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Post by robertisaacs on Feb 12, 2009 20:16:34 GMT
Having sought permission from CP i am delighted to reproduce Kevin Kirby's Thought experiments here. It is my ernest hope that any / all of you who have always wondered about what Dave and I are on about when we talk "moments" and such will step out into the wonderful world of hardcore biomechanics and play along! They'll never be a better oppertunity. I hope for questions as well as answers! If there is something you don't understand you won't be the only one so ASK! So without further ado, with full credit to Kevin and to Pod arena from whence they come, www.podiatry-arena.com/podiatry-forum/showthread.php?t=1808 here is Thought Experiment number one!
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Post by robertisaacs on Feb 12, 2009 20:19:19 GMT
1. Determine how the magnitudes of GRF change at the medial and lateral weightbearing surfaces of the foot (i.e. F1 and F2) as the axis moves from a medial to central and lateral location. 2. Why does this change in weightbearing pattern occur on the plantar foot in all three examples if the tibia is still at the same angle to the foot in all the examples? 3. If the weightbearing surfaces of the foot were movable dorsally, which surface in which foot would tend to have more dorsiflexion force on it? 4. Which foot would have more equal tendency for dorsiflexion of the weightbearing surfaces? 5. In the foot with the medially deviated STJ axis, how could we better design the leg-foot model to have reduced GRF plantar to the medial weightbearing surface by adding tensile load-bearing structures to it?
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Post by robertisaacs on Feb 12, 2009 20:24:20 GMT
Some tips to get you started.
A moment is a measure of rotational force. It can be clockwise or anti clockwise.
In a static stable system (ie a foot not moving) the rotational forces are always balanced. In other words there is the same amount of total cllockwise moments as anticlockwise. (Dave if you are reading this, suspend disbeleif for a second ok?)
A moment is always the force multiplied by the lever arm.
To resolve one of these i make two columns with clockwise moments in one and anti clockwise in the other. I write what i know and then try to resolve out the unknown quantities.
So, who's going to have a crack? (and no sneaking off to pod arena for the answers. I'll know!
Regards Robert
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podmum
Full Member
"There is no dark side of the moon"
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Post by podmum on Feb 13, 2009 10:50:01 GMT
1st confession, I did look at the 2nd of KK's postings but no further...........promise (not done any physics/ complex maths - please no laughing if it is not complex- since I was at school which was a while ago ) I will endeavour to have a go at this over the weekend Podmum
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Post by lawrencebevan on Feb 13, 2009 14:34:52 GMT
Let me see.... If I crack at the middle one to get the ball a rolling
we can assume that F1 + F2 = 400N (equal and opposite and all that) we also know F1 x 4cm (or 0.04m) = F2 x 4cm (or 0.04m) (equilibrium and no rotation and all that) so F1 = F2 x 0.04 /0.04 so F1 = F2 so if F1=F2 and F1 + F2 = 400N, F1 = 200N and F2= 200N
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Post by blinda on Feb 13, 2009 17:35:42 GMT
Hockay, I`ll have a stab.. Agree with Lawrence above. Therefore, the foot with the medially deviated STJ axis 400/8cm = 50NCM. Then F1 50NCM x2 = 100 and F2 50NCM x 6 = 300. The laterally deviated STJA F1 50NCM x 6 = 300 and F2 50NCM x 2 = 100. Cheers, Bel
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Post by robertisaacs on Feb 13, 2009 20:29:51 GMT
Gold stars for both of you. Both used different methods to reach the same conclusion. Here's how i'd do it (for the medial one) We know that total GRF (up force) is the same as the total down (400n) thats f1 + f2 = 400 We know that the two rotational forces are the same (they balance) thats f1 * 2 (thats clockwise or supinatory moments) is the same as f2 * 6 (anticlockwise or pronatory moments) So we know :- f1 + f2 = 400 2f1 = 6f2 now to solve the equation we need to get rid of one of the unknowns if 2*f1 = 6*f2 then one lot of F1 = 3* F2 so if we know that f1 + f2 = 400 we know that 3f2 (which is the same as 1*f1) + f2 = 400 3f2+f2 = 400 4f2 = 400 f2 = 100 As i say thats a different way to work it out, but it all works. Well done all! Now, the other questions....
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podmum
Full Member
"There is no dark side of the moon"
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Post by podmum on Feb 14, 2009 16:16:44 GMT
Having been advised not to get involved in BMX by a 'well wisher' I feel that maybe I should bow out gracefully - NOT ;D Am still working on it................ Podmum
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Post by robertisaacs on Feb 14, 2009 20:26:55 GMT
Keep at it! Its a great feeling when it all clicks! Its easier to demonstate with a couple of sets of scales, some 2 by 4, some rope and a fishing scale. But then, most things are
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Post by ianl on Feb 15, 2009 8:00:07 GMT
OK Rob
How about you put your Blue Peter uniform on and do that at T3. Take a couple of thought experiments and do the visual learning approach. VL works better for me so maybe it works better for some others.
Ian
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Post by robertisaacs on Feb 15, 2009 12:55:19 GMT
Visual learning you say. Well, the image of bel balanced at one end of a plank, twirly on the other, David at one end with the peroneal bit of rope and Bel at the other with the tibialis bit of rope is one which certainly clings in the mind! ;D I'll offer it up so its there if people want it. Regards Robert
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Post by blinda on Feb 15, 2009 19:32:13 GMT
Because the angle of the fulcrum (pivotal point of the tibia angle) remains in the same plane in all 3 cases, but the forces are determined by its positioning?? Hey, I don`t mind walking the plank for my biomech ignorance, but i draw the line at using a set of scales to humiliate me
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podmum
Full Member
"There is no dark side of the moon"
Posts: 169
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Post by podmum on Feb 15, 2009 20:26:51 GMT
2. Why does this change in weightbearing pattern occur on the plantar foot in all three examples if the tibia is still at the same angle to the foot in all the examples?Here is my thought on Q2....the tibia is in the same angle to the foot as it is not needed in the rotation of the STJ (Q1. not answered here as 2 gold stars already given out). Podmum
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Post by Admin on Feb 15, 2009 20:41:49 GMT
May I throw a couple of spanners into the works ? In real life the tibia rotates internally as the foot pronates. In real life the femur also rotates internally as the foot pronates (as long as the knee is flexed). This in no way affects the experiment curently under discussion, just that these actions do happen and are largely unquantifiable in our patients. BTW Podmum, I loved the comments from "a well-wisher" over on TFS. It would have been useful to have them defend said comments at T3 but it seems that it is not to be . ;D ;D ;D
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Post by blinda on Feb 15, 2009 20:52:42 GMT
Absolutely! That was my initial thought on the TE. However, i also appreciate that these are simplified diagrams designed purely to help the likes of myself understand the fundamental principles of rotational equilibrium Dippin` me toes into the unkown Bel
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