A question of mass.

Uncwilly · 2004-06-10, 17:20

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 brother-in-law (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.

Wacky · 2004-06-10, 18:45

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.

Uncwilly · 2004-06-10, 20:06

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.

Wacky · 2004-06-10, 20:31

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.

Uncwilly · 2004-06-10, 22:23

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).

rpresser · 2004-08-30, 06:06

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)

Wacky · 2004-08-30, 10:11

Quote:

It is true that with Wacky's method Lars won't know exactly what either lady weighs,

I believe that you have the names wrong. Sven is the Pilot. He will know the total weight, and to all who have this number, by subtracting his own weight, the sum of the other three passengers. However, none of the occupants can determine any more detail from the data disclosed.

R.D. Silverman · 2004-08-30, 18:30

Quote:

Originally Posted by Wacky

I believe that you have the names wrong. Sven is the Pilot. He will know the total weight, and to all who have this number, by subtracting his own weight, the sum of the other three passengers. However, none of the occupants can determine any more detail from the data disclosed.

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.

2004-06-10, 17:20	#1
Uncwilly 6809 > 6502 """"""""""""""""""" Aug 2003 101×103 Posts 2³·1,327 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 brother-in-law (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.

2004-06-10, 20:31	#4
Wacky Jun 2003 The Texas Hill Country 441₁₆ 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 2004-06-10 at 22:37

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2004-06-10, 18:45	#2
Wacky Jun 2003 The Texas Hill Country 10001000001₂ 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.

2004-06-10, 20:06	#3
Uncwilly 6809 > 6502 """"""""""""""""""" Aug 2003 101×103 Posts 2³·1,327 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.

2004-06-10, 22:23	#5
Uncwilly 6809 > 6502 """"""""""""""""""" Aug 2003 101×103 Posts 2³×1,327 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).

2004-08-30, 06:06	#6
rpresser Jul 2003 1010₂ 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)