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tpribanic57
10-13-2005, 11:46 PM
Once again, many thanks to all of you.
Best, Tomo.

Original question:

Dear All,

typical force platform data are coordinates of application point, force vector in three directions and so called free torque Tz that has only one vertical component. I apology for perhaps lack of my physics understanding, but I could not find clear explanation why is biomechanics of normal human gait such that it not possible during the normal gait to have torque also in two other horizontal directions Tx and Ty. Some people say it is absolutely impossible to have Ty and Tx, others say those two component are neglibagle. Seems to me 'absolutely impossible' and 'neglibagle are two quite distinct statements. If we can apply with the foot force in all three directions, even in situation where the foot remains still on the platform, why is it then not possible to have torque in three directions too. What is the physical/biomechanical 'guidance' that determines whether there will be torque in horizontal directions or not?

Besides some people emphasize 'free' torque? Does word 'free' implicate this torque does not depend on actual point/axis of rotation?

Last question is concerned with cases where Ty and Tx do normally exists and particular platform construction. For example Kistler platform normally outputs data from eight channels which are originally provided by four force sensors. Data from those either channels are the combined to ultimately give ground reaction force in all three directions, cop and torque in only vertical direction. However, if I am not mistaken, based on particular derivation of equations, it seems to be necessary to assume Tx=Ty=0. In other words, in cases where we cannot assume Tx=Ty=0 usual eight channel data would not allow us to compute Tx and Ty as well. Under those circumstances are there any additional force (torque) sensors needed then on platform or what?

Thank you for your time.
Best,

Tomislav Pribanic, M.Sc., EE
Department for Electronic Systems and Information Processing
Faculty of Electrical Engineering and Computing
3 Unska, 10000 Zagreb, Croatia
tel. ..385 1 612 98 67, fax. ..385 1 612 96 52
E-mail : tomislav.pribanic@fer.hr


Replay1:

As far as I understand it, for a torque about a horizontal axis to occur
requires the foot to exert some 'pull' on the force plate. For example, if
my feet were glued to the force plate with super-glue and i leant forward,
this would do the trick. Therefore, any footwear which produces any slight
suction upon the force-plate might result in such a torque being measured.
I have never measured this type of torque, so I don't actually know if is
negligble, but i don't think it is impossible.

'Free torque' refers to torque measured about an axis defined by the point
of application (cop), rather than the centre of the force plate (or some
other fixed point). To do the calculation, you simply find out where the
cop is and adjust accordingly.

I'm not sure about the third point - it seems to me that all the sensors do
exist on a kistler plate to measure torque about any axis.

Hope it's of some use

Raymond Reynolds, PhD
MRC Human Movement Group
Sobell dept. of Motor Neuroscience and Movement Disorders
Institute of Neurology
8-11 Queen Square
London WC1N 3BG

Tel: +44 020 7837 3611 ext 4120
Fax: +44 020 7278 9836

Replay2:

Dear Tomislav,

>What is the physical/biomechanical 'guidance' that determines whether there
will be torque in horizontal directions or not?

Tx & Ty SHOULD be 0. It is due to the nature of vector T. The explanation
can be found at http://kwon3d.com/theory/grf/cop.html with some diagrams.

>Does word 'free' implicate this torque does not depend on actual point/axis
of rotation?

For this reason, I started calling the free vertical torque (Tz) 'ground
reaction torque' some years ago since the term 'free' is misleading.
Normally, a torque is produced by a force so it comes with a force that
causes it. Tz does not. Tz is a pure torque with no consideration of the
force and I think it is why people call it 'free'. But, as you mentioned, a
'free' vector means a vector that does not change as the perspective
changes: force, torque, etc. A torque is already a free vector and 'free'
torque is thus confusing and misleading.

Let's go back to the first question. The interaction between the foot and
the force plate occurs through an AREA (footprint), not a point. So all the
small areas under the foot will have their own interactions with the foot.
The overall effect of the foot-ground interaction is basically equivalent to
the combined effect (force- and torque-wise) of the four forces (F1, F2, F3,
& F4) measured by the four sensors embedded in the force plate. The
foot-ground interaction can be replaced with these four forces.

If we use the plate reference frame located at the center of the surface as
our frame of reference, each force produces a torque about the origin: M1,
M2, M3, & M4. So the resultant force F is equal to F1 + F2 + F3 + F4. The
resultant torque M = M1 + M2 + M3 + M4 = r1 x F1 + r2 x F2 + r3 x F3 + r4 x
F4, where ri = position vector of sensor i. The problem is that the
resultant torque M is not necessarily equal to the torque produced by the
resultant force about the origin: M R x F. M = R x F holds only when the
foot-ground interaction occurs through a POINT, not an AREA. In reality, M =
R x F + T, where T = ground reaction torque. T is simply the portion of
torque M that can not be explained by F. Let Ri be the relative position of
the sensor to the center of pressure (CP):

ri = R + Ri

The resultant torque can then be expressed as:

M = r1 x F1 + r2 x F2 + r3 x F3 + r4 x F4
= (R + R1) x F1 + (R + R2) x F2 + (R + R3) x F3 + (R + R4) x F4
= R x (F1 + F2 + F3 + F4) + ( R1 x F1 + R2 x F2 + R3 x F3 + R4 x F4)
= R x F + T

T = R1 x F1 + R2 x F2 + R3 x F3 + R4 x F4

T is the resultant of the torques produced by the four forces about the CP.
Since CP is the balance point for the four forces, horizontal torques
produced by the forces should be balanced about the CP: T = [ 0, 0, Tz]. Tz
is the sum of the coupling effects of the forces about the vertical axis as
a result of the torsional interaction between the foot and ground.

The four-force system can nicely be replaced with a more simpler ground
reaction force + ground reaction torque system: {F1, F2, F3, F4} ==> {F, T}.
Both F and T should be considered in the inverse dynamics computations for
the foot.

>Under those circumstances are there any additional force (torque) sensors
needed then on platform or what?

Kistler plates are no exception. "Tx = Ty = 0". The only difference is that
AMTI plates give Fx, Fy, Fz, Mx, My, and Mz whereas Kistler plates give
F1x+F2x, F3x + F4x, F1y+F4y, F2y+ F3y, F1z, F2z, F3z, and F4z. The
above-mentioned webpage contains Kistler plate equations as well.

I hope it helped.

Young-Hoo
-------------------------------------------------------------------
Young-Hoo Kwon, Ph.D.
Biomechanics Laboratory
Texas Woman's University
Phone & Fax: (940) 898-2598
ykwon@mail.twu.edu / kwon3d@kwon3d.com
--------------------------------------------------------------------

Replay3:

Dear Tomislav,

The x and y torques can be calculated by the vertical force and its point of application (centre of pressure). As you say, these torques are often large - certainly larger than Tz. They are dominant components in the equations for inverse dynamics.

Chris

Replay4:

Hello Tomislav!

I'm not working on this field at the moment, but up to my understanding
of mechanics a horizontal torque in gait can not exist in theory.
Try to imagine it this way: to transduce a torque, you need at least two
fixed points. In these points you insert two forces that are of opposing
direction (or partially opposing), but not collinear (otherwise they
would cancel themselves out).
In the case of vertical torque you don't have only two points, you have
an entire contact surface. That would be more than enough. If now your
toes push left hand your heel right, you induce a vertical torque in the
ground.
Now, imagine a theoretical horizontal torque. For this, say, the toes
have to push down and the heel would have to pull up. While you will
transduce a force in the toes (downwards, ground reaction upwards), your
heel will not transduce anything but just lift off!
Therefore, unless you glue the foot to the ground, you can not transduce
horizontal torques.

That was the theory. Now, I don't know how sensitive Kistler plates are,
but I think they are really sensitive. If you would have someone walking
barefoot on them and the feet are slightly sweaty, then on lift off you
would have adhesion effects between the foot and the plate (because of
the water and due to the fact that the foot is soft, even pronouncing
this effect). Therefore it might be that the Kistler plate in fact reads
negligibly small horizontal torques.
Imagine, for example, not a human but a gecko walking. The adhesion of
them to a blank surface is pretty high (can make them walk up the
walls), so my guess is you would read substantial horizontal torques in
them.

I hope this somewhat nonscientific explanation helped.

yours

Fritz

Replay5:

Dear Tomislav

Your mail is of a great interest, I used a force platform manufactured in
France (Captels, 34270 Saint Mathieu de Treviers, www.captels.com) which
allows to measure Tz but also Tx and Ty. I think these data are very
interesting in study of pathologic gait as in case of rotational
abnormalities of lower limbs. I used Tx end Ty in accordance with Fz
specially to calculate the angle of the foot relative to the walking
direction. So, I agree, Tx and Ty can be interesting to get but their
intensity is relative to the position of the foot on the plate and so can
change greatly from one step to the next one. So its necessary to correct
to compare them.

Best bregards

Replay6:

Dear Tomislav,

The reason why Tx and Ty can be assumed to be zero is because in order
for these two torques to be produced, the foot will have to be "stuck" to
the forceplate. Imagine placing the palm of your hand on a piece of paper.
Friction between the hand and the paper will allow you to cause a rotation
in the Z direction but you will have no way to rotate the paper in the other
two directions unless your hand is stuck to the paper. (note that the
coordinate system i am using is according to the Kistler coordinate system,
not ISB).

I guess the "free" refers to the torque being a pure moment and
differentiate from an offset force about a rotating axis.

You are right. The 8 channels will not allow you to compute Tx and Ty.
However, if you are able to tap out the individual triaxial signal in each
load cells, (i.e. you will have 12 signals), you may be able to find out Tx
and Ty. But i believe the computed Tx, Ty values will be very small...

Hope this helps... Do let me know when you managed to find the Tx and Ty
values.. I will be most interested! Thanks!

Regards,
Kelvin Chan

Replay7:

You should look at the Appendix of a Paper I have published in the Journal
of Biomedical Engineering where I addressed exactly this issue.
H.-B. Schmiedmayer and J. Kastner: Enhancements in the Accuracy of the
Center of Pressure (COP) Determined With Piezoelectric Force Plates Are
Dependent on the Load Distribution.
Journal of Biomechanical Engineering, 122, pp 523-527.
If you have online access to this journal you can get a pdf under the
following link:
http://scitation.aip.org/getpdf/servlet/GetPDFServlet?filetype=pdf&id=JBENDY
000122000005000523000001&idtype=cvips&prog=search

But to answer your question in brief:
The resultant point of force application of a pressure and shear
distribution or as it is normally called the center of pressure (COP) is
exactly the point where the resultant torque vector has no components in the
x and y direction (assuming x and y lie in the plane of the force plate)

Kind regards
Heinz-Bodo Schmiedmayer

Replay8:

Dear Tomislav,
The matter is quite complicated, but is very nicely explained in
chapter 1 of Zatsiorsky, Kinetics of human motion, the section on
wrenches.
There are certainly moments, Tx and Ty, = crossproduct of GRF
with respect to the center of the force plate. (my own Bertec force
plate gives these.) From these the CoP position is calculated.
When this is done, the moments Tx and Ty w.r.t. the CoP are of
course zero, but this all depends on the point around which the
moments are calculated. In general, an additional "free" moment is
left, Tz. But again, consult Z. for details.

- -
At Hof
Center for Human Movement Sciences
University of Groningen
PO Box 196
9700 AD Groningen
The Netherlands
Phone: (31) 50 363 2645
e-mail: a.l.hof@med.rug.nl
www.ihms.nl
----------------------------------------------------

Replay9:

Pax!

Surely you can have torque in all directions. In fact this property is
used in force plates which basically have just one sensor in the middle
that measures (force components) F_i and T_i through the center of the
sensor. From this you get the COP-coordinates as

x = (z*F_x - T_y)/F_z,
y = (z*F_y + T_x)/F_z.

z is the distance form the sensor center to the force plate surface. F_z
is practically the same as the persons weight in the quiet standing
situation (save for variations due to the movement of the heart and
hemodynamics). If one uses a rectangular force plate with triaxial force
transducers at the corners, one obtain 3*4 = 12 outputs. However, since
the force transducers are in the same horizontal plane one can replace
the 8 horrizontal components with 4 sums ending up with 12 - 4 = 8
outputs which are enough for calculating T_z and COP. One can also
calculate T_x and T_y (using the above formulas in reverse direction)
but there is usually no need for those. If the transducers are named 1,
2, 3, 4, starting from left rear transucers and going counterclockwise
one may use the force sums

F23_y = F2_y + F3_y,
F14_y = F1_y + F4_y,
F12_x = F1_x + F2_x,
F34_x = F3_x + F4_x,

where F2_y is the force measured by transducer #2 in the y-direction,
etc. Using these sums we obtain

T_z = (F12_x - F34_x) *L/2 + (F23_y - F14_y)*W/2

where W is the width and L the lenght of the force plate. The COP is
calculated from the vertical components Fi_z,

COP_x = (W/2)*(F2_z + F3_z - F1_z - F4_z)/(F1_z + F2_z +F3_z + F4_z),

COP_y = (L/2)*(F3_z + F4_z - F1_z - F2_z)/(F1_z + F2_z +F3_z + F4_z).



I do not think of having ever heard of *free torque* in physics, but it
might, as you suggest, be someone's name for a *(force) couple* in
physics; that is, two forces F1 and F2 whose (vector-) sum is zero, F1 +
F2 = 0, but whose total torque is non-zero, r1 x F1 + r2 x F2 0. It
is possible to create a momentary force couple along the z-axis, due to
the friction between sole and force plate surface, by standing on the
force plate and rotating the trunk. COP is meaningless for force
couples. In order to produce force couples along x and y axis one would
have to strap the feet to the force plate (would be necessary e.g. on
the space station anyway).

I have tried to avoid *typing errors* but there is of course no
guarantee ...

Best regards

Frank Borg

Chydenius Institute, Jyväskylä University
Finland

Replay9:

Dear Tomislav!
I hope that everything is well with you!
I read your question concerning force plate data in Biomech-L.
The reason why you do not have any horisontal torque in force plate data
is that you usually walk barefoot or with "normal" shoes on the force
plate. In this case you only apply downward directed forces (push) on
the plate. In order to obtain a torque you would need a pair of forces
at different locations on the plate, one downward (pushing) and one
upward (pulling). You can do this if you are a Gecko (lizard), or if you
have shoes with suction soles.
I hopes that this helps you!
Best regards - Håkan

--
************************************************** *********************
* Håkan Lanshammar, Professor, Head of Department *
* Systems and Control, Dept. of Information Technology, Uppsala Univ. *
* P.O. Box 337, S-751 05 Uppsala, SWEDEN *
* E-mail: hakan.lanshammar@it.uu.se, Tel: +46-18-471 1087/3033 *
* Fax: +46-18-51 19 25, http://www.it.uu.se *
************************************************** *********************

Replay10:

Dear Tomislav,

At the moment I'm doing my PhD at the University of Twente, Enschede, The
Netherlands.
My research also concerns acquiring data from a force platform. The force
plate we're using, gives 6 signals (Fx, Fy, Fz, Tx, Ty and Tz) at a specific
point of the force plate (usually the center, but mentioned in the
datasheet).

In your text, you mention the application point: the point where the force
acts from, meaning no torques (Tx, Ty) there. This point is not measured by
a force plate, but derived from the signals.
The calculation of the application point is done using the rule that the
torque is the vector product between the displacement vector and the force
vector. In this case, the torque and force vector are known, and the
displacement vector (which is the distance from the application point and
the center of the force plate) can thus be calculated.

So, by calculating the application point, you get the point where there is
no Tx and Ty component. (These components are 'hidden' in the application
point).
Hopefully this will help you, otherwise you can always ask me and I'll try
to answer any questions.

Regards,
Martin Schepers

************************************************** **************************
ir. H.M. Schepers
University of Twente
Faculty of Electrical Engineering, Mathematics and Computer Science
BSS chair
Building Hogekamp (45), room 6158
P.O. box 217, 7500 AE Enschede
The Netherlands
Phone: +31 (0)53 489 2766
Fax: +31 (0)53 489 2287
Email: h.m.schepers@ewi.utwente.nl

Replay11:

Hi,

If I understand correctly, the "free torque Tz" is the amount of twisting that a limb exerts against the ground. If I attempt to twist my foot laterally, for instance, I exert a torque through my leg, and an equal and opposite torque against the Earth.

The other torques most certainly exist, but I believe they are not really thought of as "torque," but rather "pressure." For instance, when my heel strikes the ground (or a force plate), the pressure is naturally concentrated at the heel (and at the back of the force plate, which would produce a torque in the plate making it twist in such a way that the back is pushed down, and the front is pushed up). Then as I continue through the step, the pressure moves anteriorly toward the ball of the foot. If the force platform can measure this, it will indicate that the pressure is moving towards the plate's front end. Once again, a torque is being produced so that the front of the plate is now pushed down, and the back is pushed up. But it's generally not thought of as "torque." The same idea occurs in the mediolateral axis.

You indicated that this Kistler force platform measures Center of Pressure (COP). This is basically torque, but it's more practical to think of it as a pressure difference. It's measured by using the difference between front and back vertical sensors (or right and left vertical sensors, if you're measuring mediolateral pressure or torque).

Currently I'm building a force "pole," like a tree branch, for small mammals which will measure the torque applied about the long axis of the "branch." But if I built a force "platform," then I would probably say that I'm measuring pressure differences between right and left instead of torque.

Hope this is helpful.

Andy Lammers
Cleveland State University
Cleveland, Ohio, USA

Replay12:

Dear Tomislav,

let me try to answer your three questions all at once as they
really belong together.

First of all: you are right, having even only one foot on the
ground the subject in general applies three torque components
to the plate (as there is an array of sensors), and - more important -
in turn the plate applies THREE torque components to the foot. The
reason that the last statement is true is obvious: WE KNOW in advance
that there is an AREA, not only one singular point, where the foot
contacts the plate.

Now the mechanical theory of the interaction of two mass distributions
(e.g. foot and plate) tells us that you can describe their translational
(momentum) and rotational (moment of momentum) interaction by exactly
six independent variables: you need three force components and three
further numbers (e.g. torques but not necessarily torques, see below).
What you know is: the MEASURED force distribution in the sensors
underneath the plate. What you can DERIVE from this knowledge of the
mechanical effect is some knowledge of the causing distribution of
forces on the plate - you can determine exactly six unknowns.

The derivation is successful (i.e. in the best case: UNIQUELY
DETERMINED) in SPECIFIC CASES. One specific case (which applies to the
force plate problem and was formulated explicitly e.g. in old Kistler
force plate manuals back in the seventies - and eighties?) implies three
aspects: the contacting area (surface of the force plate) is known as
PLANAR, the SENSOR GEOMETRY must be KNOWN and you ASSUME
that there are NO PULLING FORCES included in the force distribution
applied to the force plate.
In this case the effect on a force plate is UNIQUELY determined by the
three summed force components (Fx,Fy,Fz: sum of all forces in the force
distribution) and exactly one combination of two components (Ax,Ay) of
the point of application of the summed force and a remaining torque
component (Tz) acting exactly vertical to the plane where the summed
force acted on.

To add it up: if you are sure that there are no pulling forces (nothing
sticks or glues to it - nowhere!) applied to the plate than the
Fx,Fy,Fz,Ax,Ay,Tz are an adequate description of the force distribution
on the plate. I have to admit, currently I do not have a quick answer to
the question what had to be done IF pulling forces were applied.

Cheers,
Michael.

Replay13:

Pax!

Thanks for the Kistler pdf. (A slightly different coordinate system is
used.) Indeed, the *free moment* refers to a force couple. I just found
there is a complete discussion of the topic in Zatsiorsky, Kinetic of
human motion (Human kinetics, 2002) pp 29-48. Forces and couples are
analyzed in term of *wrenches* and the point of wrench application (PWA).

Regards Frank B

Replay15:
Pax!

Hm, the explanations in the books may not be so straightforward, so this
is how i understands it:

The 4 x triax force plate measures the forces (vectors) Fi = (Fi_x,
Fi_y, Fi_z) at the points ri (i = 1, 2, 3, 4). From this one can
calculate the total torque M which the body in contact with the froce
plate produces,

M = sum ri x Fi (cross prodcut).

The total force F is F = sum Fi. Now the idea is to represent the
momentum M as sum of a torque R x F produced by the total force F and a
point of application R (vector), and a force couple T (*free moment*)
which is assumed to be directed (wrench axis) along the z-axis, T = (0,
0, T_z),

M = R x F + T.

If we apply the vector dot-product with F to the previous equ we get

M . F = T . F = T_z*F_z

since F. (R x F) = 0 (F is orthogonal to R x F). This is equivalent to
the Kistler-formula for T_z. So, why may one assume that T = (0, 0, T_z)
to beging with? Unless the foot is glued to the surface the ground
reaction force consists only of pressure and horizontal friction forces.
The (vertical) pressure forces produce no couple, while the hortizontal
(friction) forces can only produce torque along the z-axis.

Regards Frank B

Replay16:
Pax!

I agree (despite that i once upon a time wrote my Msc thesis on
mathematical methods in analytical mechanics) that playing with couples
and forces in statics may be confusing ...

Yet, i think the *complete* analysis goes like this. The total torque is
given by

M = sum ri x Fi ('x' = vector cross-product).

The positions ri (vectors) are the measurement points (force transducers
i = 1, 2, 3, 4) measured from the center point of the force plate (origo
of the coordinate system). Now it is a mathematical fact that this
torque can by written as

M = R x F + T

[physically: the total actions of the forces Fi acting at ri can be
counter balanced by the total force F acting at R plus a couple T]

where F = sum Fi is the total force. R is not uniquely determined by
this equation, since for R --> R + a F (a = scalar) we have, R x F = (R
+ a F) x F. However, if we demand that R be situated in the horizontal
plane, R = (X, Y, 0) -- which can be achieved by adding a suitable (a F)
to R with 'a' properly chosen -- then R is uniquely defined (= COP).
Indeed, if, as earlier assumed, T = (0, 0, T_z), then we get

M - (0, 0, T_z) = (X, Y, 0) x F = (Y*F_z, -X*F_z , X*F_y),

which implies that,

X*F_z = - M_y,
Y*F_z = M_x.

The (X, Y) determined in this way is the COP-point. This agrees with the
earlier result when we insert the expression for M in the previous
equations:

X = (W/2)*(F2_z + F3_z - F1_z - F4_z)/F_z,
etc.

Thus the COP is, as the name suggests (center of pressure), the
balancing point of the vertical forces:

R_COP_j = (sum ri_j *Fi_z)/(sum Fi_z) [analogous to the formula for
calculating the center of mass].

This is the original definition of COP. As is obvious one needs only the
z-components of force in order to calculate COP. Therefore it is
possible to use a simpler force plate that employs z-axis force
transducers -- se eg http://www.saunalahti.fi/borgbros/bisoni/bal_rapp.pdf