View Full Version : Summary: Center of buoyancy problem

02-06-1995, 07:12 AM
Dear Biomch-l colleagues:

On January 25, Scott McLean and I posted a question about center of buoyancy
and center of mass of a floating swimmer. We received many interesting
replies. Thanks to all those who responded. I have posted the individual
responses below for your reference.

Many people came up with the correct answer which is as follows:

If you support a person (whose feet would otherwise sink) with a strap
somewhere on the legs, the upward force from the strap assists the buoyant
force in keeping the person at the surface. However, the placement of the
strap (ankles vs knees, for example) will determine how much of the body
floats above the water. The larger the strap force (R), the higher the body
will rise above the water and the less the buoyant force. I had mistakenly
thought (at first) that it was possible to support the body anywhere on the
legs and have the same amount of body above water. It is pretty simple now
that I think about it.

We still have a dilemma, however, on a validation experiment we ve done with
a completely submerged weighted cylinder held by two straps with added
weights that I will post separately. Thanks again for all your insight!



=============End of responses===============

Richard N. Hinrichs, Ph.D.
Dept. of Exercise Science
Arizona State University USA
(1) 602-965-1624 (office)
(1) 602-955-8108 (fax)
Hinrichs@asu.edu (email)

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From: Hinrichs, Rick
To: biomch-l
Subject: Center of buoyancy problem
Date: Wednesday, January 25, 1995 6:18PM

Dear Colleagues:

I have a dilemma that I hope someone out there can help me with.
Scott McLean and I are trying to measure the center of buoyancy of a
human body at the surface of the water (with application to floating
and swimming). First we measure the center of mass (CM)location for
our subjects using a reaction board. Then we place them in the pool
in a either a prone or a supine floating position. For most people,
their feet tend to sink in this position, indicating that the buoyant
force (B) is less than body weight (W) and that the center of
buoyancy (CB) is cranial to the CM. So we place a strap around their
ankles and measure how much force (R) is required to keep the feet
up. A free body diagram is shown below (as best I can draw it using
7-bit ascii characters):

CB \|/------x-------|
Head ___________._____.___________________ Feet
^ CM ^
| |
|--d--| |

In the above FBD, we know W, R, and x. We would like to know d (the
distance between the CM and CB). Using equations of static
equilibrium for translation (B=W-R) and rotation (Bd=Rx), we can
arrive at B and then d (no big deal).

However, if we move the point of application of the supporting force
(R) higher up the leg, keeping the body position the same, we
decrease the distance x. Everything else about the body SHOULD have
stayed the same, i.e., the same body position in the water, same body
weight, same buoyant force, and same locations of the CM and CB. To
satisfy the rotational equilibrium condition, Bd=Rx regardless of the
value of x. So if x gets smaller then R must increase to produce the
same torque. However, [AND THIS IS WHERE I AM HAVING PROBLEMS], if R
gets larger, then the only way to satisfy the translational
equilibrium condition (B=W-R) is for B to decrease. This doesn't
make sense, however, because the buoyant force (B) is (by definition)
the weight of the displaced water which has not changed. If W, B,
and the locations of CB and CM have not changed, then we have an
impossible situation. We cannot satisfy both the rotational and
translational equilibrium equations. CAN ANYONE OUT THERE SEE THE

NOTE: We have tried this with the body totally submerged and on the
surface and for different amounts of air in the lungs. We have also
put supports at two locations (with a strap around the chest as well
as the ankles) and have come up with the same dilemma; the FBD is
only slightly more complicated).

Any help you can provide will be appreciated. Thanks in advance. I
will post a summary of responses.


Richard N. Hinrichs, Ph.D.
Dept. of Exercise Science
Arizona State University USA
(1) 602-965-1624 (office)
(1) 602-955-8108 (fax)
Hinrichs@asu.edu (email)

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Date: Wed, 1 Feb 1995 16:45:32 -0600
From: "Leonard G. Caillouet"
Subject: Re: Center of buoyancy problem

On Wed, 25 Jan 1995, Hinrichs, Rick wrote:

> gets larger, then the only way to satisfy the translational
> equilibrium condition (B=W-R) is for B to decrease. This doesn't

I think that here (below) is where your evaluation is faulty. You are
correct about the amount of bouyant force not changing. However, when the
floats differently with the force R applied the center of bouyancy has
moved, so d is what changes. You have assumed that the position of CB is
the same under two diferent rotational conditions.

The location of CB has to change if the forces creating the rotation of
the body are changed.

Hope this has been a help.

Leonard G Caillouet
L.S.U. Kinesiology
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Date: Thu, 26 Jan 1995 12:12:26 +0000 (GMT)
From: e.e.g.hekman@wb.utwente.nl (Edsko Hekman)
Subject: Center of buoyancy problem


The idea of a 'center of buoyancy' is new to me. I am assuming this
is the location where the resultant force of the distributed load due to
the water pressure on the skin can be thought.

I am assuming that what you are doing is putting the body in a certain
position with a given x and measuring R. I am also assuming constant density
of the water in the immediate surrounding of the body (otherwise a 'center of
buoyancy' would be dependant on the rotational position of the body in the
water), and suspension of the body in a 'very deep' pool, in which the density
of the water increases with depth.

I would say that since body weight does not change, a change in the position
the force R would have to be reflected in a change in B and/or its position. I
see two options for equilibrium :

1) The body is totally submerged

In this case, assuming that in the immediate surrounding of the body the
density of the water is constant, the body can theoretically rotate freely
around all axes without a change in forces.
Changing x results in a change in R (moment of force equilibrium). This in
turn means that the body will rise or go down until B (which varies with dept)
has reached the appropriate value.
Actually I think this is somewhat of an academic experiment since variation of
density of the water is small, which would make it hard to reach equilibrium
within the range of depth allowed in a regular pool.

2) The body is not totally submerged

Changing the distance x will result in either of three effects :

- The body will sink (see option 1)

- The force is kept constant. Equilibrium of forces means that B does not
change (the CM of the body will not go up or down), while equilibrium of
moments means that the moment arm has to change as well (increase of x
results in increase of d : the body rotates such that the unsubmerged part
of the body shifts towards the feet)

- The force is not kept constant. The effects on d and B depend on various
things, such as body contour, mass distribution within the body, rotational
position of the body etc.

What is seems to boil down to is that a CB can only be determined with a body
in equilibrium in complete submersion, under the assumption that the density
of the water in the immediate surrounding of the body is constant. As soon
as the body is not completely submerged any more, the location of the center
of buoyancy depends on the position of the body, how much of it sticks out of
the water and where.


Edsko Hekman
Twente University
Fac. Bio-Mechanical Engineering (WB-BW)
Postbus 217
7500AE Enschede
The Netherlands

e-mail e.e.g.hekman@wb.utwente.nl
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Date: Wed, 25 Jan 1995 22:04:35 -0700
From: l.abraham@mail.utexas.edu (Larry Abraham)
Subject: RE: Center of buoyancy problem


Here are some quick thoughts on a first pass through the problem. Maybe
they'll help. I find myself wondering what the real data look like....

Note that the fact that the feet sink does not necessarily mean that B is
less than W. My feet sink if I start in a horizontal position but they
rise if I start in a vertical position. This is (I assume) because my CB
is not only cranial to my CG but also forward (consider the relative
densities of the two halves of the body divided along the mid-coronal
plane, lungs & viscera vs vertebrae & back muscles, etc.). The upward
force R in this case is necessary to balance the moment(s). However since
it is a force it also acts to raise the body, which in turn reduces B, just
as you calculated it would. Remember that B can be expressed not only
interms of W-R, but also in terms of the volume of water displaced. Thus
your assumptions that body position is unchanged and that volume of water
displaced is constant are both faulty. The body is somewhat higher (though
still horizontal) and there is less water displaced. Of course the
situation seeks an equilibrium with the only two constants being W and d,
so for instance the moment generated by B varies as x changes because the
value of B also changes. My prediction would be that if R were moved
extremely close to CM, then B would approach zero as R increased to equal W
as the body is lifted out of the water.

I'm puzzled though by your statement that you tried this with the subject
completely underwater. In this situation B will be constant. Did you
actually determine the equilibrium level of floating, perhaps varying this
by adding weights? I don't think this experiment can be done well without
air, as it could easily take a minute or two for the body to rotate into a
stable floating position. In my experience a body which "floats" in
equilibrium suspended between the surface and the bottom of the pool is
extremely sensitive to any applied force. This experiment could most
easily be performed with a person for whom W is greater than B(submerged).
In this case the body will rest on the bottom, experiencing a force R from
the bottom equal to W-B and applied at an average value of x (considering
that it may be distributed among several points of contact) such that (B
times d) equals (R times x). If you construct an experiment such that R
is only applied at a single fixed point (using a rope, etc.) then I predict
that the body will only be stable in a suspended position (between surface
and bottom) with a single value of R, such that R = W-B, and a single value
of x, such that Rx = Bd. Please don't tell me you actually performed this
experiment with different results. But if you did, I guess I'd like to
know the exact values you found.

Let me know if this makes sense or if you see a clear error in my logic.

Larry Abraham, EdD
Kinesiology & Health Education
The University of Texas at Austin
Austin, TX 78712 USA
(512)471-1273 FAX (512)471-8914
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Date: Thu, 26 Jan 1995 10:39:22 -0500
From: yhc3@cornell.edu (Young Hui Chang)
Subject: RE: Center of buoyancy problem

This is an interesting problem. Here is my crack at it.

By adding the strap under the ankles, you are trying to resist the natural
tendency for the floating body to rotate. In the first simple case, all of
the equations you described work out with the center of rotation (c.o.r.)
being at the center of mass (c.o.m.). As you move the strap force (R) in,
are you sure that the center of rotation stays fixed? In fact, how do you
know that the center of rotation is at the c.o.m. to begin with (empirical

Perhaps as you move R closer towards the c.o.m., the c.o.r. may move toward
the head allowing body-weight (W) to act upon some moment arm, and the
buoyancy (B) to have a reduced moment arm. Then, W, R, and B stay constant
with only the moment arms adjusting to a "dynamic" center of rotation to
balance the rotational equilibrium equation.

I would love to hear what others have responded. Good luck.

Young Hui

================================================== ===========================
Young Hui Chang C C
Department of Anatomy C e-mail: yhc3@cornell.edu
College of Veterinary Medicine C U U phone: 607-253-3551
Ithaca, NY 14853-6401 C U C U fax: 607-253-3541
================================================== ===========================


Date: Thu, 26 Jan 1995 08:11:36 -0800
From: dahawkins@ucdavis.edu (David Hawkins)
Subject: RE: Center of buoyancy problem

Your FBD indicates that there is only one unique solution for static
equilibrium for that system. You know what R has to be to satisfy
translation equilibrium, and given that R you know the distance x that R
must be applied. If both these conditions are not satisfied then the body
will rotate without translating or translate without rotating or both. I
think that your analysis was correct.

David Hawkins, Ph.D.
Biomechanics Lab
UC Davis

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Date: Thu, 26 Jan 1995 18:08:23 +0100
From: bao@auto.ucl.ac.be
Subject: RE: Center of buoyancy problem

Dear Dr. Richard N. Hinrichs:

I saw your interesting problem in Biomch-L. If the real problem can be
modelled as your drawing and B,CB,W and CM are all fixed, then there will
be only one solution for R and x. however, CM and CB could be changing
since they are function of, could be sensitive to, body shape.

Huali Bao

__________________________________________________ _________________
Huali BAO
CESAME Tel: + 32 (10) 47 22 64
Avenue Georges Lemaitre, 4-6 Fax: + 32 (10) 47 21 80
B-1348 LOUVAIN-LA-NEUVE (Belgium) e.mail:bao@auto.ucl.ac.be
__________________________________________________ _________________

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Date: Thu, 26 Jan 1995 09:28:19 -0800 (PST)
From: Jack Tigh Dennerlein
Subject: RE: Center of buoyancy problem

Dear Rick:

Okay this is stab in the dark, but maybe it will give you something else
to think about.

What strikes me is your assumption that B does not change might be the
error in your logic. Actually it's more of the assumption that the
bouyance force acts at a single point, rather than a distribution of
forces along the body. I haven't thought this all the way through, but
by changing x, you are moving to different points along the bouyance
force distribution. If the distribution is uniform, your logic would be
correct (I think), but I think it's safe to say it's not uniform. I
suggest you pick a not uniform, but simple, distribution and re-do your
equations to check your assumptions. My gut says there is something going
on to the right of x that might be decreasing B or d.

I hope my morning off the cuff comments help out. Good luck.

Jack Tigh Dennerlein

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Date: Thu, 26 Jan 1995 12:01:22 -0600 (CST)
From: John Barden
Subject: Reply to C of B problem

Dear Dr. Hinrichs:

An interesting posting and problem. My first reaction and guess is that
the error in logic lies in the assumption that the C of B does not change
when you move the location of the supporting force (R) up the leg.
Possibly the amount of water that is displaced is affected in some small
way, which affects the location of the C of B and therefore of d (which is
probably fairly small anyway)? I haven't thought it through a great deal
but at first glance it seems that this would eliminate all of the diffic-
ulties that follow and meet both the translational and rotational equil-
ibrium conditions.

I hope this helps. Good luck.

John Barden
Sport Science Coordinator
Faculty of Physical Activity Studies
University of Regina
Regina, Saskatchewan


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Date: Thu, 26 Jan 1995 11:02:36 -0800
From: mmslavin@ucdavis.edu (michael m slavin)

Hi Rick. I'm a grad student in Biomechanics at UC-Davis. Your problem
interests me because I started swimming seriously last year (ie learning as
an adult). I find that although I'm a good runner, I am an awful swimmer,
though I'm fit and work hard at it. One possibility for me is that the
distance d is large. I have even experimented a little with ventilating my
lungs about a lower volume as I swim, to reduce the air buoyancy
contribution to B of points cranial to the CB, with hopes of reducing d.

As x decreases, R increases (rotational static equilib). B and CB are
constants. Thus, B+R must increase. Since W=B+R, W must increase or there
is some sort of translation or rotation. Underwater, W is constant.
However, in the horizontal position at the water surface, the body needs to
rise only slightly to move a significant % of the BM above the water line,
with a relatively dramatic increase in W and a slightly lesser absolute
decrease (in most cases) in B, with a net increase in the force acting
downward, W-B.

If the strap at the ankles were moved cranially on a COMPLETELY submerged
subject, R WOUOLD increase. In this Case A, R would keep the subject
horizontal (no rotation), but he/she would move up to the surface where the
above mechanism (increase in W-B) would counteract. At some point along
the legs, R solves both equations simultaneously, i.e. W+B+R=0 and Rd+Bx=0.
In this Case B, there is no translation or rotation. In Case C, distal to
this point, R applied upward prevents rotation due to the couple created by
B and W, but the CM sinks (i.e. translates), indefinitely (until
atmospheric pressure changes or he/she hits the bottom of the pool!).

That's my 2 cents. I'll be interested in the postings!


Michael Slavin Tel: (916) 752-9048
Biomechanics E-mail: mmslavin@ucdavis.edu
Hickey Gym, Rm. 264
Univ. of California-Davis
Davis, CA 95616

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Date: Thu, 26 Jan 1995 13:20:54 -0400 (EDT)
From: Brian Davis
Subject: RE: Center of buoyancy problem

Hi Rick

I enjoyed reading your problem. It reminded me of Heisenberg's problem of
knowing either velocity or mass, but not both.

Anyway, in your case, I think the buoyancy force does change as you
change x. I think if you impose a vertical (upwards) force on a person who
is floating in the water, he/she will become less submerged. In fact, in
the limiting case where the force is applied at the CM, you wouldn't need
any buoyancy force.

Regards, Brian

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Date: Thu, 02 Feb 1995 13:43:34 -0400 (EDT)
From: PKRONICK@arserrc.gov
Subject: buoyancy: correction

Your assumption of invariant B for a floating body is wrong. If the body is
fully im
lly immersed, you cannot vary x and still have it horizontal. If floating, you
can, but B changes as the body floats higher or lower in the water. Sorry
previous message.

Dr. Paul Kronick e-mail: pkronick@arserrc.gov
USDA-ARS-ERRC phone: 215 233 6506
600 E. Mermaid Lane fax: 215-233-6795, -6559
Philadelphia, PA 19118, USA home: 215 649 3382

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Date: Thu, 26 Jan 1995 14:16:48 -0500 (EST)
From: gagnomi@ERE.UMontreal.CA (Gagnon Micheline)
Subject: RE: Center of buoyancy problem

Dear Dr.Hinrichs:
-- Iam referring you to our work in this topic.

GAGNON,M.and MONTPETIT,R.(1981)Technical development for the measurement
of the center of volume in the human body.J.of
Micheline Gagnon
Universite de Montreal
(514) 343-7847

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Date: Thu, 26 Jan 1995 14:07:40 -0500
From: dfm106@psu.edu (Daniel Marshall)
Subject: RE: Center of buoyancy problem

> W
> |
> CB \|/------x-------|
> Head ___________._____.___________________ Feet
> ^ CM ^
> | |
> |--d--| |
> B R
>Here are a few thoughts about this problem
Look at it from the other side such that B=W-R. Since we will assume that
both B and W are constants, R must be a constant as well to assure no
rotation. What does this mean? Using the moment equation Bd=Rx, if x
changes, d must as well and your center of bouyancy is not constant as you
believed. Taking the moment about the R,
B(d+x)=Wx; or d=x(W-B)/B.
So the CB is dependent on where R is located. If R is moved to the CM, ie
x=0, the position of CM is the also the position of CB.
This seems odd that CB is dependent on x at first but bouyancy is truly just
the combination of forces acting up on the entire body.

I hope this might help out. I'm not ruling out a flaw in my reasoning so
I'll be interested in seeing what responses you recieve.
s s
Be Well, ppppp s s u u
p p s u u
Dan Marshall ppppp sssss u u
p s u u
p s s uuuuu
s s

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Date: Thu, 26 Jan 1995 16:43:41 -0700 (MST)
From: p.sinclair@cchs.su.EDU.AU
Subject: RE: Center of buoyancy problem

G'day Rick,

Good question. I like a good problem like this.

As you say, if x is decreased the person can no longer satisfy both
rotational and translational equilibrium.

If x is reduced and the body remains horizontal (rotational equilibrium)
then, as you say, R must be greater to maintain rotational equilibrium. The
person would no longer be in translational equilibrium and would rise
slightly in the water. This would reduce the size of the bouyant force and
therefore reduce the effect of the change in R.

If the person remained at the same height in the water (translational
equilibrium, R remains the same) then the torque from the reaction force
would be less and the legs would sink slightly. If this resulted in the head
coming out of the water (unlikely), this would reduce the bouyant force and
therefore reduce the effect of changing R.

I don't know if I have explained this very well. To sum up, changing the
position of the reaction force cannot result in exactly the same body
position. The force R would probably be very slightly larger which would
raise the whole body up in the water slightly. This would reduce the bouyant
force and therefore minimise any difference you may see. The person would
settle in a new equilibrium position that may be only very slightly
different to the original position.

I hope this helps. Happy Australia day for yesterday (which is why I didn't
reply yesterday). It may even still be Australia day in Arizona. I haven't

BTW. I assume you have seen the article by Page in Journal of Human Movement
Studies (1: 190-198, 1975) which uses the same technique as you. This is the
only previous measurement of centre of bouyancy that I know of.


Peter Sinclair

Division of Biomechanics E-mail: p.sinclair@cchs.su.edu.au
Faculty of Health Sciences Phone: (02) 646 6137
The University of Sydney Fax: (02) 646 6520

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Date: Thu, 26 Jan 1995 18:31:23 -0700 (MST)
From: p.sinclair@cchs.su.EDU.AU
Subject: RE: Center of buoyancy problem

Hello again,

I've been working on the same thing for a few weeks here (it is summer
holidays for students) and perhaps I've been a bit bored. I thought I would
do a bit of crude modelling to see how much change in body position you
would get for the person to remain in equilibrium. If you've already worked
out your problem satisfactorily then just ignore this.

> W
> |
> CB \|/------x-------|
> Head ___________._____.___________________ Feet
> ^ CM ^
> | |
> |--d--| |
> B R

eg lets model the body as a box 180 * 20 * 20 cm with a density of 1.02 g/cm3
volume of box is 72000 cm3
mass of box is 73.44 kg
weight of box is 719.712 N

lets assume that d is 0.1 m and, with the reaction force holding the base
up, the box is 90% submerged (the box can't be 100% submerged and still
breathe) and density of water is 1.00 g/cm3

The volume of water displaced is 90% * 72000 cm3 (64800 cm3)
therefore 10% of the box sticks up in the water (ie 2 cm)

The bouyant force is 635.04 N
Therefore, R is 84.672 N and x is .75 m

If you make x smaller then R must get larger to maintain rotational
equilibrium. If R is larger, then the body will rise out of the water
further and the bouyant force will get less.

B==VS*density/1000*9.8 (VS is volume submerged)

if these equations are solved simultaneusly

R -> W - (W*x)/(d + x),
B -> (W*x)/(d + x),
VS -> (1000*W*x)/(density*9.8*(d + x))

eg if we reduce x from .75m to .65m and the location of the centre of
bouyancy doesn't change (which it wouldn't with a box)

then R = 95.9616,
B = 623.75,
VS = 62400 (86.67% of body volume)

now the person floats with 2.67 cm sticking out of the water instead of 2 cm.

Would you notice this difference?

Note that a person is not a symmetrical box. As the body rose out of the
water more volume would become exposed at the superior end (provided they
remained horizontal). This would reduce d and further reduce the amount the
body would have to rise in the water to remain in equilibrium. (I was too
lazy to model the body as an asymetric shape)

I guess what I am trying to say here is that as you move the reaction force
closer to the centre of mass, the magnitude of this force will rise. This
will result in the body rising slightly and thus reducing the size of the
bouyant force. The error in your reasoning was in assuming that the bouyant
force would remain constant.

I hope this helps. I am going to go back to preparing lecure notes now.


Peter Sinclair

Division of Biomechanics E-mail: p.sinclair@cchs.su.edu.au
Faculty of Health Sciences Phone: (02) 646 6137
The University of Sydney Fax: (02) 646 6520

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Date: Thu, 26 Jan 1995 09:13:29 +0000
From: Stefan Schreiber
Subject: RE: Center of buoyancy problem

Hi Rick,

there is no error in your logic. The magnitudes CB and CM do not change,
neither do their location. Since both the rotational and the translational
equilibrium equations need to be satisfied, there is only ONE possible
for R and x. The thing is that you are looking at an UNCONSTRAINED system.

In a contrained system, say a beam suspended by two bearings to the left and
the right, the magnitude of the bearing forces change with magnitude F and
location x of a given force i. e. there is a response to the load. In your
system consider R as the 'load'. Now where should a 'response' come from?

There is a little more to say on statics of a rigid body. If you wish I can
elaborate a bit more.

So much for now...


__________________________________________________ ___________________________
Stefan Schreiber Technical University of Darmstadt
Institute for Mechanics I
voice: +49 6151 16 2286 Hochschulstrasse 1
Fax: +49 6151 16 6869 D-64289 Darmstadt, Germany
e-mail: stefan@mech1host.mechanik.th-darmstadt.de

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Date: Thu, 26 Jan 1995 16:30:39 +0000
From: nmadsen@Eng.Auburn.EDU (nmadsen)
Subject: RE: Center of buoyancy problem

Saw your note regarding the center of buoyancy
problem. Intrigued me and wondered what kind
of responses you got - fundamentally I believe
the answer to be that if you really implemented
the support system the way you described, there
is then actually one position where you could
attach the support and truly maintain the
floater in a horizontal position. If you
moved the support, the angle assumed by the floater
in the water would change somewhat, changing both
the force and its location. This is easier to
make clear with a picture but as you noted
that is tough over the internet. Consider a rigid rod
floating in the water with a mass center not
necessarily at its geometric centroid and I think
you will see my point. Is my
description clear? Feel free to contact me if
you have any questions. Also say hi to Rick
Hervig for me - we were high school buddies
and he is an ion microprobe nut over in geology.
Also you may remember me from long ago meetings.
Thanks, Nels.

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Date: Fri, 27 Jan 1995 03:29:07 +0100
From: deleva@risccics.ing.uniroma1.it
Subject: RE: Center of buoyancy problem

Dear Rick,

your problem reminded me the ones I had to solve in a class
given by your friend J. Dapena, that I fullhearthedly enjoied to
attend, based on a book by B. Hopper ("The mechanics of human
movement", if I remember well).

You wrote:

> ....we decrease the distance x. Everything else about the body
> SHOULD have stayed the same, i.e., [1] the same body position in
> the water, [2] same body weight, [3] same buoyant force, and
> [4] same locations of the CM and [5] CB....

Well, you almost have given your solution there. You were
perfectly right when you wrote "SHOULD" (with capital letters).
In fact, since the support force R will have to increase to
maintain an adequate torque T = R * x for obtaining static
rotational equilibrium, it is evident that the buoyant force [3]
WILL have to decrease. How will that happen? First of all,
no way you can take the body exactly in the same position in the
water [1]. More precisely, I mean that the body posture (or attitude)
can with reasonable approximation stay the same, assuming the
swimmer can stay perfectly rigid, but the orientation of the
body and position of its CM relative to the water surface will
Starting from your initial static equilibrium
condition (the one you showed in your picture), if you move the
support force R and decrease the arm x, you will need to increase
the magnitude of R, as you correctly stated. Because of this,
the linear equilibrium will be broken, and the body will receive
a net upward resultant force. The consequent slight upward
acceleration will in turn move the body in a higher
position relative to the water surface.
At the end, a new condition of static equilibrium will
be obtained, and a smaller part of the total body volume will be
submerged. This is why the bouyant force will be smaller than
initially. Notice that this also means that the torque by the
bouyant force, T = B * d, will decrease!
Thus, contrary to what you wrote, not even the torque
T = R * x that you will need to apply to obtain the new static
rotational equilibrium will be the same as before. It will have to
just reach a value equal (in magnitude) and opposite (in direction)
to B * d, and B * d will be smaller, as we have seen.
It is also likely that, since the body position relative to
the water will change, the Center of Bouyancy [5] will move,
and d will then change, in a way that I am not able to predict,
and that depends on the variable proportion by which the submerged
volume of the different parts of the body will change.
Also, it is quite likely that, when you move R, increase
its magnitude, and decrease x, you not only will obtain, for a
short time, a condition on linear disequilibrium (net upward
force), but also a condition of rotational disequilibrium
(net torque). The temporary net torque might be clockwise (CW) or
counterclockwise (CCW). For instance, if the net torque will be CW,
the upper part of the body will be lifted more than the lower
part. As a consequence, the submerged volume of the
upper part of the body will possibly decrease more than that of
the lower part (the feet might even move downward, due to the CW
rotation, and the point of application of the support force R
might even stay in the same position, working as a fixed rotation
axis around which the body will rotate, and giving you the illusion
that the body is not moving, while IT IS).
The small changes in submerged volume will bring about
a small change in the final position of the CB after the new
equilibrium condition is reached. Thus, this is another factor
that will make the distance d (between CM and CB) not constant.
Although I am not completely sure, I believe that
if the body will slightly rotate CW before reaching the new
equilibrium (as in the above example), the distance d will
decrease (CB will move caudally), and viceversa. This is what
would happen if the body had a simple shape as that in the
example below, where two spheres having different density and
volume are firmly attached to the ends of a thin rigid rod
(sorry if the spheres look like my bin bags!).

/ * * * \ ###
/ * * * * * \ /#####\
[ * * * * * * ]-----@-------[#######]
\ * * * * * / CM \#####/
\ * * * / ###

smaller density (1.0)
LARGER VOLUME smaller volume

Use this drawing to analyze your problem, and you will
possibly find it easier to find the answers you need.

Concluding, only [2] the body weight, and [4] the position
of the CM (along the body long axis), will always for sure
stay constant (assuming that the body is rigid). Notice that
there's always a solution to your problem for all values of
x ranging from 0 to the distance CM-acropodion (tip of foot).
When x = 0, a possible solution is (to have horizontal body)
R = - W (weight), and submerged volume = 0 (body lifted over
the water surface).

With regards,

__________ _________ ___________~___ ________ _________________~___
/ ~ ~ ~ ~ \
/______________~______~__________ _______~_____~______________~_____~_____\
| Paolo de Leva ~ \ Tel.+ FAX: (39-6) 575.40.81 |
| Istituto Superiore di Educazione Fisica > other FAX: (39-6) 361.30.65 |
| Biomechanics Lab / |
| Via di Villa Pepoli, 4 < INTERNET e-mail address: |
| 00153 ROME - ITALY \ deLEVA@RISCcics.Ing.UniRoma1.IT |
|_____________________~________~__________________ __________________ _____|
Panta rei :-)

================================================== =========================

Date: Fri, 27 Jan 1995 11:00:07 +0000 (KST)
From: y-hkwon@kssisun.kssi.re.kr (Young-Hoo Kwon)

Dear Rick:

Happy New Year!

Here are some of my thoughts about the buoyancy problem. They are purely

1. d(object) > d(water). where, d = density.

| R = |W-B|
| |

The object (for example, a rod) will be submerged in the water and B is
constant. You can find an ideal force application point which gives you both
translational and rotational equilibrium. Let's say the distance between CM
and this point be x(ideal) and the force required be R, where R = |W-B|. The
ideal force application point can locate either within the object or outside
depending on d(object) and the distance between CM and CB. Since CM and CB
are not identical this object tends to rotate during the downward motion to
the bottom of the pool. When it hits the bottom of the pool there will occur
two different cases.
If the ideal point lies outside the object, the torque created by R at the
point of contact with the bottom of the pool is not enough to keep the
rotational equilibrium. So the object rotates until it stands upright and
achieves the equilibrium state.
If the ideal point locates within the object, the object will get the
equilibrium state when the the vertical line drawn from the point of contact
passes through the ideal point.

2. d(object) < d(water)

Since B(max) > W, the object will float and B varies depending on the
situation. Since you are exerting force to keep the equilibrium state, R
should be the reaction force to the action the object exerted to you. In
other words, R should be equal to |W-B| in any case.
If you exert R upward to any point in the heavier side of the object or
downward to any point in the lighter side, the object will eventually get
into the equilibrium state by changing the volume merged in the water. B and
y change depending on the situation.
Going back to your problem, the volume merged in the water changes when you
decrease x. You might not notice because the change is so small. If you push
down the other side of the object to keep the object in equilibrium. The
volume merged in the water will change more.
From the figure above,

Rx = By where, R, B, x and y are scalar and positive. (1)
--> (W-B)x = By (2)
--> Wx = B(x+y) (3)

If you change x, B and y changes until anyone of the equations above
The extent of change in y and B depends on the shape and mass distribution of
the object.
From equation (3):

B = Wx/(x+y) = W/(1+y/x) (5)
--> R = W-B = Wy/(x+y) = W/(1+x/y) (6)

R is function of both x and y. From eq.(5), if y B). Place it in the
water oriented horizontally and it will also rotate as it sinks until the
the cylinder is oriented vertically. Now try to "catch" the falling and
rotating cylinder on a single knife-edge so that it will stay at a
constant depth. I think that there is only one such point, the point
such that R x + W (CMx) + B (CBx) = 0, where x, CMx, and CBx are measured
from the same end of the cylinder, for example. The value of x for
equilibrium can be obtained by taking R = W - B (positive since the
cylinder sinks) and solving for x.

In your example, you suggested moving the point of application of R. If
you do so, the system is no longer in equilibrium, because the magnitude
of R will still be determined by W-B, and the cylinder will begin to
rotate, and fall off of your fulcrum.

The W>B condition is necessary for you to have a vertically directed
reaction force under static conditions. If the body is floating, or
would float after some rotation, the R for a truly static case will be
zero (or negative).

I may have made some errors in this analysis -- I'd be glad to hear
corrections or alternative ideas. I found your problem interesting and
I'd like to hear how you've solved it.

Matt Reed
Senior Research Associate
University of Michigan
Transportation Research Institute

================================================== ========================

Date: Thu, 02 Feb 1995 10:52:03 -0500
From: Matt
Subject: Buoyancy again

I thought of one other useful illustration. Imagine the cylinder from
my previous message is not in water but rather in air. The fulcrum
would be located (approximately) under the CM. I say approximately,
because displacement of air in a gravitational field produces buoyancy,
but we normally neglect it because it's small relative to the weight of
the objects we're supporting. (But we don't neglect it in the case of
gases less dense than air at comparable pressure, for example). Now
increase the density of the medium surrounding the balancing cylinder.
As the density of the medium increases, we'll need to displace the
fulcrum to maintain equilibrium, because the buoyancy force will
increase (weight of the displaced medium). (The cylinder's CM is not
coincident with its CB). If the medium becomes sufficiently dense, B
becomes greater than W, the cylinder will float off of the fulcrum.

Matt Reed
Senior Research Associate
University of Michigan
Transportation Research Institute

================================================== =========================

Date: Thu, 26 Jan 1995 07:47:31 -0500 (EST)
From: mark@jazz.me.psu.edu (Mark Nicosia)
Subject: buoyancy response

Greetings Dr. Hinnrichs,

If the body were perfectly rigid, (i.e. if was a straight
line from head to toe, and could not bend), the diagram you
show would admit only one d for a given set of conditions,
i.e. d=Rx/B, where B=W-R. If x changes, and all else remains the same,
the body will "go unstable", and start to rotate. However,
the human body would do things to compensate (i.e. bend at
the knees, flex muscles) which would cause the situation
to deviate from a rigid rod, making the equations not
quite right for the situation.

Another thought is that the center of buoyancy may actually change
somewhat if, for example, the leg muscles are contracted (by
changing the density distribution in the body).

I'm not sure if these are the answer to your problem,
just my first impressions.
Hope this helps, and good luck with your problem.

Best Regards,

Mark Nicosia
Ph.D. student
Department of Mechanical Engineering
Penn State University

================================================== ========================

Date: Thu, 26 Jan 1995 07:58:15 -0820
From: WAITEL%MHS@mhs.rose-hulman.edu
Subject: RE: Center of buoyancy problem


At a glance, it looks like the problem is as follows. The statics
problem that you have set up looks correct. However, I believe that the
buoyant force which you are trying to measure is decreased by the
reaction force from the strap (B=f(R)). Because of that, when you change
the distance from the CG to the strap, the reaction force changes and
the actual buoyant force changes.

I think that you probably want to measure the maximum buoyant force on
the body (i.e. at it's lowest position in the water.) In your
measurement, the reaction force from the strap keeps the body from
sinking to it's lowest position. The amount of water displaced is less
as the reaction force is increased.

Sorry I don't have more time to give you a better explanation.


Lee Waite
Associate Professor
Mechanical and Biomedical Engineering
Rose-Hulman Institute of Technology
Terre Haute, IN USA

================================================== ========================
Date: Thu, 26 Jan 1995 08:47:21 -0500
From: dapena@valeri.hper.indiana.edu
Subject: RE: Center of buoyancy problem


As the strap gets closer to the c.m., R will get larger in order to
provide the same torque as dXB. However, the buoyancy force will not stay
the same as before: The body will lie higher up in the water, there will be
a smaller submerged volume, and the buoyancy force will become smaller.
That is how you will still have a net zero vertical force.

I am not sure that your problem HAS a solution. In effect, with the
strap you are forcing a horizontal position of the body. Since this is done
through the application of an upward vertical force with the strap, it means
that in all of the cases in which you use the strap the body will have a
smaller amount of volume submerged than would be the case without a strap at
all (i.e., than would be the case with the body almost vertical and with
only the top of the head and forehead sticking above water level). When the
strap is far from the c.m., the body rides lowest on the water in its
horizontal position; when the strap is closest to the c.m., the body rides
highest on the water in its horizontal position.

In either one of these two extreme cases, the LOCATION of the center
of buoyancy itself will also be somewhat different, because the "imprint"
made by the body on the water will be different in both cases --it will
depend on WHERE there is more body volume underwater in each case.
Therefore, the center of buoyancy will be in somewhat different locations,
depending on whether you are interested in the "riding low" position or the
"riding high" position (or any intermediate one). It is possible, however,
that the differences in the location of the center of buoyancy between the
two extremes will be negligible. I'd say, try it, and keep your fingers

Jesus Dapena
Department of Kinesiology
Indiana University
Bloomington, IN 47405, USA
1-812-855-8407 (office phone)
dapena@valeri.hper.indiana.edu (email)

================================================== ========================

Date: Thu, 26 Jan 1995 10:23:00 -0400 (AST)
From: Jianyu Cheng
Subject: RE: Center of buoyancy problem

Hi, Rick,
I think the solution of the two equilibrium equations,
i.e., supporting force R and the distance x
are unique for the given body. Changing R results in
translational imbalance. For the determined R=W-B, x
is then unique to keep rotational equilibrium.

Jian-Yu Cheng
St. Francis Xavier University, Canada

================================================== ========================

Date: Thu, 26 Jan 1995 10:14:51 -0500 (EST)
From: Nat Ordway - Orthopedics SUNY Syracuse
Subject: RE: Center of buoyancy problem

Hello Dr. Hinrichs,

I am responding to your question regarding the bouyant force. From your
description of the problem I don't think that the bouyant force can stay
the same if you shorten x and increase R. Lets take a simpler case of a person
just floating in the water. Their bouyant force must equal their body weight
in order to float. Now if a rope is tied to the person so they will be lifted
straight out of the water, then the bouyant force doesn't decrease BUT its
contribution does as the person is lifted out of the water. In other words the
bouyant force will have a maximum value for each individual but that doesn't
mean that the person's bouyant force is always at the maximum value. I think
dilemma in your problem has to do with the volume of displaced water. Are
you positive that the displaced volume of water hasn't changed? The specific
weight of water is 9800N/m^3 so it would seem that the change in the
amount of displaced water would be very small. How do you measure the
volume of displaced water? I hope this helps. I'll be interested to see if
this works out. Good luck.

Nat Ordway
Dept of Orthopedics Phone: (315) 464-6462
SUNY Health Science Center Fax: (315) 464-6470
Syracuse, New York USA Email: ordwayn@vax.cs.hscsyr.edu

The weather forecast for Super Bowl Sunday...
100% chance of LIGHTNING. GO CHARGERS!

================================================== ==========================

Date: Thu, 26 Jan 1995 07:37:42 -0800 (PST)
From: tburkhol@sdcc3.UCSD.EDU
Subject: RE: Center of buoyancy problem

> W
> |
> CB \|/------x-------|
> Head ___________._____.___________________ Feet
> ^ CM ^
> | |
> |--d--| |
> B R
> value of x. So if x gets smaller then R must increase to produce the
> same torque. However, [AND THIS IS WHERE I AM HAVING PROBLEMS], if R
> gets larger, then the only way to satisfy the translational
> equilibrium condition (B=W-R) is for B to decrease. This doesn't
> make sense, however, because the buoyant force (B) is (by definition)
> the weight of the displaced water which has not changed. If W, B,

It seems to me that, if you increase the force R, then you must change
the position of the body in the water (perhaps I don't fully understand
the problem). Consider the limit: where x->0 the body must be totally
supported (out of the water) by R or the body must have moved such that
also d->0.

> NOTE: We have tried this with the body totally submerged and on the
> surface and for different amounts of air in the lungs. We have also

Humans being rather floppy, you might also try making a more simple model,
like a weighted board.