20040610, 17:20  #1 
6809 > 6502
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Aug 2003
101×103 Posts
7×1,523 Posts 
A question of mass.
This was a puzzle that I heard that a thread in the forum reminded me of.
A Pilot (call him Sven) is about to fly his personal plane with his wife (Inga), her sister (Lena) and brotherinlaw (Lars). Since Sven needs to know how much the plane will be carrying (to calculate fuel and centre of gravity) he needs to know the weight of the passengers (the sum). He knows his own weight. There is a scale available and some paper and a pencil. How can he learn the entire mass of his passengers without the ladies revealling theirs (knowing how ladies are) nor Lars (who is sensative about revealling his (he is a large chap)? The simplest method is the most desired. 
20040610, 18:45  #2 
Jun 2003
The Texas Hill Country
1089_{10} Posts 
The simplest method is to have them all step on the scale at once and read the total weight. (It is a cargo scale, I presume).
Perhaps you should place some additional restrictions on the permitted weighings, resolution, etc. 
20040610, 20:06  #3 
6809 > 6502
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Aug 2003
101×103 Posts
7×1,523 Posts 
It is a bathroom scale (one person at a time).
The weight needs to be known to the nearest 0.5kg. The ideal is a single weighing of each. Only the person being weighed may look at the value on the scale. 
20040610, 20:31  #4 
Jun 2003
The Texas Hill Country
1089_{10} Posts 
Let's assume that everyone accurately knows their own weight, or uses the scale to determine it.
Sven writes some random weight, R, on the paper. He hands the paper and pencil to one of the others (eg Inga). She adds her weight, I, to the number. She retains the part with Sven's number on it, and passes the result, I+R, to another. (eg Lena). She does likewise, passing L+I+R, on to the next. Once all of the other weights have been added, Sven takes the final paper, adds his own weight, subtracts the original R, and knows the total. But note that they each need to know their weight within 125 gm to assure that the total is accurate within the 0.5 kg specification. Last fiddled with by Wacky on 20040610 at 22:37 
20040610, 22:23  #5 
6809 > 6502
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Aug 2003
101×103 Posts
7×1,523 Posts 
Assume for the occasion that the scale produces weights that are accurate to 0.06kg. Therefore, no worries about going over the tolerance. I picked the 0.5 to satisfy the Yanks (~1 pound).

20040830, 06:06  #6 
Jul 2003
A_{16} Posts 
It is true that with Wacky's method Lars won't know exactly what either lady weighs, but he will know what they weigh together, and a glance at their figures will let him make a good guess as to how it splits up. Then again, given that he needs to know the total passenger weight, there's no way around this danger, except for him to have a third party produce the original R and subtract it when done, and use the total to provide the needed fuel without telling Lars. Then again, Lars could guess from the needed fuel .... (sigh) 
20040830, 10:11  #7  
Jun 2003
The Texas Hill Country
441_{16} Posts 
Quote:


20040830, 18:30  #8  
"Bob Silverman"
Nov 2003
North of Boston
2^{3}×3×311 Posts 
Quote:
Wouldn't it just be simpler to measure their total energy and divide by C^2??? I remember taking relativity theory from Prof. Paul Bamberger in frosh physics. He used to pose problems about Sven Svenson and his 'relativistic train'. A typical problem (involving relativistic vector additions) was something like: Sven is travelling in his relativistic train, rest mass 1000 tons, at .975C. He collides inelastically with an asterod, rest mass 2 billion tons which is moving at .985C. The collision angle is 30 degrees. Describe the resulting vectors after the collision. I got full credit for the following (semi) facetious answer: PLASMA. 

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