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In my textbook, a formula for estimating error in bulk-volume measurements is derived, but I don't quite follow one step in the derivation. The book writes the following:

The bulk volume of a porous-rock core sample can be measured in two steps, by weighing the sample first in water, $m_1$ (assuming $100$% water saturation), and then in air, $m_2$.

The bulk volume is calculated as follows:

$$V_b = \frac{m_2 - m_1}{\rho_w}$$

By differentiating this equation, we obtain:

$$dV_b = \frac{\partial V_b}{\partial m_2}dm_2 + \frac{\partial V_b}{\partial m_1}dm_1 + \frac{\partial V_b}{\partial \rho_w}d \rho_w$$,

$$dV_b = \frac{m_2 - m_1}{\rho_w}\Bigg(\frac{dm_2}{m_2 - m_1} - \frac{dm_1}{m_2 - m1} - \frac{d \rho_w}{\rho_w}\Bigg)$$

If the density measurement and the measurements of $m_1$ and $m_2$ are considered to be independent, the uncertainty inherent in the bulk-volume value can be written as:

$$\Bigg(\frac{\Delta V_b}{V_b}\Bigg)^2 = 2\Bigg(\frac{\Delta m}{(m_2 - m_1)}\Bigg)^2 + \Bigg(\frac{\Delta \rho_w}{\rho_w}\Bigg)^2$$

where the error in the weighing of the two masses is considered to be identical, $\Delta m = \Delta m_1 = \Delta m_2$.

OK, so it is this last step here that I don't quite follow. I assume that we are here using that $dV_b = \Delta V_b$. Since we know that:

$$dV_b = \frac{m_2 - m_1}{\rho_w}\Bigg(\frac{dm_2}{m_2 - m_1} - \frac{dm_1}{m_2 - m1} - \frac{d \rho_w}{\rho_w}\Bigg)$$

I would then assume that we have:

$$\frac{\Delta V_b}{V_b} = \frac{\frac{m_2 - m_1}{\rho_w}\Bigg(\frac{dm_2}{m_2 - m_1} - \frac{dm_1}{m_2 - m1} - \frac{d \rho_w}{\rho_w}\Bigg)}{\frac{m_2 - m_1}{\rho_w}}$$

$$\frac{\Delta V_b}{V_b} = \Bigg(\frac{dm_2}{m_2 - m_1} - \frac{dm_1}{m_2 - m1} - \frac{d \rho_w}{\rho_w}\Bigg)$$

$$\frac{\Delta V_b}{V_b} = \Bigg(\frac{\Delta m}{m_2 - m_1} - \frac{d\rho_w}{\rho_w}\Bigg)$$

So:

$$\Bigg(\frac{\Delta V_b}{V_b}\Bigg)^2 = \Bigg(\frac{\Delta m}{m_2 - m_1} - \frac{d\rho_w}{\rho_w}\Bigg)^2$$

$$\Bigg(\frac{\Delta V_b}{V_b}\Bigg)^2 = \Bigg(\frac{\Delta m}{m_2 - m_1}\Bigg)^2 - 2 \frac{d\rho_w}{\rho_w}\Bigg(\frac{\Delta m}{m_2 - m_1}\Bigg) + \Bigg(\frac{d \rho_w}{\rho_w}\Bigg)^2$$

But this is obviously not the same.

So if anyone can please explain to me that last step in the derivation of the formula, I would be very grateful!

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1 Answer 1

up vote 1 down vote accepted

The author does not really mean that, for example, the errors in measuring the two masses are the same. (But admittedly, that is exactly what the author writes!)

What is meant is that we have a bound $\Delta m$ on the absolute value of the error in measuring a mass, and similarly a bound on the absolute value of the error in the density. The errors could be in opposite directions and add. We could have an underestimate of the mass in air, and an overestimate of the mass in water.

The squaring is traditional. What's happening is that the author, without explicitly saying so, is making a calculation of variance.

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OK. Thanks a lot for clearing this up! I haven't had statistics in three years, so the thought of this being variance did not even occur to me :). Appreciate it a lot! –  Kristian Aug 19 '12 at 16:07

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