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Author Topic: Armchair Engineering?  (Read 3899 times)

Offline Larry Fulwider

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Armchair Engineering?
« on: July 07, 2010, 11:39:47 AM »
Some modelers might enjoy this site, or already be familiar with it:
 
http://cr4.globalspec.com/blogentry/13152/

The Engineer Blog has a new puzzle each month, none with obvious answers, and the puzzles usually have hidden subtleties. Anyone who gets the puzzles correct every month is likely quite good at trimming models (and a few other things), I’d guess. Here are two recent examples:

June, 2010: You are in a sailing race on a very wide river. On land there is no wind; it is a dead calm. The race is 10 Km downstream, and the river is running 5 Km/ hr. Your opponents decide that to make the best time, they will streamline their boats to the wind and float downstream as fast as they can. You decide on a tacking strategy, sailing back and forth across the river's width. Who wins the race? What is the winning time?

July, 2010: You're in your car at a stoplight, with a helium-filled balloon pressed up against the center of the ceiling. When the light turns and you accelerate, what does the balloon do and why?


       Larry Fulwider

Offline Jim Thomerson

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Re: Armchair Engineering?
« Reply #1 on: July 07, 2010, 12:12:13 PM »
A couple of non engineer answers.  Tacking would be faster because you would get "lift'' pushing you downstream in addition to the water flow. However, you would have to stay away from the banks, because the water is flowing slower there due to friction. 

The hydrogen filled balloon would move forward as inertia would increase the air density toward the rear of the car. 

Offline Bill Little

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Re: Armchair Engineering?
« Reply #2 on: July 07, 2010, 12:50:55 PM »
Quote
Who wins the race?

Obvious answer to the *Winner* of the boat race is the one which gets to the finish line first.  What do I get for winning?

Big Bear
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Offline Larry Fulwider

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Re: Armchair Engineering?
« Reply #3 on: July 07, 2010, 02:47:07 PM »
Obvious answer to the *Winner* of the boat race is the one which gets to the finish line first.  What do I get for winning?

Big Bear

Bill --

Ahh, I should have mentioned. The staff picks a "good question" and the "correct answer" prior to publishing the question. At the end of the month, they publish the "correct answer" (as written by the person who wrote the the question).
       During the month, people like us supply "answers" and we  also get to vote on whether other answers are "good answers" or "off topic". A lot of votes for "good answer" can mean that you have a lot of loyal friends, or that your peers judged your answer to take into account all the variables and you did a good job of explaining your thinking. All of this voting is without access to the "official" answer.

Anyway, so nobody really "wins", although sometimes a few responses are a tad more complete than the "official answer".
       Yes, you would get a lot of votes for "off topic" (or, in effect, not a serious, but enjoyable, answer)  ::)

As an aside, Jim's density gradient of the air within the car has been pretty well shot down as being trivial compared to the larger buoyancy forces.

       Larry Fulwider


     

Offline Larry Fernandez

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Re: Armchair Engineering?
« Reply #4 on: July 07, 2010, 02:48:17 PM »
  What do I get for winning?

Big Bear
[/quote]

How about a New RO Jett .65 ?!?!?!?!?!?

Larry, Buttafucco Stunt Team


Offline Bill Little

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Re: Armchair Engineering?
« Reply #5 on: July 07, 2010, 03:35:21 PM »
  What do I get for winning?

Big Bear


How about a New RO Jett .65 ?!?!?!?!?!?

Larry, Buttafucco Stunt Team



Hi Larry,

I'll take it!!!!!!!  LL~  You need my address??? 

Mongo
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Offline Randy Powell

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Re: Armchair Engineering?
« Reply #6 on: July 07, 2010, 05:22:40 PM »
This reminds me way too much of: two trains are traveling from LA to San Francisco, one with 4 passengers .....
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Offline Bill Little

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Re: Armchair Engineering?
« Reply #7 on: July 07, 2010, 10:39:20 PM »
This reminds me way too much of: two trains are traveling from LA to San Francisco, one with 4 passengers .....

Or, a plane crashed on the border of Canada and the USA (4 or 5 paragraphs of explanation then the question) where do you bury the survivors.....

But, you'uns gotta admit, my answer is 100% correct! LOL!! 

Big Bear
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Offline Clayton Berry

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Re: Armchair Engineering?
« Reply #8 on: July 08, 2010, 06:28:52 AM »
The sailboat thing is won by the guy with the gun to his head.

The balloon is snatched by the first kid that comes along with a pocket knife to remove it from whatever it is tied to.

There really is a simple answer to most of life's little questions.
Clayton - forever busy committing random acts of coolness

Offline FLOYD CARTER

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Re: Armchair Engineering?
« Reply #9 on: July 08, 2010, 03:13:37 PM »
I used to love the "sailboat"-type problems while in college.

All these similar problems use D=RT as the basic equation.

It is always necessary to write a system of simultaneous equations based on D=RT, and then solve in the usual way, by substitution.

Floyd
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Offline FLOYD CARTER

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Re: Armchair Engineering?
« Reply #10 on: July 09, 2010, 09:57:51 AM »
And here's one of my favorites.  Simply stated, and elegant.

"It is exactly 4 o'clock.  What time will it be when the hands are exactly together."  (give answer to 3 decimal places)

Floyd
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Offline Bill Little

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Re: Armchair Engineering?
« Reply #11 on: July 09, 2010, 12:17:48 PM »
And here's one of my favorites.  Simply stated, and elegant.

"It is exactly 4 o'clock.  What time will it be when the hands are exactly together."  (give answer to 3 decimal places)

Floyd

First answer, no thought: 1200:000 and 2400:000

Big Bear
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Offline Bruce Reynolds

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Re: Armchair Engineering?
« Reply #12 on: July 09, 2010, 01:14:55 PM »
And here's one of my favorites.  Simply stated, and elegant.

"It is exactly 4 o'clock.  What time will it be when the hands are exactly together."  (give answer to 3 decimal places)

Floyd

I get 4:21.818  ;D

Online john e. holliday

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Re: Armchair Engineering?
« Reply #13 on: July 10, 2010, 09:28:28 AM »
So who's correct.   I never could get into formulas and such.  My uncle would leave Kansas City headed for the folks farm.  I would leave at the same time and get there before him.  He was known for exceeding the speed limit most times.  With me it was max 65 mph with the old 48 flat hea Ford.   D>K
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Offline Don Hutchinson AMA5402

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Re: Armchair Engineering?
« Reply #14 on: July 10, 2010, 10:15:48 AM »
Try this one: I have a sphere with a hole drilled through it. The length of the hole is exactly 6 inches. What is the volume of the remainder of the sphere?
Don

Online john e. holliday

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Re: Armchair Engineering?
« Reply #15 on: July 10, 2010, 10:19:29 AM »
Depends on the size of the hole. D>K
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Offline FLOYD CARTER

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Re: Armchair Engineering?
« Reply #16 on: July 10, 2010, 04:55:27 PM »
Maybe that was too easy.  Answer for elapsed time is 20+20/11 , a repeating fraction. So

T=21.818181818.....

a curious result!
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Offline John Cralley

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Re: Armchair Engineering?
« Reply #17 on: July 10, 2010, 05:34:08 PM »
Floyd,

Good math except you dropped a 4 someplace!!  n1

Make it 4:21.8181818... o'clock!!!! ;D
John Cralley
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Offline FLOYD CARTER

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Re: Armchair Engineering?
« Reply #18 on: July 11, 2010, 09:06:26 AM »
Remember I said "elapsed time".

Floyd
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Offline Ward Van Duzer

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Re: Armchair Engineering?
« Reply #19 on: July 11, 2010, 09:23:45 AM »
Typically, most sailboats have a hull speed of roughly six knots (or make it MPH for simplicity). The tacking boat will take way more distance doing the zig-zag routine than the boat travelling in the straight line. Even though the boat being assisted by the wind should be going faster he'll have to cover twice(?) the distance to finish. Therefore the boat riding with the current wins...

(Did it?)


W.  ::)
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They are easier to handle than dumb mistakes!  Ward-O AMA 6022

Offline Larry Fulwider

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Re: Armchair Engineering?
« Reply #20 on: July 11, 2010, 11:01:50 AM »
Typically, most sailboats have a hull speed of roughly six knots (or make it MPH for simplicity). The tacking boat will take way more distance doing the zig-zag routine than the boat travelling in the straight line. Even though the boat being assisted by the wind should be going faster he'll have to cover twice(?) the distance to finish. Therefore the boat riding with the current wins...

(Did it?)


W.  ::)

Ward --

I've generally avoided discussion (in this forum) of the two specific example problems in the original post, for the simple reason that if anyone wants to discuss / debate them, they can go to: http://cr4.globalspec.com/blogentry/12742/Sailing-Newsletter-Challenge-06-01-10 for the sailing comments and "Answer".
       Or, the site in my original post http://cr4.globalspec.com/blogentry/13152 for discussion or debate of the "Balloon Problem". Any discussions here are mostly likely echoes of those in the cr4blog.

However, I will say here that the sailboat problem is a subtle "relative frame of reference" shift. After the initial start, and the boats are floating downstream at a steady pace, the problem can be re-stated from the boat's frame of reference as:
"You are in a wide, but still, waterway with no current. You have a steady 5 km/hr headwind. The walls of the waterway are also moving "backwards" at a steady 5 km/hr. To move your boat forward (or keep it from moving backwards) are you better off tacking into the steady wind, or streamlining your boat to minimize the effects of the headwind?"


As an aside, the reasons I think the Cr4Blog problems (in general) are far, far, more interesting than, say, the clock hand, DRT, algebraic / trigonometry problems, or misleading wording problems is that there are no "trick problems" or "obtuse results" problems.
       The Cr4Blog problems are straightforward examples of applied science and mathematics -- akin to " . . . At what angle to the horizon is the fuel in a model airplane fuel tank flying at 45 degree elevation , , ,".  Some might say, "Well, mostly it sloshes around".
       But for those with the innate curiosity as to "how the world fits together", the fuel level question (and the typical Cr4Blog problem) can an interesting exercise in how various forces interact to produce some specific result.
       Those who favor "schoolbook type puzzles" -- puzzles more like crosswords, conundrums (word trickery), or problems resembling "story problems" in arithmetic will find little joy in the Cr4Blog puzzles.

       Larry Fulwider


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