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Multidisciplinary
Design, Analysis, and
Optimization Branch
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EDUCATIONAL ACTIVITIES: THE NASA AEROQUIZ
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Week of 2/7/00:
Q:
Here's a followup to last week's Super Sunday helium-filled football question
(Click
here to see it)! The weight savings of inflating a football with
helium may be small, but how about inflating bicycle tires with it?
In cycling, why is reducing wheel weight more important than reducing the
weight of the bicycle frame, your body, or other non-rotating masses?
A:
The tires and wheels undergo angular acceleration in addition to
linear acceleration. The helium allows for a lighter total mass
for the tire and wheel.
Congratulations to Joe Slayden.
As Joe says, not only do cyclists need to get wheels moving forward,
but they also need to get them spinning. If you calculate the energy
requirement to set a bike/wheel system in motion, you'll find that the
wheels need more energy per unit mass than the frame and rider do. Exactly
how much more depends on the mass and moments of inertia of the components.
This effect leads to the cyclists' rule of thumb: "A pound off the
wheels is worth two off the frame," which is calculated by analyzing
the wheels as homogeneous disks. Using an approximate tube volume (yes,
a lot of road racers still use tubes!) of about 690 cm3
and a pressure of 125 psig, I calculate a weight savings of about 7 grams.
For typical performance road racing tire, tube, and hub/wheel weights of
250, 70, and 1200 g, respectively, using helium makes the wheel assembly about
half a percent lighter. Percentage-wise, this is less advantageous than
filling a football with helium. However, the wheels need to be turned
over the entire distance of a road course. Lighter wheels, even slightly
lighter wheels, can be significant.
- The Aeroquiz Editor.
Week of 2/14/00:
Thanks to Andrew White for submitting this
week's Aeroquiz!
Q:
A group of airline executives were discussing the proposed
purchase of several new wide-bodied jet aircraft, larger than
any the company currently owned. On the boardroom table were
plans for modification of their maintenance hangar. Although the
floor was strong enough to withstand the increased pressure of the large
tricycle landing gear, the existing doors were three feet too low
for the giant airliner's fin tip. The plans showed that the internal
hangar headroom was adequate, but that additional mini-doors in the gable
ends would be needed for the tailfin. After lunch, an elderly
attendant who was clearing plates could not help looking at the plans,
and offered her own advice on this problem. Her solution was simple
and cheap, earning her a tidy bonus. What did she suggest?
A:
If three feet is all they need to lower the plane, they could lower
it by depressurizing the landing gear struts before they tow the
plane in. (Congratulations to Frank Brown).
Since it has tricycle gear, one possibility would be to raise
the nose of the airplane, thus lowering the tail and allowing it to clear.
(Congratulations to Mark Witt).
Week of 2/21/00:
Thanks to Alan Nies for submitting this
week's Aeroquiz!
Q:
Six aerospace industry executives visit a jet engine component
warehouse. Each executive represents their company. Three of them
are from aircraft manufacturing companies; the other three represent
the engine manufacturing companies that are each partnered with one of
the airframers. Each of them buy as many components as they pay in
dollars for each component (that is, if an executive buys 5 components,
they cost 5 dollars each; if an executive buys 17 components, they
cost 17 dollars each). Each of the aircraft company executives
spends exactly 63 more dollars than their engine company partners.
The airframers are Boeing, Lockheed-Martin, and Gulfstream. The engine
companies are General Electric, Pratt & Whitney, and Honeywell.
Boeing buys 23 more components than Pratt & Whitney, and Lockheed-Martin
buys 11 more components than General Electric. What engine company
is partnered with each airframer?
A:
Boeing is with Honeywell, Lockheed-Martin is with Pratt & Whitney,
Gulfstream is with GE.
Congratulations to Brian Wallenfelt.
Brian didn't show his work, but I'll cut him some slack due to his
Illinois pedigree. (By the way, Brian, it's tough for Illini
alumni here in Buckeye country!) The solution goes something like this:
The amount spent by each executive is a square number, and each
aircraft rep spends $63 more than their engine company partner. First,
we must find three sets of squares that differ by 63:
A12 - E12 = 63 or (A1 + E1)(A1 - E1) = 63
A22 - E22 = 63 or (A2 + E2)(A2 - E2) = 63
A32 - E32 = 63 or (A3 + E3)(A3 - E3) = 63
As seen above, we must find three sets of two factors (integers) whose
product equals 63. Listing the larger factor on the left, and since
(A+E) must always be larger than (A-E), the only possibilities are:
63 x 1 = 63, or A1+E1 = 63; A1-E1 = 1
21 x 3 = 63, or A2+E2 = 21; A2-E2 = 3
9 x 7 = 63, or A3+E3 = 9; A3-E3 = 7
Solving the three sets of simultaneous equations:
A1 = 32
A2 = 12
A3 = 8
E1 = 31
E2 = 9
E3 = 1
These numbers represent the number of items that each rep purchased (as well
as the number of dollars paid for each item).
Since Boeing bought 23 more items than Pratt & Whitney, it is easily
seen that Boeing must be A1 (32 items), and Pratt & Whitney must be E2.
Similarly, Lockheed-Martin must be A2 (12 items), and General Electric
must be E3 (1 item).
And that leaves Gulfstream as A3, and Honeywell as E1.
Solution complete!
Boeing is partnered with Honeywell,
Lockheed-Martin is partnered with Pratt & Whitney,
Gulfstream is partnered with General Electric.
- The Aeroquiz Editor
Week of 2/28/00:
Thanks to Alan Nies for submitting this
week's Aeroquiz! Alan sent another question along with the one he sent
last week It has absolutely nothing to do with aeronautics, but
it's too good to pass up!
Q:
Find the fifteenth number (in ascending order) that is both square
and triangular.
Note: A triangular number is the sum of all consecutive integers from
unity up to a given integer. For example, the triangle of 5 = 1+2+3+4+5 = 15.
A triangular number satisfies the equation n(n+1)/2 where
n is any integer [5 x (5+1)/2 = 15]. Interestingly, Gauss
discovered this relationship while in elementary school. Basically,
the problem is to find the fifteenth number that satisfies the
following equation: N2 = n(n+1)/2, where N
and n are integers (but not necessarily the same integer).
To get you started, here are the first three:
1 (Square for N = 1, Triangular for n = 1)
36 (Square for N = 6, Triangular for n = 8)
1225 (Square for N = 35, Triangular for n = 49)
Alan asks for the fifteenth answer to prevent us from using
a brute-force machine solution (Computers quickly run short of
precision when manipulating numbers this large!). Have fun, and good luck!
A:
The answer is: 2,893,284,510,173,841,030,625.
This is the square for: 53,789,260,175
And it is the triangle for: 76,069,501,249
After using a brute force program to get the first few
terms of the series, I began to analyze the series of
squares and triangles separately. I chose to study
the squares first, since they seemed to have a more
discernable pattern.
After iterating my equations to get the 15th term in
each sequence, I then used a spreadsheet to break
down the multiplication for me and give me the exact
integer value of the square and triangular number in
question.
Congratulations to Joshua Chambers.
Several people got this quiz, but Joshua answered first and best.
He also provided the patterns he noticed, but I had the following, similar,
description prepared, so I'll just paste it here.
Like Joshua, the way we approached it was to write the smaller, easier answers
out and try to find a pattern. A short computer program (eight or so lines)
can help determine this pattern by supplying solutions for a while, until
the machine runs short of precision. Here are the solutions:
N n N2 n(n+1)/2 n =
1 1 1 1 1 12
2 6 8 36 36 32-1
3 35 49 1225 1225 72
4 204 288 41616 41616 172-1
5 1189 1681 1413721 1413721 412
6 6930 9800 48024900 48024900 992-1
7 40391 57121 1631432881 1631432881 2392
This is the point where my Sun Ultra II running double precision Fortran
(yes, Fortran!) quits giving accurate results. But by now, we've recognized the
pattern shown in the last column. n exhibits the pattern
bk = 2bk-1 + bk-2 - C, where C
is 0 (for odd k), and -1 (for even k). There are other patterns as well,
such as a relation between n and N (Ni = ni
- ni-1 - Ni-1). If there is a way to
analytically derive these patterns, it's beyond us! The series continues:
8 ..... ..... .......... .......... 5772-1
9 ..... ..... .......... .......... 13932
10 ..... ..... .......... .......... 33632-1
11 ..... ..... .......... .......... 81192
12 ..... ..... .......... .......... 196012-1
13 ..... ..... .......... .......... 473212
14 ..... ..... .......... .......... 1142432-1
15 ? ? ? ? 2758072
The 15th n is 2758072 = 76069501249. And the triangle of this
number is (get ready): 2893284510173841030625!
- The Aeroquiz Editor
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