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Tesseract, v. 2, issue 1, January 1937

4     tesseract

A PECULIAR PHENOMENON OF SUPER SPEED TRAVEL 
by Raymond Van Houten
(Author's note. This article has been chocked over by my science prof. who states that it is all right scientifically, providing the reader doesn't believe too strongly in the Fitzgerald contraction theory. It is purely hypothetical, however, and should antagonize no one's [set] theories.)

From A to E - - 1.00 seconds
"   A to C - - 1.74   "   "
Also, since OB is a light second, we find that these times are true:
Light goes from G to 0 in 1.45 sec.
"   "   " D to 0 in 1.23   "
"   "   " F to 0 in 1.20   "
"   "   " B to 0 in 1.00   "

C   E   B   F   D   G   A
2   12   10   12   123   145   2
O

To an observer who was quick enough, an object, say a space ship, traveling past him at twice the speed of light would look very peculiar, indeed. He would not see a space ship go tearing across his line of vision, but rather a broken strip of shimmer that at first would be unintelligible. There would be no sequence in the motion of the ship, for in certain parts of its course it should appear to move backwards. For proof of this refer to the above diagram.
Along the straight line AC our space ship is traveling at twice the speed of light. An observer, O, is placed 1 light second, represented by 4 centimeters, from the course. A and C are each 2 light second from O, CO and AO each being 8 cm. On AC locate the following points: G, 2.8 cm. from A; D, 4 cm. from A; F, 6 cm. from A; B is already determined, OB is the perpendicular from O to AC and is 4 cm. long; E, 8 cm. from A. AC will be 13.95 cm. long.
So much for the construction, now for the proof. All times are from the instant that the ship is at A. On the scale that we are using, 4 cm. to the light second, our space ship would travel twice that or 8 cm. per sec. from A to E. Therefore, using the above measurements, the following times are calculated:
From A to G - - .35 seconds.
"   A to D - - .40   "   "
"   A to F - - .75   "   "
"   A to B - - .87   "   "

Light goes from E to O in 1.20 sec.
"   "   "   A to 0 in 2.00   "
"   "   "   C to 0 in 2.00   "

Adding the adjacent times we see that the observer will see the space ship at:
G   1.80 sec. after it was at A
D   1.63   "   "   "   "   "   "
F   1.95   "   "   "   "   "   "
B   1.87   "   "   "   "   "   "
E   2.20   "   "   "   "   "   "
C   3.74   "   "   "   "   "   "
In other words the observer will see the ship first at D, then at G, then at B, then at F, then at A, then at E, and last at C. Probably the plotting of other points along AC will bring to light other peculiarities, but these will do for a start. Try finding the point at which he will see the ship first. It will be located very close to D, if it is not D. There is a point between B and E where the total time will be 2.00 sec. Then the observer will see the ship at this point and at A simultaneously. Curious facts. Nevertheless they are facts, upheld to be simple calculations that are intelligible to everyone.

We thank the following for their letters and comments: R.A. Madle, R.A. Squiers, H. P. Lovecraft, Willis Conover, Jr., William S. Sytora, Henry Lemaire, J. Harvey Haggard, and L. A. Ware.

SEASON'S GREETINGS
SFAA Trustees and Tesseract Staff members.

Saving...

4 tesseract A PECULIAR PHENOMENON OF SUPER SPEED TRAVEL by Raymond Van Houten (Author's note. This article has been chocked over by my science prof. who states that it is all right scientifically, providing the reader doesn't believe too strongly in the Fitzgerald contraction theory. It is purely hypothetical, however, and should antagonize no one's [set] theories.) From A to E - - 1.00 seconds " A to C - - 1.74 " " Also, since OB is a light second, we find that these times are true: Light goes from G to 0 in 1.45 sec. " " " D to 0 in 1.23 " " " " F to 0 in 1.20 " " " " B to 0 in 1.00 " C E B F D G A 2 12 10 12 123 145 2 O To an observer who was quick enough, an object, say a space ship, traveling past him at twice the speed of light would look very peculiar, indeed. He would not see a space ship go tearing across his line of vision, but rather a broken strip of shimmer that at first would be unintelligible. There would be no sequence in the motion of the ship, for in certain parts of its course it should appear to move backwards. For proof of this refer to the above diagram. Along the straight line AC our space ship is traveling at twice the speed of light. An observer, O, is placed 1 light second, represented by 4 centimeters, from the course. A and C are each 2 light second from O, CO and AO each being 8 cm. On AC locate the following points: G, 2.8 cm. from A; D, 4 cm. from A; F, 6 cm. from A; B is already determined, OB is the perpendicular from O to AC and is 4 cm. long; E, 8 cm. from A. AC will be 13.95 cm. long. So much for the construction, now for the proof. All times are from the instant that the ship is at A. On the scale that we are using, 4 cm. to the light second, our space ship would travel twice that or 8 cm. per sec. from A to E. Therefore, using the above measurements, the following times are calculated: From A to G - - .35 seconds. " A to D - - .40 " " " A to F - - .75 " " " A to B - - .87 " " Light goes from E to O in 1.20 sec. " " " A to 0 in 2.00 " " " " C to 0 in 2.00 " Adding the adjacent times we see that the observer will see the space ship at: G 1.80 sec. after it was at A D 1.63 " " " " " " F 1.95 " " " " " " B 1.87 " " " " " " E 2.20 " " " " " " C 3.74 " " " " " " In other words the observer will see the ship first at D, then at G, then at B, then at F, then at A, then at E, and last at C. Probably the plotting of other points along AC will bring to light other peculiarities, but these will do for a start. Try finding the point at which he will see the ship first. It will be located very close to D, if it is not D. There is a point between B and E where the total time will be 2.00 sec. Then the observer will see the ship at this point and at A simultaneously. Curious facts. Nevertheless they are facts, upheld to be simple calculations that are intelligible to everyone. We thank the following for their letters and comments: R.A. Madle, R.A. Squiers, H. P. Lovecraft, Willis Conover, Jr., William S. Sytora, Henry Lemaire, J. Harvey Haggard, and L. A. Ware. SEASON'S GREETINGS SFAA Trustees and Tesseract Staff members.