Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP Posting-Version: version B 2.10.2 9/5/84; site mnetor.UUCP Path: utzoo!utcs!mnetor!clewis From: clewis@mnetor.UUCP (Chris Lewis) Newsgroups: net.origins Subject: Re: more on large animals and gravity Message-ID: <1942@mnetor.UUCP> Date: Sun, 1-Sep-85 12:18:00 EDT Article-I.D.: mnetor.1942 Posted: Sun Sep 1 12:18:00 1985 Date-Received: Sun, 1-Sep-85 13:30:42 EDT References: <382@imsvax.UUCP> <1905@mnetor.UUCP> Reply-To: clewis@mnetor.UUCP (Chris Lewis) Organization: Computer X (CANADA) Ltd., Toronto, Ontario, Canada Lines: 78 Summary: In article <1905@mnetor.UUCP> clewis@mnetor.UUCP (Chris Lewis) writes: >Yeah, and the Verranzo Narrows bridge collapsed due to an incomplete >understanding of aerodynamics and resonance. Even more modern bridges and >structures are collapsing. Oops, "Tacoma Bridge". The book I mentioned about planets in very close orbit is: > > The Flight of the Dragonfly - Robert L. Forward PhD 1984 > greg >-- >Gregory J.E. Rawlins, CS Dept., U.Waterloo, Waterloo,Ontario N2L3G1,Canada Thanks Greg. (lsuc!msb got your reply and forwarded it to me. How it was sent to him is beyond us). Another interesting tidbit: If, Venus was orbitting with a radius of 9800 miles as I calculated as a requirement to make the gravity on the side of Earth facing Venus approx .5G (assuming Venus the same mass as the earth to a first approximation), the gravitational vector at 90 degrees to the Earth-Venus axis at point X would be: ----- / \ | A | Venus \__|__/ | --|-- / | \ | B--X Earth \_____/ 1 G towards centre of Earth + gravity from Venus on A-X direction. Let's see: A-B = 9800 mi. B-X = 3500 mi. A-X = sqrt(A-B*A-B + (B-X)*(B-X)) = 10406 mi.; Gravity at 3500 mi. from Venus centre is 1G, => at 10406 mi. is: (3500 / 10406) squared = .11 G on A-X axis B-X-A angle = atan(9800 / 3500) = 70 degrees A-B component of A-X gravity = arcsin(70) * .11 = .1 B-X component of B-X gravity = arccos(70) * .11 + 1 = 1.03 Result vector angle = arctan(.1 / 1.03) = 5.7 degrees. Hence, the gravitational vector would be 5.7 degrees off of the B-X line, towards A. Since water always assumes a surface perpendicular to the gravitational vector, we can then see that water would rest 5.7 degrees off of the "normal" horizontal plane. Just to give you an idea of how BIG the tides would be in such a situation, consider how high 5.7 degrees IS after a 100 miles: (in a 100 miles the Earth hasn't curved much, and the vector equations would still hold approximately) = 100 * tan(5.7) = 9.9 MILES! Tides under such a circumstance would be so high as to mash EVERYTHING flat. So much for Velikovskian theories about reduced gravity from other planets allowing larger animals. We've already established that the moon cannot orbit close enough to have any appreciable effect (if it was orbitting with a separation of 0, the effective gravity would only be 5/6 G.) If the planet was MUCH larger than the Earth - so it could be farther away to have the same effect of reducing the gravity by .5, it is easy to see that the tidal effects would be worse. Eg: assume a mass big enough to be far enough away that 3500 miles would have no appreciable effect in attenuating gravity. Then, not only would the gravity on the side of the Earth facing it be a sum of 1G + .5G the other way (to halve gravity), the vector at the X position would be approximately 1G down plus .5G at 90 degrees. Then you'd have tide angles approaching 20-30 degrees! I sure wouldn't want to live there!!!!!! Anybody want to do the full Calculus to figger out how high the tides would be? -- Chris Lewis, UUCP: {allegra, linus, ihnp4}!utzoo!mnetor!clewis BELL: (416)-475-8980 ext. 321