tag:blogger.com,1999:blog-29179089.post1817683257368488576..comments2017-07-27T02:07:38.725+10:00Comments on Planetary Vision: Space reentry vehicles, part 7Paul Clarkhttp://www.blogger.com/profile/11326150646151735913noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-29179089.post-4858527798685958632014-01-16T18:07:52.293+10:002014-01-16T18:07:52.293+10:00Thanks for the info. I haven't considered thos...Thanks for the info. I haven't considered those angles before.Paul Clarkhttps://www.blogger.com/profile/11326150646151735913noreply@blogger.comtag:blogger.com,1999:blog-29179089.post-87572605371712072082014-01-04T10:20:41.653+10:002014-01-04T10:20:41.653+10:00Sir,
It is rather easy to code up Allen and Eggars...Sir,<br />It is rather easy to code up Allen and Eggars formulas (which are very simple and elegant, BTW) for reentry deceleration and compute quantities such as the dynamic pressure acting on these sorts of reentry vehicles. I have done so and arrived at values that max out (for ICBM warheads) at multiple (>10X) hundreds of Kilo-Pascals. However, atmospheric pressure at sea level is itself around 100 Kilo-Pascals, so this is not a horribly high value after all. Again, when you uses these same formulas for computing deceleration of warheads, you get generally at most about 120 g's, but for reentry angles and ballistic coefficients that are much higher than employed for manned reentry. Prompted by your line of inquiry here, I lowered both of these quantities in some of my tools for implementing these formula in order to be more representative of manned reentry (even ignoring lift) and was able to lower the maximum level of deceleration g's to around 8-10 g's - suitable for manned flight. --><br />Then, if you multiply the 10 g's = 100 m/s^2 by the command module mass of ~5000 Kg, you get a maximum dynamic pressure level of ~500,000 Pascals = 500 kPascals, which is only 5X sea level atmospheric pressure. I would encourage you to investigate Allen's wonderful formulas (If my memory serves correct, I believe the reference you consulted for your plots above has some of the formulas documented.), and you might begin to see how reentry was a dicey proposition, but within the realm of engineering solution. It turns out that heat loads were probably more worrisome than the dynamic loads to NASA engineers, once they turned to blunt RVs (meaning low ballistic coefficient values) reentering at low reentry angles.<br />Regards,<br />RayManfred Arcanehttps://www.blogger.com/profile/03095676652874371048noreply@blogger.comtag:blogger.com,1999:blog-29179089.post-49288241073905895562014-01-02T18:02:53.600+10:002014-01-02T18:02:53.600+10:00Manfred, well spotted. I thought about including t...Manfred, well spotted. I thought about including that consideration initially in the post but got lazy.<br /><br />This is a fair criticism, and I have updated the post to reflect that. I appreciate the effort you made to find and point out that flaw, and I hope I have addressed your concerns in the update above.Paul Clarkhttps://www.blogger.com/profile/11326150646151735913noreply@blogger.comtag:blogger.com,1999:blog-29179089.post-67911672715602718922013-12-31T09:47:34.138+10:002013-12-31T09:47:34.138+10:00Sir,
your calculation for dynamic pressure on Apol...Sir,<br />your calculation for dynamic pressure on Apollo makes a colossal error. By the time Apollo is at 10 km, its velocity is less than 1 km/sec (refer to: http://www.aulis.com/images_reentry/apollo-8_re-entry.jpg). It will have slowed down by close to a factor of 30 from 10 km/sec, in fact. Hence your projection of pressure is off by three orders of magnitude.<br />Regards.<br />PS. The two figures you present I have seen before, and am convinced are really screwy. The only way to get those altitudes for maximum deceleration is for a ballistic coefficient to be extremely high, much, much higher than for even a reentering warhead. Don't know why they made that choice in their document, but maximum deceleration for a manned spacecraft reentering would be in excess of 30km altitude. The reentry flight path angle would be very shallow, so as to cause reentry deceleration not to exceed about 8 G's, the upper bound on what humans can endure. This drives the pressure down considerably, because it is the pressure that causes the deceleration that must be bounded above to be tolerable. The 100's of G's shown in those two plots clearly does not correspond to manned reentry.Manfred Arcanehttps://www.blogger.com/profile/03095676652874371048noreply@blogger.com