin at'all) once did a pops concert in which he counted them as the orchestra played it, but I don't know what number he came up with.

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[Peter]

Originally posted by batman8100:
Does anybody know, how many times the mofit at the beggining of the symphony 5(tha, tha, that tham), repeats itself alone the whole symphony? I try to count them and I get different numbers.

I don't know the answer without spending half a morning counting from the score - however it just shows the amazing genius that can produce such a powerful original piece based on a repeated rhythm of 4 notes!

[Michael]

Originally posted by robert newman:

I estimate that Beethoven used this motif 147 times in the 1st movement alone - this based on the assumption that the square of the longest side of a parallel rhomboid is 0.453 of the circumference of a sphere that can be drawn within it muliplied by the tangent of its shortest side. This, plus the date of Beethoven's birth gives us, roughly, 147 or (as Bruckner would say) about 9 seconds of tree leaf counting.

I think it's 0.452, Robert, but I am wide open to correction on this. Another good method is to multiply the number of C sharps in the first movement of Symphony No. 8 by the number of curses scribbled in the margins of the original manuscript of the Ninth.
I had to pound the keyboard 238 times to send this.

Michael

[robert newman]

I stand corrected !

[Michael]

Originally posted by Peter:
I don't know the answer without spending half a morning counting from the score - however it just shows the amazing genius that can produce such a powerful original piece based on a repeated rhythm of 4 notes!

He does deviate from the four-note rhythm a few times - in the bridge passage (the horn-call theme leading into the second subject) and also in the coda.

Michael

[batman8100]

Since I place this question, I had the chance to hear several times, I just love it and I counted 152 times, reading your results and I think I am closed.
Thanks to All