Any value for Pi shall give 'generally' simmilar results. An aid to memory for the *Chinese value 113355 was to keep the first three digits in denominator*. For the value I propose, a verse was reported in:
Vedic Numeral Code; ISI Bulln.;V27 N10; p.381 (1975) that read: Kos-shunyam Bhagam Lat-jaloye: to mean *KEEP THE HEAD COOL BY WETTING THE HAIR WITH WATER AND THUS REDUCE DISTANCES TO NOTHING KEEPING AWAY THE FEELING OF TIREDNESS*.
The value could be remembered as: 'One followed by FIVE zeros on division by thirty-one thousand eight hundred and thirty-one' -the denominator has mirror image of 138.
Believe-it-or NOT, NO OTHER VALUE fits to define the value of Pi(�)or that for Radian in 'exact expression' - no truncation! As I mentioned earlier, a normal man of height 2m shall see horizon, at beach, about 5 kilometre; exact distances for shipping can be calculated.
Regards,
Brij B. Vij TIME: to think Metric!<[EMAIL PROTECTED]> <And Calendar too>
From: John Nichols <[EMAIL PROTECTED]>
Reply-To: [EMAIL PROTECTED]
To: "U.S. Metric Association" <[EMAIL PROTECTED]>
Subject: [USMA:24709] Fwd: RE: Odd numbers 1357
Date: Sat, 01 Feb 2003 09:56:22 -0600
Dear Pat,
Some of us just look at the ship and think "that is a nice scene " and then go back to drinking the coke. Where was that beach umbrella?
John
User-Agent: Microsoft-Entourage/9.0.2509John Nichols BE, Ph.D. (Newcastle), MIE (Aust), Chartered Professional Engineer
Date: Sat, 01 Feb 2003 17:30:23 +1100
Subject: [USMA:24708] RE: Odd numbers 1357
From: Pat Naughtin <[EMAIL PROTECTED]>
To: "U.S. Metric Association" <[EMAIL PROTECTED]>
Reply-To: [EMAIL PROTECTED]
Sender: [EMAIL PROTECTED]
Dear Joe,
Thanks for your views. You took a different path but you arrived at the same
point. Let me explain my approach.
1 I made the same assumptions as you did. 'The above results assume that
the earth is a smooth sphere without an atmosphere instead of a spheroid
with mountains and an atmosphere'.
2 I drew a circle showing a radius to the point where the horizon could be
seen and I labelled this 'r' for 'radius of the earth'. I drew another
radius to the location of the observer and extended it by an additional
height labelled 'h' for 'height of observation'. I then joined the ends of
these two lines and labelled the third line of the triangle 'd' for distance
to the horizon.
3 Knowing that the line of sight to the horizon is a tangent, I knew that
the triangle that had been formed was a right angle triangle with the right
angle at the point of the horizon.
4 I then used Pythagoras' theorem to calculate the following results:
h sqrt (h) horizon distance
sqrt (h) x 3.57
1 m 1 3.570 km
2 m 1.4142 5.049 km
3 m 1.7321 6.184 km
4 m 2 7.140 km
5 m 2.2361 7.983 km
.........................
10 m 3.1622 11.289 km
..........................
100 m 10 35.700 km
...........................
1 000 m 31.6228 112.893 km
.............................
10 000 m 100 357.000 km
..............................
100 000 m 316.228 1 129 km
............................
1 000 000 m 1000 3 570 km
5 000 000 m 2236.1 7 983 km
10 000 000 m 3162.3 11 289 km
We are in fairly close agreement on all of these except for your last value
(for a height of 10 000 000) that I suspect is a slip of the calculator.
>From my point of view the number sequence 1-3-5-7 gives me a great 'Rule of
thumb' as it is easy to remember and it gives tolerably accurate results.
As an example, I was standing on a cliff top near Geelong, looking out to
sea, when I saw a vessel on the horizon. I estimated its distance as
follows:
1 I guessed that the height of the cliff was about 20 metres
2 I guessed that the square root of 20 is about 4.5
3 I multiplied 4 by 3.5 to get 14 (actually I doubled 3.5 to 7 and then
doubled that again to get 14
4 I added half of 3.5 to get 15.75
5 I estimated that the vessel was 16 kilometres from where I was standing.
Thanks for your detailed calculations. It gives me confidence to know that
you have independently confirmed this 'Rule of thumb'. I'll add it to my
collection.
Cheers,
Pat Naughtin LCAMS
Geelong, Australia
on 2003-02-01 08.42, Joseph B. Reid at [EMAIL PROTECTED] wrote:
> Pat Naughton wrote in USMA 24669:
>>
>>> From the easily remembered 1357 we have established the basic rule for
>> finding the distance to the horizon and we can use it for any other heights.
>> Let's (for ease of calculation) consider 4 metres.
>>
>>> From a height of 4 m the horizon is about 7 kilometres (sqrt (4) x 3.57).
>> But knowing this requires that we remember the 'constant' 3.57 that I think
>> is readily remembered as part of the sequence 1357.
>>
>> Cheers,
>>
>> Pat Naughtin LCAMS
>
>
>
> This intrigued me. The formyla that I came up with was
> a = arccosine (r/h+r),
> where a = angle subtended by horizon distance at earth center
> r = radius of the earth
> h = height of observation
>
> This formula led to the following results:
> observation height horizon distance 3.57 X sqrt (height)
> 1 m 3.570 km 3.570 km
> 2 m 5.048 km 5.049 km
> 3 m 6.182 km 6.183 km
> 4 m 7.139 km 7.140 km
> 5 m 7.982 km 7.983 km
> .........................
> 10 m 11.288 km 11.289 km
> ..........................
> 100 m 35.696 km 35.700 km
> ...........................
> 1 000 m 112.875 km 112.893 km
> .............................
> 10 000 m 356.732 km 357.000 km
> ..............................
> 100 000 m 1 121 km 1 129 km
> ............................
> 1 000 000 m 3 357 km 3 570 km
> 5 000 000 m 6 218 km 7 983 km
> 10 000 000 m 7 461 km 11 289 km
>
> The above results assume that the earth is a smooth sphere without an
> atmosphere instead of a spheroid with mountains and an
> atmosphere.
Assistant Professor
Texas A&M University
Department of Construction Science
Langford AC
Rm: A414 MD 3137
College Station, TX 77843-3137
Electronic mail: [EMAIL PROTECTED]
Telephone: 979 845 6541
Facsimile: 979 862 1572
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