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Mechanics

Chapter 5

Spreadsheet Applied to Model the Kittinger Descent 

The Kittinger Descent

Motivation

A motivation for constructing a spreadsheet that included a representation of the standard atmosphere and took into account the variation of gravitational attraction with altitude was a desire to model the 1960 descent from a balloon of Captain Joseph Kittinger of the United States Air Force.

Briefly, his balloon reached an altitude of about 31,330 metres at the point when he exited his gondola.  After about 12 seconds of fall, he opened a 1.8 metre stabilization parachute and continued falling until his main parachute, 8.5 metres, deployed at a height of about 5500 metres.  He landed about eight minutes later.

A picture of his gondola, below, taken at the Air Force Museum can be found here.



Some Details

For details of the man, and of the records that were then set, see  here.

Two excerpts from the foregoing link follow:

1) Kittinger remained at his peak altitude, over three times higher than a commercial airliner typically flies, for about 12 minutes before he stepped off the "Highest Step in the World" to begin his fall to the Earth's surface. In addition to his pressure suit, Kittinger carried instruments and safety gear that weighed as much as he did. He also wore several layers of clothing to help protect him against the extremes of his high-altitude environment. During his fall, Kittinger experienced air temperatures as low as -94°F (-70°C)!

2) A final question to consider is why so many sources mistakenly claim that Joseph Kittinger broke the sound barrier during his 1960 skydive. The most likely explanation comes from the fact that these sources typically claim the maximum speed Kittinger reached was 714 mph, just one digit off from the actual speed he attained of 614 mph. If Kittinger had reached the higher speed, he would have indeed been supersonic and achieved approximately Mach 1.05. It seems probable that the author of some official source accidentally miswrote the maximum speed and the mistake has been replicated in numerous locations ever since.

Nevertheless, the actual value of 614 mph comes from Joseph Kittinger himself in a 1960 article he wrote for National Geographic and a subsequent book he authored describing the jump. A direct quote from the former states:

"An hour and thirty-one minutes after launch, my pressure altimeter halts at 103,300 feet. At ground control the radar altimeters also have stopped-on readings of 102,800 feet, the figure that we later agree upon as the more reliable. It is 7 o'clock in the morning, and I have reached float altitude... Though my stabilization chute opens at 96,000 feet, I accelerate for 6,000 feet more before hitting a peak of 614 miles an hour, nine-tenths the speed of sound at my altitude."

What can be Taken from these Excerpts for Use in a Model?

The weight, or mass, of Kittinger and his equipment is not given. Excerpt 1) tells us that the equipment weighed as much as he weighed.  He was about 32 years of age and likely quite fit. We will examine cases based on three possible weights for Kittinger, 134, 154, and, 174 pounds.   Doubling these weights to allow for the equipment and converting to kilograms provides the approximate mass values of 121.8, 140, and, 158.2 kg.

From Excerpt 2) it seems that the peak of his acceleration was reached prior to reaching 90,000 feet of altitude. Measured by his altimeter? Has he corrected, in this statement, for its offset from the ground radar altimeters?  Is his altimeter referenced to sea level or earth surface level?

Is the ground radar altimeter reading referenced to sea level or earth surface level?

What is his method of measuring velocity?  Could the method be the rate of change of his altimeter observations?

In order to Proceed with a Model of the Kittinger Descent

We will presume that the ground radar altimeter reading is referenced to the local earth surface.  Converted to metres this will be ~ 31,333 metres.  What was the altitude of the radar set?  We will use the altitude of Albuquerque New Mexico, ~1509 metres.  Thus the altitude of the balloon at the start of the fall will be taken as ~ 32,842 metres above sea level.

We presume that he is referring to his altimeter with the statement "
he experienced no noticeable change in acceleration until he fell to 90,000 ft ".   We interpret this as relative to his initial reading of 103,300 ft. for the balloon height.  Presuming the statement to be correct, we can expect that maximum velocity was reached after dropping 4,054 metres from 32,842 metres, that is, some time after falling to an altitude of 28,788 metres. 

Objective of the Model: We will attempt to model the fall to obtain the altitude at which he reached the peak velocity of 614 miles per hour, 274.48 metres per second, and the time that was taken to reach that peak velocity.

His stabilization chute had a radius of ~ 0.9 metres.  This chute would be expected to be only partly full until some time after maximum decent speed was reached. The effective radius of the combination of chute and Kittinger must be found, as must a value for the drag.

A density is also required for the combination of chute and man. 

Reaching An Answer

The spreadsheet, using Exper = 2, that was introduced in Parts A  and B of this topic provides vertical velocity and the corresponding altitude and time.

We need to solve for values of the sphere radius, density, and drag coefficient  given a maximum fall rate of 274.48 metres per second and three suggested mass values.

This can be done by adding a steepest decent routine, introduced in Chapter 2, to the spreadsheet.  For those readers interested in adding the routine to the downloadable spreadsheet it is seen next.



Although iteration to find the solutions is simple enough to be done "by hand" the process is tedious.  Moreover, a more capable steepest decent routine will be employed in a later Topic and this application serves as an introduction to its use.

The three solutions, one for each suggested mass value, 121.8, 140, and 158.2 kg  are seen next.



For all three solutions the low radius values seem more representative of Kittinger and his equipment than of the chute.  This suggests that the stabilization chute is streaming out above Kittinger but not open to any significant extent. The most significant drag may stem from Kittinger and his pack of instrumentation. 

All three solutions show the time taken to reach the maximum fall rate of ~ 274.5 m/s as ~ 41 seconds at an altitude of about 25,778 metres.

Intuitively, this answer does not seem to be far out of line.  It is undoubtedly not true in the light of all the presumptions that that were needed and the unknown precisions that were involved.

A by-product of obtaining the three solutions was a range of possible drag coefficients, 1.49, 1.56, and 1.62 for the assemblage of a streaming chute, and Kittinger with his equipment.

How well did the actual atmosphere over the New Mexico desert on August 16, 1960, correspond to the 1976 U.S. Standard Atmosphere?  Although it is readily conceivable that seasonal adjustments could be made to our atmospheric model, this has not been attempted.

For interest, the web-based calculator linked in Chapter 6 and also found on the upper row of tabs is used to provide graphs of altitude and velocity during the fall, to ~25,000 metres for the 140 kg case, seen following:

 

Notwithstanding the uncertain parameters used in modeling the Kittinger decent, the spreadsheet, with its inclusion of a model of the standard atmosphere, should be quite useful for estimating the behaviour of projectiles and falling bodies.

The Parameter Table


As has been mentioned, the macro varies some of the labels on the Par Sheet in accord with the value, 0 through 4, of the Parameter Exper, which in this case has the value 2 as is seen in the next table for modeling Kittinger's descent.

The ">", greater-than, character indicates that the value to its right is to be provided by the user. These values show in bold type.



Three values, B4:B6, are permanent reminders of underlying assumptions.

The values B2 and B3 are calculated immediately when the platform altitude is entered as possibly of interest to the user. Similarly, on entry of r in D2, D3:D4 and D6:D7 are calculated and on entry of d in D5, D6:D7 are immediately recalculated.

Some cell labels are applicable to other Experiments. Their associated values will remain blank.  Some labels and their values are common to all or most experiments. 

The value for Target Altitude in D8 causes calculation to terminate when that altitude is reached. Similarly calculation will terminate when the number of calculation steps given in D10 is reached.

Notice the value 2.0 in B7 and 45 in B8. These are there to represent Kittinger's little jump, up and out, to exit his gondola.

The Data Table

The first two rows of the data table are shown, in two sections, following:

Row 0 shows the initial conditions.

Row 1 is special in that is the culmination of calculating eleven sub rows consisting of: two rows calculated using 1/1024 of the step size, followed by 9 rows each employing twice the step size used in the previous step.  This is equivalent to a whole step at the specified step size. 

The reason for doing this is that projectiles may have a quite large initial velocity that drops quite quickly when an atmosphere is encountered.  The precision of the calculation benefits from smaller step sizes in that initial period of encountering the atmosphere when deceleration is most rapid. 



Data rows around the maximum fall rate are shown, in two sections, next:



One might note that the acceleration, Dv(y)/Dt, changes sign after the maximum fall rate is reached,

Next

Chapter 6 will present applications of the spreadsheet and calculator to non-pressurized and pressurized ballooning, high altitude research, and ballistics, and will provide a link to the web-based version of this calculator.


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