Raising Bookworms: Getting Kids Reading for Pleasure and Empowerment by Emma Walton Hamilton
Room by Emma Donoghue
So look for those reviews coming up.
A science news preview of 2011They chose five main topics:
A Bacterium That Can Grow by Using Arsenic Instead of Phosphorus (full text PDF)During and right after the press conference, your humble correspondent believed that the claims were likely to be true - partially because Ms Iron Lisa Wolfe-Simon was cute (I am not 100% immune against such things) and there was nothing "obviously wrong" with it. Arsenic is right beneath phosphorus in the periodic table and there was no reason why a random young passionate yet naive scientist and a former performance oboist couldn't have discovered such a new life strategy.
4. If a researcher claims to have found something that he or she has planned and wanted to find from the very beginning, one should realize that such a researcher was biased and the probability that the evidence supporting such a predetermined conclusion is wrong, cherry-picked, or even fabricated substantially increases.Needless to say, virtually all of the officially funded "climate research" is based on the plan that the researchers want to find something shocking - some new kind of a threat or mechanism. That's surely the case of Alexander Ač and 149 other crackpots who are now funded by EUR 25 million "CzechGlobe" project of the European Union. For a person who is looking at the field from a distance, it's another reason to think that the claims about the climate disruption are analogous to the claims about the arsenic life forms.
5. As long as a new scientific discovery remains incompatible with some previously established insights, i.e. until all the inconsistencies in the overall picture of science including the new discovery are cured, one has to appreciate the possibility that the empirical evidence supporting the new discovery is flawedAnd that's the memo.
6. Unless a rock solid proof of a new hypothesis has already been obtained, one should only decide about its validity after a fair comparison with alternative hypotheses; in this fair comparison, one shouldn't be biased in favor, or against, the alternatives that look more sensational
Burt Rutan on CAGW v4.0:We appreciate Gene's viewpoints on science but one can say that Gene has been a scientist. It's comparably refreshing to see how an engineer who actually worked with engines :-) looks at the difference between science and engineering, especially when it comes to the responsibility and the frequency of errors.
PDF, PDF to HTML, PowerPoint
“The Alarmist (scientist, journalist, politician etc.) chooses to huddle with other alarmists inside an echo chamber, attacking messengers who arrive, but spends no time to carefully inspect the data that forms his opinions, nor to notice the reporting of fraud”Thanks to Bill ZajcBurt Rutan, 2009
Draw a square consisting of four equally long connecting line segments hinged at the vertices. Such a structure may degenerate into a rhombus if you apply some pressure. How many additional interlinks of the same length must be supplemented to prohibit this excessive degree of freedom and to prevent the square from being tilted? The interlinks must belong to the same plane as the quad and each one may only be pegged to others at the endpoints.So far, I can't link to the original blog now because you would find a solution. The owner of the blog and his fastest reader found a solution with 43 extra linkages. After I was told what it was, your humble correspondent found a generalized solution of the same kind that only uses 29 extra linkages, if you allow me to draw frictionless intersections, or 31 extra linkages, if you don't.
finds = {{0, 0, 0, 0, 0, 0, 0}};
(* next, separated line with input *)
Dynamic[{a, b, lenVsq, c, d, e, f}]
Dynamic[MatrixForm[finds]]
(* next, separated line with input *)
For[a = 1, a <= 7, a++,
For[b = -7, b <= 7, b++,
v = {a, 0} + b*{1/2, Sqrt[3]/2};
lenVsq = v[[1]]*v[[1]] + v[[2]]*v[[2]];
For[c = -5, c <= 5, c++,
For[d = -5, d <= 5, d++,
For[e = -5, e <= 5, e++,
For[f = -5, f <= 5, f++,
u = {d/2 + c, d*Sqrt[3]/2}
- {f*Sqrt[3]/2, f/2 + e};
lenUsq = u[[1]]*u[[1]] + u[[2]]*u[[2]];
finds =
If[(c*c + d*d)*(e*e + f*f) != 0 &&
Abs[lenUsq - lenVsq] < 0.00001,
finds~Join~{{a, b, lenVsq,
c, d, e, f}}, finds];
]
]
]
]
]
]
I guess that if you're not told the main idea, chances are that even the code above will be useless for you. Their "solution 43" as well as my "solution 29/31" have already been posted to the web but I won't give you coordinates before some of you try to solve it - and perhaps find an even more economical solution?
Bonus: a no-go theorem
I am sure that I was not the only one who got stuck for some time with tests of pictures like this one:
Well, JollyJoker is another victim because the picture above was offered by him.
Note that there is a whole class of pictures where all the linkages are either vertical, or horizontal, or have another azimuthal angle that is a multiple of 30 degrees. Correspondingly, all internal angles of all faces are multiples of 30 degrees as well - 30, 60, 90, 120, or 150 degrees.
With some help from TRF, it's straightforward to prove that no picture of this kind can be rigid.
First, use (0,0), (1,0), (0,1), (1,1) for the vertices of the square. If all edges have one of the allowed azimuthal angles mentioned above - 0, 30, 60, 90, 120, or 150 degrees (or the opposite-direction edges whose azimuthal angle differs by 180 degrees), then all vertices or hinges indirectly connected to the square have coordinates that are integer linear combinations of the following four vectors:t = (1, 0)Note that the nontrivial numbers are, up to the last sign, sines and cosines of 30 or 60 degrees. However, if you have a collection of points whose coordinates are
u = (1/2, sqrt(3)/2)
v = (0, 1)
w = (sqrt(3)/2, -1/2)
at + bu + cv + dw,then you can see that both coordinates of the points are integer combinations of 1/2 and sqrt(3)/2. However, you may deform the picture by preserving vectors t,u and rotating v,w into v',w':
a, b, c, d are integers,
v' = (sin(phi), cos(phi))All edges whose length was equal to 1 in the old diagram - composed of points with coordinates point - have to have the length 1 in the new diagram composed of points with coordinates point'. That's because
w' = (sin(120°+phi), cos(120°+phi))
point' = at + bu + cv' + dw',
a, b, c, d are integers.
t, u, t-u; -t, -u, -t+u;are the only integer combinations of vectors t,u,v,w whose length is (or was) exactly equal to one (see below). However, if you replace v,w by v',w' in the expressions above, it's still true that all the vectors will have length equal to one. That proves that the diagram may be tilted.
v, w, v+w; -v, -w, -v-w
{ (A+B sqrt(3))/2, (C+D sqrt(3))/2 }Four times the squared length of this vector should be equal to 4 but it is
+ A2 + 3B2 +where A,B,C,D are integers. This can only be equal to 4 if (AB+CD) vanishes - it's the coefficient of the square root of three, an irrational number that can't cancel against others because of the integrality of the coefficients. Moreover, the four positive terms on the first two lines have to add up to four.
+ C2 + 3D2 +
2 sqrt(3) (AB+CD)
t, u, t-u; -t, -u, -t+u;You see that the combinations of t,u are decoupled from the combinations of v,w, so you may rotate v,w separately from t,u (that you may keep fixed, for example), and it won't break any of the linkages because all linkages that existed will continue to have length equal to one.
v, w, v+w; -v, -w, -v-w
(3.5, sqrt(3)/2)The squared length of this vector - the distance between the two points of the triangular truss - equals
3.52 + 3/4 = 12.25 + 0.75 = 13just like previously. So three pieces of the triangular bridge construction may be connected just like in the case of the 3-4-5 Pythagorean triangle and the right angle of the square is ensured in this way.
0 NE steps: 0, 1, 4, 9, 16, 25, ...The last line will be referred to as the white list.
1 NE step: 1, 3, 7, 13, 21, 31, ...
2 NE steps: 3, 4, 7, 12, 19, 28, ...
3 NE steps: 7, 9, 13, 19, 27, ...
4 NE steps: 12, 13, 16, 21, 28, ...
union: 0, 1, 3, 4, 7, 9, 12, 13, 16, ...
At + Bu + Cv + Dwhas the squared length equal to
A2 + B2 + C2 + D2 +There are no AC and BD terms because the pairs t,v and u,w are orthogonal: even in the upper part of the text, I have switched the sign of "1/2" in w to "-1/2", apologies to old readers. ;-)
+ AB - CD + sqrt(3) (AD + BC)
A,B,C,D = K,L,K,-Land their equivalents with sign flips and permutations that don't change the essential geometry (and the new angles at the special hinges). The apperance of K,L = 3,3 or more or K,L = 2,4 or more would already produce the total squared length above 31. We must also check
A,B,C,D = K,K,L,-L
K,L = 1,1 or 1,2 or 2,2 or 2,3
A,B,C,D = 1,2,2,-4.However, it produces the squared length of 1+4+4+16+2+8 = 35 or 1+4+4+16-2-8 = 15 (for -1,2,2,4) which are not on the "white list" and are getting too high, anyway. Similar small values of A,B,C,D that are not pairwise equal may be seen to produce too big a squared length, or a squared length that is not in the allowed list. For example, for A,B,C,D=1,2,3,-6, one gets AD+BC=0. However, the squared length is 1+4+9+36+-(2+18)=50+-20 which is 30 or 70, too large.