Goodwill Estate Pipes (Pic Heavy)

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spartan

Lifer
Aug 14, 2011
2,963
7
These are the before pictures of some pipes I got off of the Goodwill Auction website thing. Got these pipes for $10 + $13 shipping. I bought this lot because I wanted the Bulldog and the Ventilator.
I was lazy and took most these indoors.
Here's a quick shot of all of them.

From top to bottom:
Honeymaster

Kaywoodie Billiard (I'm still learning all the shapes. So if it's not a billiard then please correct me)

Kaywoodie Bulldog

Medico Ventilator
GoodwillPipes.jpg

Honeymaster

This looks to have one of those stingers that I can just rip out. I look forward to doing so, but will smoke it with the stinger first.
Honeymaster1.jpg


Honeymaster2.jpg


Honeymaster3.jpg


Honeymaster4.jpg


Honeymaster5.jpg

Kaywoodie Billiard
KaywoodieBilliard1.jpg


KaywoodieBilliard2.jpg


KaywoodieBilliard3.jpg


KaywoodieBilliard4.jpg

Kaywoodie Bulldog

You'll see that this pipe has an overclocked thing-a-ma-jiger, but it does not affect the stem. When it's all screwed in it's perfect. So I hope I can cran it into position without affecting the stem alignment.
KaywoodieBulldog1.jpg


KaywoodieBulldog2.jpg


KaywoodieBulldog3.jpg


KaywoodieBulldog4.jpg


KaywoodieBulldog5.jpg


KaywoodieBulldog6.jpg

Medico Ventilator
MedicoVentilator1.jpg


MedicoVentilator2.jpg


MedicoVentilator3.jpg


MedicoVentilator4.jpg


MedicoVentilator5.jpg

Will post cleaned picks later, Still need to purchase some wax & Micromesh.

 

cynyr

Part of the Furniture Now
Feb 12, 2012
646
113
Tennessee
Nice deal on those items, and the price was certainly right!
I dig the Bulldog most - I'm gonna have me a Rhodesian or a 999 or bust.

 

spartan

Lifer
Aug 14, 2011
2,963
7
On a side note that marble chess board I was taking photos on is currently being repaired by me. There was a panel that came "unglued"
It's currently receiving an epoxy treatment.
I've never used it. /sigh

 

spartan

Lifer
Aug 14, 2011
2,963
7
They're just pipes not a complex math problem.
Find real values of the number a for which a.i is a solution of the polynomial equation

z4 - 2z3 + 7z2 - 4z + 10 = 0.

Then find all roots of this equation.
Solution
Since a.i is a solution of the equation, we have
(a.i)4 - 2(a.i)3 + 7(a.i)2 - 4(a.i) + 10 = 0
a4 + 2.i.a3 - 7a2 - 4.i.a + 10 = 0
a4 - 7a2 + 10 = 0 and 2a3 - 4a = 0
a2 = 2
a = sqrt(2) or a = - sqrt(2)
Now, we know that sqrt(2).i and - sqrt(2).i are roots of z4 - 2z3 + 7z2 - 4z + 10 = 0 .

This means that z4 - 2z3 + 7z2 - 4z + 10 is divisible by (z - sqrt(2).i)(z + sqrt(2).i) = z2 - 2.

The quotient is (z2 - 2 z + 5) . The roots of this polynomial are 1 + 2 i and 1 - 2 i.
The four roots of the given equation are

sqrt(2).i -sqrt(2).i 1 + 2 i 1 - 2 i.
I think I'll be fine.

 

spartan

Lifer
Aug 14, 2011
2,963
7
And you have made a mistake in your spelling; it's SPARTAN!!! :twisted:
And I just copied that math problem from a google search lol. I can barely do Pre-Cal. Couldn't wrap my head around the concept. I'm sure now as a 24 year old I would have many more questions about everything in class than I did in my teens...
I'll be going back to school soon............-ish.

 

1robert

Lurker
Dec 25, 2011
13
0
Good catch there pipeinhand! We all know that 42 is the answer to the ultimate question of Life, The Universe, and Everything...well, that and a good pipe filled with your favorite tobacco.

 
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