Hands-On Math

Last Revision March, 2011

Last Revision March, 2011

The Differential Analyzer, V. Bush,

He began with a means for providing solutions to the linear form of an equation and then modified the linear form by entering variable coefficients, v and g, using

For example, where gravity might previously be assumed to be constant through variations in height, it could instead be input as some chosen function of height by using an input table.

It is the Bush method, adapted for spreadsheet use, that is employed herein for finding solutions to non-linear differential equations.

For ease of discussion it will be assumed throughout this topic that falling bodies

In the linear form the acceleration acting on the body will be g less some multiplier of velocity.

This form is derived from Newton's second law, the force on the body F = m * a = m * g less a factor due to the resistance presented by the medium. That factor is termed a drag force,

As numerical methods are presumed, employ finite but small steps of change that can be represented with symbols such as ∆, or D. We then may write:

a = Dv/Dt

with the understanding that step size can be chosen small enough to meet the precision requirements of an application.

Presume that one wishes to calculate the values that falling velocity, v, and fallen distance, s, take on at successive small intervals in time, t, after a 1 kg body is dropped and that we know the values g ~= 10 metres per second and k =2. Use a pen and paper for the calculations.

Decide to calculate at 1/10-second intervals. Divide the paper into 5 columns headed in the first row by t, Dv, v, Ds, and s. To begin, populate the second row with the initial values, all zeros. Next populate the remainder of the first column with the successive values of the time t at which results are desired.

After fall has begun the body would normally arrive at the beginning of a calculation step with a non- zero velocity. That velocity happens to be zero in the first step. The equation for acceleration, (1),

(g - k * v) = Dv/Dt

provides that:

Dv = Dt * (g - k * v). Then, for the first step Dv = 1.0.

From here it is easy going. v is formed as the accumulation of the small steps Dv. Each value of v in a step moves the body an incremental distance Ds = v * Dt. Values of Ds are accumulated, negatively, to form s.

t | Dv | v | Ds | s | ||

0 | 0 | 0 | 0 | 0 | ||

0.1 | 1.0 | 1.0 | 0.1 | -0.1 | ||

0.2 | 0.8 | 1.8 | 0.18 | -0.28 |

For the next row and successive rows of calculation the procedure is unchanged.

The foregoing method of solving a differential equation using pen and paper is fundamentally the same as was used by early astronomers to calculate planet orbits, planet masses and planetary distances. Some of those calculations took a great many days.

Click the cell addresses following for their expressions.

. | . |

C7 | D7 | E7 | F7 | G7 |

Since

The aforementioned reference provides values for the ratio μ/

In this case the equation for the acceleration of the body becomes:

a = g - k * v

All that is needed to accommodate this form of the equation is to change the expression in column D from =(D$3-F$3 * E6

Exploration of the differences ascertained that full agreement with the 20 pairs of values, (velocity, fall distance), given in the TI calculator reference could be attained by halving the Dv contribution of the first step, i.e., changing (D$3-F$3 * E6

To determine a reasonable step size, trial calculations for the range 0 to 10 seconds were made with eight different step sizes Dt = 2, 1, 1/2, 1/4, 1/8, 1/16, 1/32, 1/64. Values attained on each trial for 0, 2, 4, 6, 8, and 10 seconds were then plotted on the same graph to observe their convergence. These plots for v and s are shown next.

Only the plots for the four or five largest step sizes appear to be distinguishable. The fifth of these, 0.125, was chosen for the following wider range plot of v and s. The 20 Texas Instruments pairs of values were superimposed for comparison.

m * a = m * g - k * v

Terminal velocity is reached when g = (k * v

For the previous example, we used m=1, g = 9.8 and k = 0.005. For these values vt = 44.271887... metres per second. The velocity attained at 30 seconds in the previous example was 44.2717799, a tiny bit short of vt.

Then using the force equation:

m * a = m * g - (1/2 *

When terminal velocity is reached:

2 * m * g =

Return to the acceleration equation:

a = g - [1/2 *

Substitute 2 *m *g / vt

an interesting form of the equation. Of course it is no more than a rearrangement.

Under the approximation, the k seen in equation (2) is:

k = 1/2 * 1/m *

The Rayleigh approximation requires

The web-based, 1D, calculator given at the end of the next chapter and available in the upper row of tabs will be used to provide an example for a buoyant application. As that calculator does not it self account for buoyancy, a method for causing its results to include the effect of buoyancy is now given.

Buoyancy is an upward force due to the mass of the atmosphere that the object displaces. The force equation for a buoyant object is:

F = m * g -

Dividing the latter force equation through by m, the acceleration equation is then:

a = g * (1 -

Thus using

A latex spherical toy balloon inflated with air has a mass of about 0.0145 kg. Its radius is about 0.13 metres. Guess

Press Calculate, and the resulting charts are:

Tape a metal weight to the neck of the balloon such as a machine screw with washers added to closely match the balloon's mass thus doubling its mass. Note that when weighted, the balloon falls faster.

Plug this new value for m into the spreadsheet and employ a new value for

The light balloon appears to fall more slowly. Air resistance appears to slow the heavy balloon less than it does the light balloon both for high and low velocities. But this may not be the full story at the lower velocities. Recall that at very low velocities we should use the Stokes model. Would that make any substantial difference to the plotted results?

Use the 1D calculator to see how the falling rates of the two balloons compare when the value for the density

The story of Galileo finding that an iron ball and a same size wooden ball arrived at the ground at the same when dropped together from the leaning tower is believed by historians to be fiction. Nonetheless the story has inspired a great deal of controversy over the years.

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