How to stay alive...underwater (Part 2)

As a follow-up to our previous post we will now work through the minimum gas calculations we discussed. The specific worst-case scenario that we must be able to handle is this following: one of us is cut off from our gas when we are at the deepest part of the dive. If this happens then we need enough gas in the remaning buddy’s cylinder to support a safe ascent including all safety stops.

Let us quickly review how we do our safety stops for dives up to 100 feet deep. In this case we perform what’s known as “minimum deco”, with the ascent as follows: If the maximum depth of our dive is $d$, we first ascend to depth $\mathrm{max}(d/2, 30)$ with an ascent rate of 30 ft/min. At depth $\mathrm{max}(d/2, 30)$ we do a 30 second stop, then ascend 10 ft in 30 seconds, then do another 30 second stop, ascent 10 ft in 30 seconds etc until we reach the surface. Dives deeper than 100 feet require a different algorithm and different considerations, but that’s a topic for a different blog post.

Calculation: between 60 and 100 feet

In order to simplify our calculation we will approximate a 30 second safety stop at depth $x$ followed by a 30 second ascent to $x-10$ by a 60 second ascent from $x$ to $x-10$. In other words, while we are alternate between 30 second safety stops and 30 second ascents during the very last part of our dive, we will assume a 10 ft/min constant ascent rate (because of our simplifying assumption).

Also note that the pressure $p(x)$ at a depth $x$ in ATA can be computed as $p(x) = (1 + x/D)$, where $D$ is the depth at which the pressure doubles, i.e. $D = 33$ if our unit is ft. The gas consumption at a depth $x$ is $s\cdot p(x)$ where $s$ is the SAC rate.

When ascending a small distance $\Delta x$ with an ascent rate $v$ the amount of gas consumed during this small ascent can be calculated as $s\cdot p(x) \Delta x / v$. The amount of gas consumed when ascending from a depth $d$ can thus be computed as the sum of two integrals

$$\int_{d/2}^{d} \frac{s\cdot p(x)}{v_1} dx + \int_{0}^{d/2} \frac{s\cdot p(x)}{v_2} dx$$

where $v_1$ is the ascent rate from $d$ to $d/2$ and $v_2$ is the ascent rate from $d/2$ to the surface. For us, $v_1$ is 30 ft/min and $v_2$ is 10 ft/min.

We are almost to our answer. However, we need to add one more thing. Given an out of air emergency, we need a small amount of time $t$ (normally about one minute) in order to communicate with each other and begin the process of air sharing. During this time interval at the deepest part of the dive the gas consumed is $s \cdot p(d) \cdot t$. With this additional time built into the model, the total amount of gas we use is:

$$\begin{align} g(d) &= s \cdot p(d) \cdot t+ \int_{d/2}^{d} \frac{s\cdot p(x)}{v_1} dx + \int_{0}^{d/2} \frac{s\cdot p(x)}{v_2} dx \\ &= s \cdot p(d) \cdot t + \frac{s}{v_1} \int_{d/2}^{d} \left( 1 + \frac{x}{D} \right) dx + \frac{s}{v_2} \int_{0}^{d/2} \left( 1 + \frac{x}{D} \right)dx \\ %&= s \cdot t\left(1 + \frac{d}{D} \right) + \frac{s}{v_1} \left( \frac{d}{2} + \left[ \frac{x^2}{2 D} \right]_{d/2}^{d} \right) %+ \frac{s}{v_2} \left( \frac{d}{2} + \left[ \frac{x^2}{2 D} \right]_{0}^{d/2} \right) \\ &= s \cdot t\left( 1 + \frac{d}{D} \right) + \frac{s}{v_1} \left( \frac{d}{2} + \frac{ 3 d^2}{8 D} \right) + \frac{s}{v_2} \left( \frac{d}{2} + \frac{d^2}{8 D} \right) \\ &= s \cdot t + \frac{s}{2} \left( \frac{1}{v_1} + \frac{1}{v_2} + \frac{2t}{D} \right) d + \frac{s}{8 D} \left( \frac{3}{v_1} + \frac{1}{v_2} \right) d^2. \end{align}$$

We see that the consumed gas $g(d)$ can be expressed as a quadratic polynomial:

$$g(d) = A + Bd + Cd^2.$$

where the coefficients are

$$\begin{align} A &= s \cdot t,\\ B &= \frac{s}{2} \left( \frac{1}{v_1} + \frac{1}{v_2} + \frac{2t}{D}\right), \\ C &= \frac{s}{8 D} \left( \frac{3}{v_1} + \frac{1}{v_2} \right). \end{align}$$

Now we just plug in the numerical values for $D$, $v_1$, $v_2$, $t$ and $s$. In our previous post we determined our SAC rate to be 1.0 cubic feet per minute, but since we are two people breathing we double this to 2.0 cubic feet per minute. The numerical values are then

$$\begin{align} D &= 33\mathrm{\ feet}\\ v_1 &= 30\mathrm{\ feet/min}\\ v_2 &= 10\mathrm{\ feet/min}\\ t &= 1\mathrm{\ min}\\ s &= 2\mathrm{\ cubic\ feet/min} \end{align}$$

Plugging in these values we obtain

$$\begin{align} A &= \left( \frac{2 \mathrm{\ feet}^3}{\mathrm{min}} \right) \cdot \left(1\mathrm{\ min}\right) = 2\mathrm{\ feet}^3\\ B &= \left(2\frac{\mathrm{\ feet}^3}{\mathrm{min}}\right)\cdot \left( \frac{1\mathrm{\ min}}{30\mathrm{\ feet}} + \frac{1\mathrm{\ min}}{10\mathrm{\ feet}} + \frac{2\mathrm{\ min}}{33\mathrm{\ feet}} \right) = \frac{32}{165}\mathrm{\ feet}^2 \approx 0.194\mathrm{\ feet}^2 \\ C &= \frac{2\mathrm{\ feet}^3}{8 \cdot 33\mathrm{\ min\ feet}} \left( \frac{3\mathrm{\ min}}{30\mathrm{\ feet}} + \frac{1\mathrm{\ min}}{10\mathrm{\ feet}} \right) = \frac{1}{660}\mathrm{\ feet} \approx 0.00151 \mathrm{\ feet} \end{align}$$

Note that $A$ has units of cubic feet, $B$ has units of square feet, $C$ has units of feet, so that $g(d)$ has units of cubic feet, which is exactly what we expect!

We now convert cubic feet to pressure. By the ideal gas law the gas amount is directly proportional to the pressure in the cylinder. A standard aluminum 80 cubic foot cylinder will contain around 77 cubic feet of gas at 3000 psi. Therefore the amount $g(d)$ corresponds to a pressure of

$$g(d) \cdot 3000/77 \approx 38.9 \cdot g(d).$$

Finally then we have the following formula for the minimum tank pressure $m(d)$ that we need to safely ascend:

$$m(d) = \frac{3000g(d)}{77} = \frac{3000}{77}\left( 2 + \frac{32}{165} d + \frac{1}{660} d^2 \right) = \frac{6000}{77} + \frac{19200}{2541} d + \frac{50}{847} d^2.$$

As an example, when $d = 60$ feet we obtain $g(d) \approx 19.09$ cubic feet and $m(d) \approx 743.8$ psi.

Calculation: 30 to 60 feet

The calculation is similar to the one above, except that we begin our safety stops at $u = 30$ feet instead of $d/2$ feet. Thus the integral becomes

$$\begin{align} g(d) &= s \cdot p(d) \cdot t+ \int_{u}^{d} \frac{s\cdot p(x)}{v_1} dx + \int_{0}^{u} \frac{s\cdot p(x)}{v_2} dx \\ &= s \cdot p(d) \cdot t + \frac{s}{v_1} \int_{u}^{d} \left( 1 + \frac{x}{D} \right) dx + \frac{s}{v_2} \int_{0}^{u} \left( 1 + \frac{x}{D} \right)dx \\ %&= s \cdot t\left(1 + \frac{d}{D} \right) + \frac{s}{v_1} \left(d - u + \left[ \frac{x^2}{2 D} \right]_{u}^{d} \right) %+ \frac{s}{v_2} \left( u + \left[ \frac{x^2}{2 D} \right]_{0}^{u} \right) \\ &= s \cdot t\left( 1 + \frac{d}{D} \right) + \frac{s}{v_1} \left(d - u + \frac{d^2 - u^2}{2 D} \right) + \frac{s}{v_2} \left(u + \frac{u^2}{2D} \right) \\ &= s \cdot \left(t - \frac{u}{v_1} -\frac{u^2}{2Dv_1} +\frac{u}{v_2} + \frac{u^2}{2Dv_2}\right) + s\left( \frac{1}{v_1} + \frac{t}{D} \right) d + \left( \frac{s}{2Dv_1}\right) d^2. \end{align}$$

and our values for $A$, $B$, and $C$ become:

$$\begin{align} A &= \frac{86}{11} \approx 7.82 \mathrm{\ feet}^3\\ B &= \frac{7}{55} \approx 0.127\mathrm{\ feet}^2\\ C &= \frac{1}{990} \approx 0.00101 \mathrm{\ feet} \end{align}$$

Note that when $d = 60$, this gives us $19.09$ cubic feet or $743.8$ psi, which is exactly the same as the previous formula for $d = 60$. We also have for $d = 30$ that the amount of gas used is $12.55$ cubic feet, corresponding to a tank pressure of $488.8$ psi.

Calculation: less than 30 feet

Diving at less than 30 feet we will approximate our ascent by going directly to the surface at 10 ft per minute. Our integral then becomes:

$$\begin{align} g(d) &= s \cdot p(d) \cdot t+ \int_{0}^{d} \frac{s\cdot p(x)}{v_2} dx \\ &= s \cdot p(d) \cdot t + \frac{s}{v_2} \int_{0}^{d} \left( 1 + \frac{x}{D} \right)dx \\ &= s \cdot t\left(1 + \frac{d}{D} \right) + \frac{s}{v_2} \left(d + \frac{d^2}{2 D} \right) \\ &= s\cdot t + s\left(\frac{t}{D} + \frac{1}{v_2}\right)d + s\left(\frac{1}{2Dv_2}\right)d^2 \end{align}$$

and our values for $A$, $B$, and $C$ become:

$$\begin{align} A &= 2 \mathrm{\ feet}^3\\ B &= \frac{43}{165} \approx 0.261\mathrm{\ feet}^2\\ C &= \frac{1}{330} \approx 0.00303 \mathrm{\ feet} \end{align}$$

Plugging in the value $d = 30$ gives us $12.55\mathrm{\ feet}^3$ or $488.8$ psi, which matches our second formula.

Summary

Using the above formulas we can compute the minimum gas requirements in psi for various depths. To recap, these numbers represent the tank pressure where we need to start ascending in order to be able to handle an emergency out of gas situation. Note that these values represent the values for our particular SAC rate and is not a general guideline. The reader needs to determine her own SAC rate in order to compute the corresponding numbers, as well as adjust the computations if she is diving with a different type of cylinder.

Depth	Minimum psi for a 80 cubic foot aluminum cylinder.
20	328
40	566
60	744
80	1060
100	1424

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Written by

Christian Lundkvist

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• scuba 2