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The Voyage of the Mako (Posted on 2011-04-05) Difficulty: 3 of 5

The space freighter ‘Mako’ trades out of Antares. It travels at constant speed through hyperspace and its route is as follows:

ITINERARY
Antares-Alniyat ...........7 days
Alniyat- β Normae .......4 days
β Normae-1 Scorpii .....8 days
1 Scorpii-Antares ........5 days
Antares- π Scorpii .......7 days
π Scorpii-1 Scorpii.......3 days
1 Scorpii-Alniyat .........5 days
Alniyat-Dschubba ........3 days
Dschubba-Akrab .........6 days
Akrab-π Scorpii ..........4 days
π Scorpii- β Normae .....6 days
(Parts of days ignored throughout)

How many days would it take the ‘Mako’ to travel back home from β Normae to Antares?

See The Solution Submitted by broll    
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Some Thoughts an attempt at a solution | Comment 4 of 14 |

The distances from pi to Akrab, Akrab to Dschubba and Dschubba to Alniyat seem superflouous, as they add up to 13, which is much looser than the constraint imposed on the Alniyat-pi distance imposed by Alniyat-1 Sco-pi total of 5+3=8.

As seen from 1 Scorpii, the angular separations in degrees in the sky that matter, as determined by plane trigonometry, are:

Aln to Beta: 24.14684799650239
Beta to pi: 39.57121945723089
pi to Antares: 120
Antares to Aln: 88.8540080016114

These form a quadrilateral on 1 Scorpii's celestial sphere, which allows some leeway in its angles.

We are interested in the number of decrees separating Antares from Beta Normae in 1Sco's sky, from which, knowing their respective distances of 5 and 8 respectively, using plane trig, we can figure their mutual distance.

Let's try varying the Antares-Alniyat-Beta angle of the quadrilateral to see what its effects are.

DEFDBL A-Z
pi = ATN(1) * 4
dr = pi / 180
DEF fnac (x) = pi * ABS(x < 0) + ATN(SQR(1 - x * x) / x)

antAln = 88.8540080016114# * dr
alnBeta = 24.14684799650239# * dr
betaPi = 39.57121945723089# * dr
piAnt = 120 * dr


FOR angle1 = 91 TO 179 STEP 2
    cAntBeta = COS(antAln) * COS(alnBeta) + SIN(antAln) * SIN(alnBeta) * COS(angle1 * dr)
    distAntBeta = SQR(25 + 64 - 5 * 8 * cAntBeta)
    cosPi = (cAntBeta - COS(piAnt) * COS(betaPi)) / (SIN(piAnt) * SIN(betaPi))
    PRINT USING "### ###.#### ###.####"; angle1; distAntBeta; cosPi
NEXT

The program uses spherical trig to find the angular distance from Antares to Beta Norma in 1Sco's sky, and then uses plane trig to find the linear distance from one to the other. As a check, the cosine of the angle in the 1Sco's sky quadrilateral at pi Scorpii to see if it's a reasonable cosine. In all cases it does seem to be that:

varied    sought   check
angle    distance  cosine
 91       9.4104   0.7188
 93       9.4407   0.6929
 95       9.4708   0.6671
 97       9.5007   0.6413
 99       9.5304   0.6157
101       9.5599   0.5902
103       9.5891   0.5649
105       9.6179   0.5398
107       9.6464   0.5149
109       9.6745   0.4903
111       9.7022   0.4660
113       9.7295   0.4420
115       9.7562   0.4184
117       9.7825   0.3951
119       9.8082   0.3723
121       9.8334   0.3499
123       9.8580   0.3279
125       9.8820   0.3065
127       9.9053   0.2855
129       9.9280   0.2651
131       9.9500   0.2453
133       9.9713   0.2261
135       9.9919   0.2075
137      10.0117   0.1895
139      10.0308   0.1722
141      10.0491   0.1556
143      10.0666   0.1396
145      10.0832   0.1244
147      10.0990   0.1099
149      10.1140   0.0962
151      10.1281   0.0833
153      10.1413   0.0711
155      10.1537   0.0598
157      10.1651   0.0493
159      10.1756   0.0396
161      10.1852   0.0307
163      10.1939   0.0227
165      10.2016   0.0156
167      10.2084   0.0093
169      10.2142   0.0040
171      10.2190  -0.0005
173      10.2229  -0.0041
175      10.2258  -0.0068
177      10.2278  -0.0086
179      10.2287  -0.0095

So it would seem that 9 or 10 would work as an answer, though there might be other constraints that I overlooked. And of course, I'm here treating the numbers as exact, even though they are rounded down.


  Posted by Charlie on 2011-04-06 14:50:41
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