Anybody there solved problem 7.3 of the last years exam? its about delta(L) pennate vs. delta(L) parallel muscle if the muscle shortens to 50% of it’s original size…
thanks in advance [-o<
do you want to compare your results or do you need an explanation?
I need an explanation, please.
the pennate muscle shortens much less than a parallel, because the shortening of a muscle depends on the effective parallel length. In our case we want to look at the fiberlengths itself, so interpretated in that way that i simply said the length of my parallel muscle shortens to the half which is 3cm, the fiber length of the pennate is defined as l= a/cos(tetha) and for shortening i divide it by 2. so i get a factor of pennate to parallel of 0.385.
but i don’t think that this is the same like you say the muscle shortens in this way, especially for the pennate, this is only if yu look at the fiber i think!
Oh there is a seperate thread for that…
I c&p my thoughts:
My considerations are, that the parallel muscle will simply shorten by the amount of each fiber, e.g. \Delta l_{muscle,parallel} = \Delta l_{fiber} \cdot L while the pennate muscle will only shorten \Delta l_{muscle,pennate} = \Delta l_{fiber} \cdot sin(\alpha) \cdot L where L is the number of Elements in Series (or l in the slides e.g. the height of the section), e.g the ratio will be \frac{\Delta l_{muscle,parallel} }{\Delta l_{muscle,pennate}} = \frac{1}{sin(\alpha)}
@tola99: How do you get to l = a/cos(\theta)? You used the CSA for that, right? In the tutorial i wrote down that he said, the movement is only l \cdot sin(\alpha) but maybe I got that wrong… I looked into the slides of Angeli’s Motor Muskel where a model is given for muscles and there it says that the length is also cos(\alpha) but they write the angle on a different location (comparing to the slides, it would be \frac{1}{2}\pi - \alpha)…