Hydrostatic Weighing

Lawrence Davis

31 Hydrostatic Weighing

The method of allows us to determine the average ( $\rho$ ) of a any object without any need for a (V) measurement by measuring only its weight ( $W_0$ ) and , also known as ( $UWW$ ). To see how we arrive at this useful result, follow the steps in the at the end of this chapter.

(1) $\begin{equation*} \rho = \frac{W_O}{W_O-F_A}\rho_W \end{equation*}$

Reinforcement Exercises

Calculate the density of an that object that weighs 12 N, but has an apparent weight of only 8.5 N.

The previous equation is very similar to the body density equation used for , but you will notice a slight difference. The previous equation determines the average density of the object including any hollow parts containing trapped air, but the body density equation is designed to determine the average density of body tissues only, not including trapped air. Therefore, the body density equation is modified to account for a volume of air trapped inside the body, known as the residual volume (RV). Also different standard symbols are used to designate body density, apparent weight, and water density.

Equations for residual volume are given for men and women. For men: 0.0115 x age (years) + 0.019 x height (cm) -2.24. For women: 0.009 x age (years) + 0.032 x height (cm) -3.90. An arrow shows where these values are used in an equation calculating body density: Db = BW/[(BW-UWW)/Dh2o –(RV +0.1)]. Arrows indicate where the body density is used in computing body fat percentage by two methods. Siri: BF% = 495/Db -450. Shutte: BF% = 437/Db -393 — Formulas used in calculating residual lung volume, body density, and body fat percentage. Image Credit: Measure Body Fat Via Under Water Weighing by MattVerlinich via Instructables

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We arrived at equation (1) by starting with the definition of as divided by its :

$\begin{equation*} \rho = \frac{m_O}{V_O} \end{equation*}$

We can find the mass of an object if we divide its weight by g:

$\begin{equation*} m = \frac{W_O}{g} \end{equation*}$

Inserting that result for mass into the density equation we have:

$\begin{equation*} \rho = \frac{W_O}{gV_O} \end{equation*}$

For a completely submerged object the volume of water is equal to the volume of the object, so we can replace $V_O$ with $V_D$ .

$\begin{equation*} \rho = \frac{W_O}{gV_D} \end{equation*}$

Using the definition of density again, we can replace the volume of water displaced with the displaced water mass ( $m_W$ ) divided by water density ( $\rho_W$ ).

$\begin{equation*} \rho = \frac{W_O}{g(m_D/\rho_W)} = \frac{W_O}{g m_D}\rho_W \end{equation*}$

We can look up the density of water, but it depends on the water temperature, which is why its important to measure the water temperature when . Notice that we happen to have the mass of displaced water multiplied by g in the previous equation. That is exactly how we calculate the weight of the displaced water ( $W_D$ ), so we can make that substitution:

$\begin{equation*} \rho = \frac{W_O}{W_D}\rho_W \end{equation*}$

which tells us that the pushing upward on objects in a fluid is equal to the weight displaced fluid. Therefore we can replace $W_D$ with $F_B$ .

$\begin{equation*} \rho = \frac{W_O}{F_B}\rho_W \end{equation*}$

We have learned that the difference between an object’s weight ( $W_0$ ) and apparent weight ( $W_A$ ) tells us the size of the buoyant force ( $F_B$ ), as long as the body is in (holding still):

$\begin{equation*} F_B = W_O - F_A \end{equation*}$

Making that replacement in our density equation we have:

$\begin{equation*} \rho = \frac{W_O}{W_O-F_A}\rho_W \end{equation*}$

We now have an equation that allows us to calculate the density of an object by measuring only its and , as long as we know the of the fluid we are using.

Specific Gravity

The ratio of the of a substance to that of water is known as the . Specific gravity can be determined by . If we simply divide both sides of our density equation by the density of water we will have a formula for the specific gravity:

(2) $\begin{equation*} SG = \frac{\rho}{\rho_W} = \frac{W_O}{W_O-F_A} \end{equation*}$