The above figure gives a

schematicrepresentation of the 23 June 1997 Minuteman II missile launch, drawn from Hayne Palmour's beautiful picture (138 kB) from the NCT of 6/24/97, B1. All colors are true colors in that Figure, except that the color black was used instead of the white in the contrail. The hatching in the lower left part of the main emission feature at the top indicates a large area of diffuse white light. The upper ring is incomplete in Figure 1 - the inset shows the emission knot seen at the extreme left of the ring.I don't have any accurate measurements of the size of the contrail, but can estimate its size in 4 different ways. If I can get more information about when Hayne took his picture and the field of view of the image, the numbers can be much improved.

Estimate 1. My recollection is that the width of the main part of the contrail was about the width of my fist at arm's length, which is about 1/6 radian or ~10^{o}. At a distance of 200 miles, parts of the contrail have been blown ~200 miles / 12 ~ 20 miles away from the rocket's path. Since my observations occurred roughly 30 minutes after launch, which was at 8:39 p.m., the wind speed needed for this motion is about 40 mph.The column is about as high as it is wide, giving a height of the ring and the large emission knot at the top of the contrail of about 25 miles above the Earth's surface, taking into account that the Earth's surface is 6 miles below the horizon at a distance of 200 miles.

Estimate 2. The SDUT reported that the contrail was seen from Phoenix, 482 miles away from Vandenberg. The Earth's surface drops 32 miles in that distance, so the top of the contrail must conservatively have been greater than 40 miles above Vandeberg. So it is clear that my recollection underestimates the height and width of the contrail.

Estimate 3. Another way to estimate the height of the contrail uses the location of the red to white transition in the contrail in Hayne's picture. That point gives the location of sunset at that point above the Earth's surface. Time of sunset for Vandenberg Air Force Base, assumed to be at a latitude of 34.5^{o}and east longitude of 120^{o}25', was 8:18 p.m. on 23 June. When the sun is 1/2 hour below the horizon, full sunlight occurs at a height of about 20 miles above the Earth's surface (for detailed formulae and assumptions used, see below. A naive calculation gives 40 miles!) This would be a height of 15 miles above our horizon. The height of the top of the column is 2-3 times higher than the sunset point in Hayne's picture above the horizon, giving an estimated apparent height of the top of the contrail of 30-45 miles, or 35-50 miles above Vandenberg.

Estimate 4. Here is a quick summary of the Earth's atmosphere:

Altitude

(miles)Temperature profile

with increasing heightName 0-8 decreasing Troposphere (jet planes fly at 6 miles) 8-30 increasing Stratosphere 30-50 decreasing Mesosphere >50 increasing Thermosphere (noctilucent clouds at around 50 miles) Using the numbers from the previous estimates, it is clear that the contrail is within the stratosphere and the mesosphere, and below the altitude of noctilucent clouds. The beginning of the mesosphere at 50 miles is the coldest region of the atmosphere, and water vapor forms ice clouds readily there. It is quite likely that the top of the contrail, the source of the "twilight phenomenon", is at about 50 miles, as estimated above.

Summary. The top of the contrail is most likely at an altitude of 50 miles, implying that parts of the contrail were blown around 25 miles from the center of the contrail, requiring wind speeds of 50 mph.Finally, at what appeared to be the same altitude as the top of the contrail, the same altitude as the location of the twilight luminescence, there was a very large ring, part of which can be faintly seen in the picture in the NCT. The diameter of the ring appeared larger than the apparent length of the entire contrail from top to bottom, about 30

^{o}. There was a strong "limb brightening" effect, in that the leftmost part of the ring, where I was looking tangential to the ring, was much brighter and appeared wider than the rest of the ring. The bright patch may have been due to the presence of something similar to a "sun dog". The rightmost part of the ring showed a similar, but much fainter, effect. The ring appeared to go through the top of the contrail.The contrail is indeed the source of the large ring, although I didn't at first believe it. I watched a later contrail develop from the beginning, and it formed the same ring.

Detailed Formulae and Assumptions Used in Calculations

Formula for drop in Earth's surface as a function of distance from the observerdue to curvature of Earth:

h = (1 / cos(w) - 1) * R _{Earth},(1)

where w is the angle subtended by the distance between the observed point and the observer, measured from the center of the Earth. At a distance of 200 miles, the subtended angle w is

(200/25,000) * 360^{o}= 2.9^{o},

where 25,000 miles is of course the circumference of the Earth, giving h = 5.6 miles.

Formula for height above Earth's surface of sunset when sun is below the horizonby a given amount of time. This formula is not just a simple application of the above formula except for locations along the equator during the spring and fall equinoxes! I have derive the formula appropriate for this launch where the earth's axis is directly pointed toward the sun.

Derivation: Define coordinate system such that sun is along x axis, and the earth's axis is tilted by angle d from the x axis toward the sun. Then the sunrise/sunset locus is the cylinder

y ^{2}+ z^{2}= R^{2}_{Earth}(2)

as long as one is not too far from the time of sunrise/sunset.Let z

^{'}and x^{'}be a rotation of the x-z plane such that z^{'}points along the rotation axis of the earth. Then:

z = z ^{'}* cos(d) - x^{'}* sin(d)

x = x^{'}* cos(d) + z^{'}* sin(d)

y = y^{'}(3)

The location of the sunrise/sunset locus is obtained in the prime coordinate system by substituting for z as above. The Earth's rotation is easily described in the prime coordinate system, as:

z ^{'}= R * sin(lat)

x^{'}= - R * cos(lat) * sin(wt)

y^{'}= R * cos(lat) * cos(wt)(4)

using an arbitrary origin for time t and letting R stand for R_{Earth}.Sunset occurs at x = 0, which is when

sin(wt _{sunset}) = tan(lat) * tan(d)(5)

The formula checks out in the limiting case of d = 0, where the sunrise/sunset is then independent of latitude (12 hour days everywhere on Earth) and for the case of lat=0, where sunrise/sunset is 12 hours independent of the Earth's tilt.To find the sunrise/sunset location above a given point on the Earth's surface, one simply adds a height h to R in equations (4) and uses those values in equation (2). The resulting equation for h is:

( R / (R+h) ) ^{2}=cos ^{2}(lat) * cos^{2}(wt) + sin^{2}(lat) * cos^{2}(d)

+ cos^{2}(lat) * sin^{2}(wt) * sin^{2}(d)

+ 2 * cos(lat) * sin(wt) * sin(lat) * cos(d) * sin(d)(6) To find h, set t equal to t

_{sunset}plus the desired time after sunset, calculate the right hand side of equation (6), and then solve for h.

*Go to Fallbrook Information: Views: Vandenberg Air Force Base Missile Contrails
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Copyright © 1997-2000 by Tom Chester.
Permission is freely granted to reproduce any or all of this page as long as credit is given to me at this source:
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Comments and feedback: Tom Chester
Last update: 16 August 2000.
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