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Observation on TPS damage on Orbiter
NASA photos
| 2-3-03
| BoneMccoy
Posted on 02/04/2003 1:34:19 AM PST by bonesmccoy
In recent days the popular media has been focusing their attention on an impact event during the launch of STS-107. The impact of External Tank insulation and/or ice with the Orbiter during ascent was initially judged by NASA to be unlikely to cause loss of the vehicle. Obviously, loss of the integrity of the orbiter Thermal Protection System occured in some manner. When Freepers posted the reports of these impacts on the site, I initially discounted the hypothesis. Orbiters had sustained multiple impacts in the past. However, the size of the plume in the last photo gives me pause.
I'd like to offer to FR a few observations on the photos.
1. In this image an object approximately 2-3 feet appears to be between the orbiter and the ET.
2. In this image the object appears to have rotated relative to both the camera and the orbiter. The change in image luminosity could also be due to a change in reflected light from the object. Nevertheless, it suggests that the object is tumbling and nearing the orbiter's leading edge.
It occurs to me that one may be able to estimate the size of the object and make an educated guess regarding the possible mass of the object. Using the data in the video, one can calculate the relative velocity of the object to the orbiter wing. Creating a test scenario is then possible. One can manufacture a test article and fire ET insulation at the right velocity to evaluate impact damage on the test article.
OV-101's port wing could be used as a test stand with RCC and tile attached to mimic the OV-102 design.
The color of the object seems inconsistent with ET insulation. One can judge the ET color by looking at the ET in the still frame. The color of the object seems more consistent with ice or ice covered ET insulation. Even when accounting for variant color hue/saturation in the video, the object clearly has a different color characteristic from ET insulation. If it is ice laden insulation, the mass of the object would be significantly different from ET insulation alone. Since the velocity of the object is constant in a comparison equation, estimating the mass of the object becomes paramount to understanding the kinetic energy involved in the impact with the TPS.
3. In this image the debris impact creates a plume. My observation is that if the plume was composed primarily of ET insulation, the plume should have the color characteristics of ET insulation. This plume has a white color.
Unfortunately, ET insulation is orange/brown in color.
In addition, if the relative density of the ET insulation is known, one can quantify the colorimetric properties of the plume to disintegrating ET insulation upon impact.
Using the test article experiment model, engineers should fire at the same velocity an estimated mass of ET insulation (similar to the object seen in the still frame) at the test article. The plume should be measured colorimetrically. By comparing this experimental plume to the photographic evidence from the launch, one may be able to quantify the amount of ET insulation in the photograph above.
4. In this photo, the plume spreads from the aft of the orbiter's port wing. This plume does not appear to be the color of ET insulation. It appears to be white.
This white color could be the color of ice particles at high altitude.
On the other hand, the composition of TPS tiles under the orbiter wings is primarily a low-density silica.
In the photo above, you can see a cross section of orbiter TPS tile. The black color of the tile is merely a coating. The interior of the tile is a white, low-density, silica ceramic.
TOPICS: Breaking News; Editorial; Extended News; Front Page News; Government; Miscellaneous; News/Current Events; Your Opinion/Questions
KEYWORDS: columbiaaccident; nasa; shuttle; sts; sts107
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To: Budge
Good job Budge- I don't know how you did it- but kudos just the same.
1,721
posted on
02/14/2003 12:17:41 AM PST
by
freepersup
(And this expectation will not disappoint us.)
To: freepersup; bonesmccoy; XBob; wirestripper; All
1708 - BTW, I have everything except coverpage there.
1,722
posted on
02/14/2003 12:21:46 AM PST
by
Budge
(God Bless FReepers!)
To: bonesmccoy
I'm not saying with absolute certainty that it was the MLG door but I am saying it could have been. Suffice to say that something significant came off of the bird and I would venture that the shuttle responded accordingly.
1,723
posted on
02/14/2003 12:23:53 AM PST
by
freepersup
(And this expectation will not disappoint us.)
To: bonesmccoy
OK...here's my verticle ballistics formula
Calculating time to impact based upon verticle acceleration of 32 feet per sec per sec
Previous data point is
13:56:24 GMT
36.48154, -110.4165
H= 220235 ft
V= 21.17 mach
In 6 seconds the vehicle went from 220235 to 219820 feet.
That's about 415 feet.
I'm going to assign an initial verticle acceleration of 0.
Final verticle acceleration of the object is equal to g = 32 feet/sec^2.
If
y = 219820 feet
v(iy) = 0 ft/sec
a(y) = 32 ft/sec/sec
-219820 ft = (0 ft/sec)(t) + 0.5*(-32 ft/sec/sec)*(t^2)
first term on right cancels due to zero, simplifying
-219820 ft = (-16 ft/sec/sec)*t^2
(-219820/-16) = 13738.75 = t^2
square root of 13738.75 = 117.2124 seconds
1,724
posted on
02/14/2003 12:26:11 AM PST
by
bonesmccoy
(Defeat the terrorists... Vaccinate!)
To: Smokin' Joe
It is fairly open. There are some links to drawings posted here but I don't know the exact posts. However, check what I have at Fotki. You may find what you're looking for there. The link is
here.
1,725
posted on
02/14/2003 12:27:49 AM PST
by
Budge
(God Bless FReepers!)
To: bonesmccoy
The first big problem is coming up with density vs. Altitude equation I can put in Excel. I entered values every 10,000ft and came up with a table that I plotted in Excel and did a trend line.
I'll use D=.0024e^(-.00004*Altitude) until I have I have time to make it a little better.
D is in slugs/ft^3 and altitude is in feet.
I'll use CD = one for now.
Drag=Area*CD*AirDensity*V^2
Area would be the average of the six sides.
John
To: bonesmccoy
I'm afraid aero drag plays a big role here! That's much to simple.
To: Budge; All
I managed to get some free space for a limited time.
Here it is:
http://mhga.ca/upload/shuttle.zip
Currently at 180.8K in size
Budge, if you can, please find a more permanent location as I don't know how long this will be up for.
As I mentioned earlier: With this version you can press F1 and it will create a "capture.tga" in the working directory. The image is fairly large in size but you could then use some jpg compressor to reduce if you want to.
To: bonesmccoy
When are we conjecturing that the "object" came off the orbiter?
At the 1st Roll Reversal initiation point?
That's datapoint 46 on NASA's pdf.
i initially estimated about 5 seconds after roll reversal, however we have better info now, so when ever it was about 50 miles north of flagstaff.
1,729
posted on
02/14/2003 12:33:58 AM PST
by
XBob
To: John Jamieson; bonesmccoy
Sitting in recliner marveling at the math.
This is why I had to get dirty to make money! LOL
To: halfbubbleofflevel
Can we put this on the server where my web page is? That would be a lot more permanent. BTW, the link didn't work.
1,731
posted on
02/14/2003 12:37:18 AM PST
by
Budge
(God Bless FReepers!)
To: Budge
what are you using to view it, or download it.
internet explorer.
LOL - I just went to try to open real player and found out I dont even have it installed anymore.
i guess that anti-virus ate it, when i cleaned up the klez virus.
1,732
posted on
02/14/2003 12:39:51 AM PST
by
XBob
To: wirestripper
Sitting in recliner marveling at the math. Me too. When they get outside the 1+1=11 range... :)
1,733
posted on
02/14/2003 12:41:39 AM PST
by
Budge
(God Bless FReepers!)
To: bonesmccoy
Horizontal Ballistics Calculation for object departing at Time of First Roll Reversal
x = v(ix)*t + 0.5*a(x)*t^2
Defined
v(ix) = initial velocity of object when departing vehicle
If v = Mach 21.13 at 13:46:30GMT,(46)1st Roll Reversal initiation
Assuming velocity= 21.13 * (761 mph) = 16079.93 mph
16079.93 mph * (1/3600 hour/sec) * (5280 ft/mile) = 23583.9 ft/sec
a(x) = 0 (is that correct guys?)
Time = 117.2124 seconds
x = (23583.9 ft/sec)*(117.2124 sec) + 0.5*(0 ft/sec/sec)*(117.2124 sec)^2
Since the left most factor drops out,
x = 2764326 feet = 523.5465 miles = 523 miles and 2885 feet
Of course, I am not accounting for any atmospheric factors, drag, etc.
I'll try and plot a map.
1,734
posted on
02/14/2003 12:42:50 AM PST
by
bonesmccoy
(Defeat the terrorists... Vaccinate!)
To: Budge
cute- I am off to bed. Good night all. tighten the screws-
1,735
posted on
02/14/2003 12:43:08 AM PST
by
freepersup
(And this expectation will not disappoint us.)
To: XBob
1732 - IE then ust the other day I got the RP pos just to view it.
1,736
posted on
02/14/2003 12:44:55 AM PST
by
Budge
(God Bless FReepers!)
To: John Jamieson
Yes, I am aware of the fact that drag plays a big role here.
However, I do not have differential equations to relate drag at various altitudes, the prevailing winds on the morning in question, and the ground terrain.
So, the best I can surmise is the basic college ballistics equation and use that as the initial point of reference.
We can at least see where the impact point could have been and how close it is to a populated area.
1,737
posted on
02/14/2003 12:45:55 AM PST
by
bonesmccoy
(Defeat the terrorists... Vaccinate!)
To: Budge
You should be able to do that. You would either give a person the URL or have a link to it on a web page to download it. Strange, I just saw your post and immediatly tried the link - and it worked for me.
To: halfbubbleofflevel
You need to enable sharing. The page comes up as unauthorized to view.
To: freepersup
yeah, me too... gotta get some sleep.
1,740
posted on
02/14/2003 12:50:13 AM PST
by
bonesmccoy
(Defeat the terrorists... Vaccinate!)
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