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Scott Mclean
02-21-1995, 01:03 AM
Greetings:
I want to thank everyone for their input over the past few weeks
into the discussion on buoyancy. I would like to summarize the responses
for the latest posting. These responses all pointed to a measurement error
as a prime suspect for the strange results. One message I received from Hans
Mattes noticed that a systematic addition to R1 and R2 seemed to solve the
problem. After pulling a few hairs out looking at the old data I decided to
re-run the validation. The numbers I produced agreed quite well (within 5
mm) with the predicted CB regardless of the positioning of the 2 supporting
forces. Thus there was an error in the first set of data. Upon further
inspection of the original data I noticed that the force data were almost
identical to the new validation. However, there were some errors in the
recording of the moment arms in the first set of data. With the moment arms
corrected these data provided a validation similar to the new set of data
(i.e., within 5 mm). I appreciate your time (and patience) with this matter.
A summary of the responses is appended.

---
Scott Mclean
smclean@iastate.edu

ORIGINAL POSTING

Greetings:

Dr. Hinrichs and I were pleased by the response to our previous
posting concerning the center of buoyancy. We now have a problem which may
be more interesting. The method of locating the center of buoyancy was
validated using a 180 cm long piece of PVC pipe of approximately 2 inch
diameter. One end was filled with concrete. The ends were sealed so the
other end was filled with air. The CM of the pipe was measured to be 65 cm
from the heavy end. The CV was at 90 cm or half way along the length of the
pipe due to the symmetry of the pipe.

The measurement of the CB was validated by supporting this pipe with
two straps (one on each side of the CB) while the pipe was fully submerged.
The fully submerged CV would be coincident with the CB. A 4 kg mass was
added to the strap on the light end and a 2 kg mass was added on the heavy
end to keep the pipe submerged. The free body diagram is pictured below.

|--------------dB--------------|
/|\B
/|\ R1 | /|\ R2
|--dR1--| | |
heavy | CM | | light
||================================================ ===========||
end | CB end
|---------dW---------|
|
|
\|/W
|--------------------------dR2--------------------------|

where
R1 = supporting force at the heavy end
R2 = supporting force at the light end
B = buoyant force
W = weight
dR1 = distance from heavy end to R1
dR2 = distance from heavy end to R2
dB = distance from heavy end to CB
dW = distance from heavy end to CM

The translational and rotational equations of statis equilibrium for this
system are

sum of forces = 0

eqn (1) R1 + B + R2 - W = 0.

sum of moments = 0

eqn (2) (R1*dR1) + (B*dB) + (R2*dR2) - (W*dW) = 0.

Solving these equations for the location of the CB in terms of measured
forces and distances gives

eqn (3) (W*dW) - (R1*dR1) - (R2*dR2)
dB = ------------------------------.
W - R1 - R2

The net forces (i.e., with the added weight due to the added masses
subtracted out), R1 and R2 were measured using calibrated load cells. R2
was found to be negative (implying that the application of a downward force
was required to keep this end of the pipe submerged).

The measurement of dB was approximately 90 cm (within 3 or 4 mm) when R1 and
R2 were positioned symmetrically about the CB regardless of the distance
between the load cells. However, when the heavy end support remained
stationary and the light end support was moved towards the middle, the
calculated dB grew linearly from 89.5 cm to 109.8 cm (over 9 locations).
The opposite was found when the light end supporting force remained
stationary and the heavy end supporting force was moved. dB decreased
linearly from 89.5 cm to 69.1 cm over 9 locations.

We suspected that there were errors in the measurements and perhaps caused
the strange results. But upon inspection, the buoyant force was predicted to
be approxmately 3380 g (* 9.81/1000 N). This calculation held for every
condition. This compared well with the calculated volume of the pipe
(3309 cm~3). The systematic (linear) nature of the deviation suggests that
the results were not due to poor data.

The system contains two unknowns, the magnitude of the buoyant force (B) and
the point of application of the buoyant force (dB). If the data we have
collected are real then moving the load cells asymmetrically changes either
the magnitude of the buoyant force (B) or the point of application of this
force. The fact that the measured buoyant force agrees so well with the
theoretical buoyant force would indicate that dB changes. This makes no
sense based on the definitions of the buoyant force and the CB.

Can anyone see an error in the derivations we have made or the logic we are
using? Any help would be greatly appreciated. We will post a summary of
all responses we receive.

Thanks again,

Scott
---
Scott Mclean
smclean@iastate.edu
================================================== =========================
RESPONSE #1

Scott,

I don't have the solution for your problem, but I have some
suggestions that might help you to solve it.

I think we can take it for certain that the CB is exactly in the
middle of the bar and cannot move. In my opinion, we can always
rest sure of that. No way it can be or move somewhere else.

I suspect that the CM could have moved, or the Weight could have
changed. Did you try to weigh the bar at the end of the experiments?
Some water might have leaked into the cilinder. Assuming that the light end
of the bar will always tend to be slightly higher (closer to the surface)
than the heavy end, the water inside the cilinder would both move
the CM closer to the heavy end and make W larger.

Are you sure that the scale you used to weigh the cilinder was
calibrated correctly? The best thing would be to use the load cells to
sure W. Did you?

Your method was correct, so the error must be searched in the
collected data. Since you wrote that the load cells were calibrated,
the only data that can be wrong seem to be W and dW.

Something could be said also about possible couples of forces
generated by the straps, which would change your equation for
static rotational equilibrium.

Also, the point of application of the forces exerted by
the straps might not coincide with the center of the straps. How wide
were the straps? Did the masses suspended below the straps deform them?

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Panta rei :-)
================================================== ==========================
RESPONSE #2

Hi Scott:

When you say "... the added weight due to the added masses was subtracted
out", do you mean that the upthrusts on the two masses have also been
addressed?

Bryan Finlay, PhD
Director, Orthopaedic Research Laboratory
University Hospital 519-663-3063
P.O. Box 5339 519-663-3904 FAX
London, Ontario, CANADA, N6A 5A5 BFinlay@uwovax.uwo.ca
================================================== ==========================
RESPONSE #3
Scott:

Here are my wild guesses:

Maybe you have not taken into account the weight forces of the 2 Kg
and 4 Kg masses? They don't show up in your equations, but maybe you
included them already as part of R1 and R2 --I don't know.

Another (although unlikely) possibility is that you need to take
into account the buoyancy forces exerted on each of the two masses. But I
suppose you used metal weights, so the volume would be small enough to make
the buoyancy forces negligible.

---
Jesus Dapena
Department of Kinesiology
Indiana University
Bloomington, IN 47405, USA
1-812-855-8407 (office phone)
dapena@valeri.hper.indiana.edu (email)
================================================== ==========================
RESPONSE #4

Hi Scott

Just a thought--have you checked the accuracy of your load cells over a
range of loads. When they are positioned at an equal distance to the left
and to the right of the center, you presumably get the same readings. Now,
if you moved one towards the center the readings at both locations would
change--one would go up and the other would go down. (I have not checked
this, but I think this is the case.) What would happen if one reading
increased too much and the other dropped too much? Your net buoyancy force
could stay the same, but the calculated CB would change.

Regards, Brian Davis
================================================== ==========================
RESPONSE #5

Hi, Scott,

I believe this is the same question as that you proposed before. If you feel
no problem with that simple one, then this one dose not exsist. Please see
below.

> The translational and rotational equations of statis equilibrium for this
> system are
>
> sum of forces = 0
>
> eqn (1) R1 + B + R2 - W = 0.
>
> sum of moments = 0
>
> eqn (2) (R1*dR1) + (B*dB) + (R2*dR2) - (W*dW) = 0.
>

Now denoting

eqn(4) R=R1+R2
and
eqn(5) dR=(R1*dR1+R2*dR2)/(R1+R2)

then eqn(1) and (2) become

eqn(6) R+B-W=0
and
eqn(7) R*dR+B*dB-WdW=0

The eqn(6) and (7) are same as yours in the previous posting. That means no
matter how many supporting forces (eg. R1, R2) you would apply, their sum
(R) should balance the difference of the body weight and buoyancy, and the
position (dR) at which the sum of all supporting forces is applied is
determined by the net moment due to the weight and buoyant. For a given body,
R, W, dB, dW are all fixed, so you can only find one value for R and one
value for dR to reach a static equilibrium. You can not change dR, moving the
support will destroy the rotational balance.

Back to your purpose, if you want to measure the buoyancy and CB for a given
body, I do not think your approach is very good particularly for the CB. You
will have to move the support to locate a point at which the rotational
balance is realized, this point gives dR (or dR1 when dR2 is fixed, or dR2
when dR1 is fixed). Then the equations can be used to find out CB. The
problem is, as I understand, this kind static equilibrium may be unstable.
This increases the difficult of the measurement. If you have any rough idea
about CB in advance, your method might work well.

Jianyu Cheng
================================================== ==========================
RESPONSE #6

Rick:

I enjoyed our conversation the other day, but I haven't gotten back to
you because, until now, I haven't had time to look at the problem. Now
that I do look at the problem it isn't clear to me that there is any
structural issue with the model that's being used for analysis. What
does appear, however, is a clear skewing of the numerical results whenever
the weights are moved assymetrically. Could this be an experimental
measurement error rather than a conceptual one? It's interesting to look
at the data from the "CB Calculations" page of the Excel spreadsheet you
gave me. I plotted "d" versus "R Position." The first third is flat
(consistent COB with symmetrical weights) while the second and third
thirds trend up and down (with assymetrical loadings). If, however, the
"CB tare" values for "LC1" and "LC2" are increased by 870 (grams?) and
890 grams respectively (to 5023 and 2864) the values which compute for
"d" and virually constant for ALL positions of the weights!!!

Is it possible that cables or brackets or clamps or something else
were not accounted for? Might bouyancy of some of the apparatus be
accountable? Not having seen the equipment, it's hard for me to make any
assessment, but its a REMARKABLE situation that corrections of about
880 in both tare weights totally remove the anomolous results.

I seems worth thinking about.

Hans Mattes
(201) 386-3266
================================================== ==========================
RESPONSE #7

Scott, Hi...
Greetings to you all. Due to e-mail problems I missed your first problem and
the responses to it, so I might be repeating what someone else has already
thought. But, here it goes:
a) If the pipe rotates when submerged, then you need to apply a couple
(torque), that is R1 and R2 must be equal.
b) Now, the pipe will not rotate, but it may sink or float. In those cases
you should apply a force, let's say RSTAR, through the point around which you
perform your torque equilibrium (ie. CM). This RSTAR should always be used
(even when you add the couple mentioned above for to prevent rotation!) and
is the one that gives you the force equilibrium (B=W+RSTAR). Note R1 and
R2 always cancel at the force equilibrium.
For simplicity, you may want to apply R1 also at the CM. But, If you vary R2
without varying R1 you do not have force equilibrium. Your system is either
coming out of the water (thus B has changed) or is unbalanced (thus moving
up or down).
Well that is what I could come up with from here! I hope it helps. Good luck
and thanks for the problem...