Aquatic Exercise and Buoyancy
Date: Winter 2011-2012
A person, In neck-high water, in a swimming pool, feels much lighter (weight-wise) when water walking and exercising, than out of the water. It is a known fact there is less force on the knees, etc., and this is why elderly people, who cannot exercise by walking or in a gym, can exercise successfully in the water without injury to the joints. What is the ratio of a person's water-weight vs. out of water weight. Also, what is the percentage of a person's water weight compared to out-of-water weight?
Before addressing your question, I would like to make a comment. Be very
careful about using phrases like, "It is a known fact..." All observations
have statements limiting the scope of the observation. For example, the
"answer" you get in your inquiry assumes "fresh" water vs. "salt" water. So
the "known fact" is not so well known fact is different in the two cases.
The principle of buoyancy is this: A body immersed in a fluid is
buoyed up by a force equal to the WEIGHT of the VOLUME of the FLUID
DISPLACED by the IMMERSED BODY. Consider this statement very carefully.
Mentally munch on it.
How is it possible for a ship to float? Why does a balloon filled
with helium rise in the atmosphere, but the same balloon filled with the
same volume of air not rise in the atmosphere? Search the Internet for the
term "buoyancy". Study it carefully. Your misunderstanding of the term is
very common, even by many teachers.
Buoyancy is the reduction in the weight of a body submerged in a
surrounding fluid. The fundamental principle is this: The reduction in the
apparent WEIGHT of an object is equal to the WEIGHT of the VOLUME of the
surrounding fluid displaced by the object being submerged. In this form it
is not always so easy to measure the VOLUME of the displaced surrounding
fluid. In most applications the principle is used in reverse. The weight of
the object is determined in and out of the surrounding fluid. Knowing the
density of the surrounding fluid lets one then compute the volume of the
weight of the object, even if object has a complicated shape.
This is an example of Archemedes' Principle. This principle states
that the weight of an object is buoyed up (that is, its weight is reduced
by) an amount equal to the weight of the volume of the fluid (think water,
but that is not necessarily so) displaced by the immersed object. So the
weight is reduced by the weight of the volume of the part of the object that
is immersed. This sounds more complicated than it actually is. Determine
the volume that is immersed. Multiply that volume by the density of fluid
(e.g. water). The largest reduction in the weight of the object is when the
object is completely immersed.
Consider the weight of the water pushed out of the way of by the
person. This weight of water is equal to the buoyant force acting on
the person. The more the person is immersed, the greater the buoyant
force. The ratio you ask for depends on how much of the body is in
the water, which parts of the body (legs are denser than lungs, for
example), the body type, amount of fat, etc. There is no general
percent that is true in all cases.
A good conceptual physics source that is easy to read and will help you is:
Paul Hewitt "Conceptual Physics" Addison-Wesley
---Nathan A. Unterman
Sorry, Wanda, I do not think you can define an exact percentage.
It varies with how much of the person is submerged at each instant.
If your feet are not touching the bottom, you weigh 0% of your normal weight.
I have heard that a head is about 12 pounds,
out of say 150 pounds, so if you are up to your neck,
add a couple more pounds for the neck,
then the ratio is roughly 10:1 and the percentage is about 10% or less.
I think many standing moves would need shoulders to be out of the water,
and then it might be about 4:1 and 25%.
If submerged up to the balance-point near the waist, it would be 2:1 and 50%.
I can imagine a physical therapist maybe using this as some kind of sequence.
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Update: June 2012