## Business hours

**Weâ€™re OPEN Most days about 9 or 10, occasionally as early as 7, But some days as late as 12 or 1. Weâ€™re CLOSED About 5:30 or 6, occasionally about 4 or 5, But sometimes as late as 11 or 12. Some days or afternoons we arenâ€™t here at all But lately, weâ€™ve been here just about all the time, Except when weâ€™re someplace else. But we should be here then, too.Â …Â sign seen in Henniker, New Hampshire.**

## One my fav cartoons ever.

## Martin Gardner: Mathematical Games : A Cipher I broke that Defeated Poe

Peter Bakke : I solved this cipher in 1987 while working on my master’s degree in systems science at SUNY Binghamton (University of Binghamton, now).

“Ge Jeasgdxv,

Zij gl mw, laam. xzy zmlwhfzek

ejlvdxw kwke tx lbr atgh lbmx aanu

bai Vsmukkss pwn vlwk agh gnumk

wdlnzweg jnbxw oaeg enwb zwmgy

mo mlw wnbx mw al pnfdcfpkh wzkex

hssf xkiyahul. Mk num yexdm wbxy

sbc hv wyx Phwkgnamcuk?”

In 1839, in a regular column Edgar Allan Poe contributed to a Philadelphia periodical, Alexander’s Weekly Messenger, Poe challenged readers to send him {cryptograms (monoalphabetic substitution ciphers), asserting that he would solve them all “forthwith.” One G. W. Kulp submitted a ciphertext in longhand. It was printed as shown above in the issue of February 26, 1840. Poe “proved ” in a subsequent column that the cipher as a hoaxâ€””a jargon of random characters having no meaning whatsoever.” In 1975 Brian J.Winkel, a mathematician at Albion College, and Mark Lyster, a chemistry major in Winkel’s cryptology class, cracked Kulp’s cipher. It is not a simple substitution â€” Poe was right â€” but neither is it nonsense. Poe can hardly be blamed for his opinion. In addition to a major error by Kulp there are 15 minor errors, probably printer’s mistakes in reading the longhand. Winkel is an editor of a new quarterly, Cryptologia, available from Albion College, Albion, Mich. 49224, at $16 per year. The magazine stresses the mathematical and computational aspects of cryptology. The first issue (January, 1977) tells the story of Kulp’s cipher and gives it as a challenge to readers. So far only three readers have broken it.

## Symphony of Science / A Glorious Dawn / Carl Sagan / Stephen Hawking

## Symphony of Science / Carl Sagan

## Famous computer science problems / algorithms

## The width of the Veil Nebula is 3 fingers

So how does one calculate the width (in degrees) of any distant object – from say the Veil Nebula 2100 light years away to the width (in degrees) of that mountain just 21 miles away? Simple. But you will need a scientific calculator. Here’s one:Â http://web2.0calc.com/

Using the information given to you, say in an article about the Veil NebulaÂ , calculate a ratio of distance over width.

In this case, we are told the Veil Nebula is 2100 light years away and it is 110 light years in width. It is very convenient that they gave us distance and width in the same units (light years), else we’d have to do that ourselves.

So, 2100 / 110 = 19.1

Using your scientific calculator (see above), enter 19.1 and hit the COT function (COTANGENT).

The answer of 5.9 (degrees), or about the width of your three fingers extended at arms length towards the fabulous night sky – in this caseÂ in the direction ofÂ the Swan Constellation.

Happiness is Astronomy !

(Conversely, of course, if you estimate the distance of an object and also estimate its width (in degrees), then you can determine the approximate width ofÂ the object {in feet, miles, light years, etc.}).

###### Related articles

## The Veil Nebula. What’s a few trillion miles between friends?

Article Source:Â http://www.chron.com/news/nation-world/space/article/NASA-s-Hubble-images-of-Veil-Nebula-supernova-6530074.php

I believe someone’s math is incorrect in this article. I’m not spoiling the fun, just correcting the record. The image is truly breathtaking and wondrous! But Numeracy counts. :~)

63,000 just didn’t seem big enough, and I was right. Somebody’s calculation re: the width of the Veil Nebula is off by a factor of 100.

Given:

A light year is approx 6 trillion miles, or 6,000,000,000,000 miles.

At a width of 110 light years, the Veil Nebula is therefore about 660,000,000,000,000 miles across.

The sun is about 93 million miles away, or 93,000,000.

If you do the math (660 trillion / 93 million), 110 light years across is not about 63,000 times the distance between earth and the sun as claimed in the article, but instead is about 7,000,000 (7 million) times! An even more incomprehensible number than 63,000, I know.

But, what’s a few trillion miles between friends?

###### Related articles

## Cool English Town names

*Bognor Regis
Beachy Head
Royal Turnbridge Wells
Snave
Smarden
Great Mongeham
Foulness Island
Pratfield Mortimer
Bumbles Green
Pilgrims Hatch
Shellow Bowells
Great Oxney Green
Messing
Cressing
Walton on the Naze
Preston Bissett
Maids’ Moreton
Hinton on the Hedges
Rotherfield Peppard*