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Post by johnsmith on Jul 21, 2009 16:18:45 GMT -4
According to the video at nssdc.gsfc.nasa.gov/planetary/lunar/apollo_15_feather_drop.html, h = 1.5 m, g_moon = 1.622 m/s^2, then the free-fall time interval should be t = sqrt(2h/g_moon) = 1.36 sec. However, the hammer is in free fall for at most 1.1 seconds. Is there an explanation concerning this matter? Note: According to the NASA web page, the height was even 1.6 m. nssdc.gsfc.nasa.gov/planetary/lunar/apollo_15_feather_drop.htmlQuote: "During the final minutes of the third extravehicular activity, a short demonstration experiment was conducted. A heavy object (a 1.32-kg aluminum geological hammer) and a light object (a 0.03-kg falcon feather) were released simultaneously from approximately the same height (approximately 1.6 m) and were allowed to fall to the surface." David Scott is 1.83 m (6 feet) tall. One should also add the thickness of the lunar boots (several centimeters). So the height of 1.5-1.6 m seems to be the correct estimate. |
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Post by blackstar on Jul 21, 2009 16:40:02 GMT -4
The video has been converted from its original form to Quicktime, which would involve a degree of data compression and loss of information, you cannot take any useful measurements from this sort of online video.
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Post by johnsmith on Jul 21, 2009 16:52:49 GMT -4
One can assume that the video rate has been altered after converting the video from analog tape to digital format.
However, this does not explain why the audio rate remains unaltered.
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MarkS
Earth
Why is it so?
Posts: 101
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Post by MarkS on Jul 21, 2009 16:54:32 GMT -4
Is there an explanation concerning this matter? I would suspect error in the values for time and/or height. The Moon's value for the acceleration due to gravity is well understood, as is behavior of objects in free fall. |
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Post by johnsmith on Jul 21, 2009 17:09:36 GMT -4
Well, the height info originates from NASA, approx. 1.6 m.
Concerning the time interval (not exceeding 1.1 seconds), the unaltered voice rate is synchronous with the (allegedly altered?) video rate.
Thus the free fall duration is justified by the audio synchronicity.
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Post by blackstar on Jul 21, 2009 17:09:41 GMT -4
One can assume that the video rate has been altered after converting the video from analog tape to digital format. However, this does not explain why the audio rate remains unaltered. Sorry but on what do you base the assumption that the audio rate is unaltered? Also if you accept the video rate has been altered then why try and make detailed measurements from this, find a high quality uncompressed source and take measurements from that, surely the sensible approach? There is nothing alleged here. The whole point of formats such as Quicktime is to compress video for the web, information is lost in the process, that's just the way it works, find a high res source and take your measurements there. If on the other hand you have an alternate theory about the 'discepancy' then please just come out and state it. |
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Post by johnsmith on Jul 21, 2009 17:16:58 GMT -4
If someone changes the audio rate by
((1.36/1.1)-1)x100% = 23.6%
this would be quite noticeable, isn't it?
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Post by drewid on Jul 21, 2009 17:33:21 GMT -4
One can assume that the video rate has been altered after converting the video from analog tape to digital format. However, this does not explain why the audio rate remains unaltered. Depends how the compression is done and the video/audio streams are interleaved. Really it would be best to find a higher quality source. |
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Post by johnsmith on Jul 21, 2009 17:58:21 GMT -4
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Bob B.
Bob the Excel Guru?
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Post by Bob B. on Jul 21, 2009 18:20:24 GMT -4
Well, the height info originates from NASA, approx. 1.6 m. I don't care if it's posted on a NASA website or not, the height estimate of 1.6 m is clearly not right. The hammer head is no higher than Scott's mid-torso. I'd estimate it to be maybe 1.1 to 1.2 meters, certainly not 1.5 or 1.6. |
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Post by blackstar on Jul 21, 2009 18:31:09 GMT -4
Well that's 15 minutes of my life I'm never getting back; I downloaded the 8Mb and the 80Mb files to view. Yes the 80Mb file is better quality but its still not very good, you seriously can't be trying to take accurate timings for the drop from that can you? The original was clearly a TV broadcast and not great quality to start with and after its been through Quicktime conversion its nothing more than a nice illustration of Gallileo's observation. Again if you have an alternate explanation please just save a lot of time and just lay it out. |
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Post by johnsmith on Jul 21, 2009 18:50:34 GMT -4
This statement contradicts the video observation.
If 1.83-meter-tall person with lunar boots measures the height to its navel, the distance would already be approximately 1.2 meters.
However, the astronaut clearly raises his hands at chest-level before dropping both hammer and feather.
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Post by BertL on Jul 21, 2009 18:53:54 GMT -4
I distinctly remember discussing this a couple of years ago. I also distinctly remember using better quality footage at the time. EDIT: Wait a sec, I just found it. Let me find a place to upload it real soon (this should take five minutes). EDIT 2: Direct link: poopcake.netau.net/tings/a15v_1672206.mpg |
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Post by johnsmith on Jul 21, 2009 19:01:45 GMT -4
I present the contradicting data asking for an explanation.
I would prefer not to speculate providing possible answers to the reported contradiction.
So far, the only 'explanation' I obtained as a feedback is that the QuickTime video posted by NASA might be allegedly altered after conversion.
Once again, there is no noticeable alteration of the audio.
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Post by blackstar on Jul 21, 2009 19:03:41 GMT -4
You can argue about the exact level on the suit at which Scott drops the hammer and feather from, and to me he looks slightly bent forward but its subjective, I also think the pauses in his speech before he drops and after the landing renders issues of audio synchonisation impossible to determine, again my subjective take.
You have a piece of video that has been compressed for web use and wasn't especially hi res to begin with I suspect. It's also hard to determine the drop height with the precision needed to calculate the drop time. All you can say with any certainty is that the two are released at the same time and land at the same time, and that given the limitations of the video that time is broadly within what one would expect for 1/6th g and a vacuum.
If you find the explanations you have been offered unsatisfactory then please, please, give us your take on what actually happened.
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