|
Post by robertisaacs on Feb 16, 2009 16:45:43 GMT
No. ;D Please see the rules for this section of the forum. This is the MISS bit of the forum. (MAKE it simple) We can beat that bit up on Biomech DISCUSSIONS if you like though! Although of course you are entirely correct. We could make a LOOOOOOONG list of "In real life"s. As you say this is all unquanitfiable in vivo, even if we could design a sufficiently complex model (Which would be damn near impossible even with a cray) cos even the swedish, who are very good about experimenting on real patients, would baulk at the sheer number of strain guages needed to get the data. Plus of course there is tensegrity to consider... I feel a thread coming on.... Not here though Regards Robert.
|
|
|
Post by robertisaacs on Feb 16, 2009 18:44:49 GMT
OOOOOOhhh la, get her with her fulcrums . I think thats about right. The Angle may not have changed but the Position of contact with the foot surely has! For eg. Stand on your wii fit (foot). Without leaning (changing the angle of the body / tibia) put a little more weight on the front of your feet. The wii fit will detect changes in forces. Same angle, different position. Well done all. The questions are the really relevant bit, otherwise its just maths (and flawed maths as David points out! Who's gonna be brave with the others? R
|
|
|
Post by davidsmith on Feb 17, 2009 13:29:00 GMT
All
In his introduction to students reading the course books for MSc Biomechanics Professor Sandy Nichol writes,
" It must be clearly understood that in a mechanical system, any mechanical system, whether static or in motion, there will always be equilibrium of forces and moments"
Therefore when considering a mechanical problem all equations of equilibrium must sum to zero. If we consider rotational equilibrium in terms of moments then all equations of moments must sum to zero.
Therefore using the convention of anticlockwise (A/cw) and clockwise (C/w) moments (M) A/cw M + C/w M = 0. Where C/w by convention is a negative rotation.
Since a moment is defined as force x Lever arm (or moment arm) the equation for solving the thought experiment are
(F1x L1) + (F2 x L2) = 0
Using the known forces and lever arms (where L3 = total length of both lever arms i.e. 0.08m in this case) and Fulcrum point is (c)
IE 400N Fg (Force gravity)
find the moments about point F2 by Fg = Fg x L2
Next find the force at point F1 required to balance moments about point F2 = F1x L3 - Fg x L2 = 0
which becomes F1=[(Fg x L2)/L3]
To find F2 then we need now to consider moments about c (fulcrum) and C/w Mc = F1 x L1 so:
C/w Mc + A/Cw Mc = 0 and
If F1 x L1 + F2 x L2 = 0 then F2 = [(F1 x L1)/L2]
This way will always reveal the forces and moments for equilibrium regardless of the force vector or position. Other methods become difficult where the applied force vectors are not applied at a convenient and simple orthogonal or perpendicular angle.
Try this method on all the examples, you should find it quite simple once you have done it a couple of times. If you add horizontal forces to the equation then this method will still work fine.
Cheers Dave
|
|
|
Post by blinda on Feb 17, 2009 16:35:53 GMT
The scales are slowly but Shirley falling from my eyes....I will try that method when I (himself) have sorted out my laptop, the hard drive is no more Thanks for your collective patience Robert & Dave
|
|
podmum
Full Member
"There is no dark side of the moon"
Posts: 169
|
Post by podmum on Feb 17, 2009 17:26:14 GMT
Dave Thanks for the breakdown of the formula . Podmum
|
|
|
Post by dtt on Feb 23, 2009 9:56:05 GMT
Hi Rob Just browsing this thread If its medially deviated would it not be the first ray ? Cheers D
|
|
|
Post by robertisaacs on Feb 23, 2009 18:30:32 GMT
S'right!! So now. We have our foot and the 1st ray has gone done dorsiflexed. Or perhaps some passing orthopod came along and hacked an inch off the 1st met. Either way we now have 1. Do the maths. How much grf on the 2nd met head. 2. What callus pattern do you tend to see here? 3. Seen much of this? R
|
|
|
Post by davidsmith on Feb 27, 2009 9:16:38 GMT
Robert
As there's not much action here:
Equ1) F2 = (Fg x L1)/L3
Equ 2) F1 = Fg - F2
Mp1by Fg = 400 x 0.01 = 4Nm (Eq1)
For equilibrium Mp1 by F2 = 4Nm
F2 = 4/0.07 = 57.14N
F1 = Fg - F2 => 400 - 57 .14 = 342.86N (Eq2)
Compare this to 300N on the 1st MPJ in the previous example when it eas not dorsiflexed. This equals an greater force sub 2nd MPJ by some 43N.
Now consider the surface area of the 2nd compared to the 1st. Lets assume 9cm^2 1st and 4cm^2 2nd.
300/9= 33.33N/cm^2 or 0.33333 MPa ( Mega-Pascal) 1 pascal = 1N/m^2 Atmospheric pressure is 10N/cm^2 or 0.1Mpa
343/4 = 85.75N/cm2 or 0.8575 MPa
85.75 / 33.33 = 2.57 times the pressure on 2nd MPJ compared to 1st in this example.
You can see why this might cause come pathology such as callus or necrosis or capsulitis.
Cheers Dave
|
|