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April 28

Grade of Shade Cloth

Is the percentage of sunlight 'blocked' by a weave of material. How is this calculated using widths and gaps? --124.182.160.177 (talk) 08:59, 28 April 2010 (UTC)[reply]

Let's assume that the no light passes through the threads and that the same threads are used in both directions, and the same gaps are also present. Then we get a square cell, the distance between the start of each thread and the start of the next thread:

  /* *       \
 / * *        G
W_c  * *       /
 \ * * * * * \ W_t
  \* * * * * /

So, the width of a cell is the width of a thread plus the gap:

W_c = W_t + G

The percentage of light that gets through is:

100×G²/W_c²

The percentage of shade, then, is:

100 - (100×G²/W_c²)

Alternatively, you could find the percentage of shade by adding up the components:

100 × (W_t² + 2G×W_t)/W_c²

However, this doesn't account for "fuzz" outside the threads, so actually measuring the reduction in light would be better. Also note that it may lose "fuzz" over time, and a more threadbare cloth likely provides less shade. StuRat (talk) 12:27, 28 April 2010 (UTC)[reply]

You've missed the

/W_{c}^{2}

in the last line. -- Meni Rosenfeld (talk) 13:48, 28 April 2010 (UTC)[reply]

Thanks, I added it now. StuRat (talk) 19:22, 28 April 2010 (UTC)[reply]

Cross ratio

I don't understand how the cross ratio answers when 4 points in projective line are projectivly equivalent. According to my book, "if z_1,z_2,z_3,z_4 and w_1,w_2,w_3,w_4 are two sets of 4 points, the are projectivly equivalent if their cross ratios are the same". The explanation is "since the cross ratio of z_1,z_2,z_3,z_4 is the image of the (unique) projective transformation carrying z_1 to 1, z_2 to infinity and z_3 to 0". How can the cross ratio, a scalar, be the image of a map on the projective line, a point defined by two coordinates? Thanks m uchly. —Preceding unsigned comment added by 122.109.239.224 (talk) 10:01, 28 April 2010 (UTC)[reply]

I think it must mean that the cross-ratio is the image of z_4 under this projective tranformation. From its description, the transformation is

f:z\mapsto {\frac {(z_{2}-z_{1})(z_{3}-z)}{(z_{3}-z_{1})(z_{2}-z)}}

This makes the image of z_4

f(z_{4})={\frac {(z_{2}-z_{1})(z_{3}-z_{4})}{(z_{3}-z_{1})(z_{2}-z_{4})}}

which fits the standard definition of cross-ratio up to a permutation of terms. Gandalf61 (talk) 10:31, 28 April 2010 (UTC)[reply]

Any triple of distinct points on the line is "equivalent" to any other, so the shape of a tuple of four points is just a matter of where you put the fourth point, and that is expressed by a scalar. Michael Hardy (talk) 03:28, 29 April 2010 (UTC)[reply]

Thanks but how exactly? I thought that may be the scalar represents the point in affine coordinates, but what exactly does it mean to multiply and divide two points in the projective line? Could you give a proof that the cross ratio is the image of z_4? There isn't a proof in Wikipedia, so I'm not sure. All I got is if z_3=az_1+bz_2, then if z_1 is sent to (x,0) and z_2 is sent to (0,y) (just looking at the transformation on affine space) z_3 is sent to (ax,by). Thanks muchly. —Preceding unsigned comment added by 122.109.239.224 (talk • contribs)

When you calculate a cross-ratio, you aren't multiplying and dividing points, you are multiplying and dividing distances between points (or, to be completely accurate, signed distances). If you are working in the real projective line, then you have four points P_1, P_2 etc. represented by four extended real numbers z_1, z_2 etc. (they are extended reals because we have to assign a value to the point at infinity), and the signed distances between pairs of points is z_2 − z_1 etc.
The projective transformation doesn't multiply or divide points either - it turns points into extended reals, multiplies and divides the reals, then turns the result back into another point (but the invertible mapping between points and extended reals is not usually spelled out).
A projective transformation is defined by the images of three points, not two. Given any two sets of three points {P_1, P_2, P_3} and {Q_1, Q_2, Q_3}, there is a unique projective transformation that maps P_1 to Q_1, P_2 to Q_2 and P_3 to Q_3 - we say that the group of projective transformation is "triply transitive" on the projective line.
The cross ratio of four points is an extended real number and a property of the set {P_1, P_2, P_3, P_4} - and if you want to think of it as a point, you can regard it as the image of P_4 under the projective transformation given above.
Projective transformations preserve the cross-ratio - indeed, you can define projective geometry as the study of the group of transformations that preserve the cross-ratio. Gandalf61 (talk) 10:30, 29 April 2010 (UTC)[reply]

Math Shortcuts

I couldn't think of a better title. I thought this article was very interesting considering I am a mathmetically challenged. Does anyone have any more to add to this list? --Reticuli88 (talk) 12:59, 28 April 2010 (UTC)[reply]

I suppose there's the fairly obvious dividing by powers of 10 rule: just shift the decimal to the left by the number of zeros:

123/10 = 12.3
123/100 = 1.23
123/1000 = .123
123/10000 = .0123 <- Add leading zero
123/100000 = .00123         "

StuRat (talk) 13:17, 28 April 2010 (UTC)[reply]

Mental multiplication of large numbers can be done as follows. First consider the rather contrived case of two numbers close to a round number N. Let's write the numbers as x = N+x' and y = N + y'. Then we have:

x y = (N+x')(N+y') = N^2 + (x'+y') N + x' y' =

N(N + x' + y') + x' y' =

N(x + y') + x' y'

Let's consider a numerical example. What is 974*986?. If we take N = 1000, then x' = -26 and y' = -14. Then x' y' is easily computed mentally like e.g. 2*13*2*7 = 4*91 = 4*(90+1) = 360 + 4 = 364. Computing x + y' is even easier, this is 974 - 14 = 960. So, the answer is 960*1000 + 364 =

960364.

Next, suppose that we have two numbers, x and y, that are close to two different round numbers, N and p N where p a small integer. Then with x = N + x' and y = p N + y', we have:

x y = (N + x')(p N + y') = p N^2 + N y' + p N x' + x'y' =

N(p N + y' + p x') + x' y' =

N(y+px') + x'y'

Example, let's compute 982*2113. We can then take N = 1000,

x' = -18, y'=113, p = 2. We have x' y' = -18*113 = -(20-2)*113= -2*(10-1)*113 =-2(1130 - 113) = -2*1017 = -2034

Then y + p x' = 2113 - 2*18 = 2077, so the product is

2077*1000 -2034 = 2074966

So, clearly a piece of cake to compute mentally this way. But this is still a somewhat contrived case. In general, you won't be able to find a round N and a pN such that subtracting these numbers from the numbers you want to multiply yields small easy to multiply numbers. However, you can iterate this procedure. With some practice you can learn to multiply numbers with ten digits or more this way in your head.

Note that from a computational POV, the above method is not efficient at all, it uses far more steps than the most efficient multiplication method. But those efficient methods are worthless for mental multiplication. The reason why mental arithmetic is difficult has nothing to do with a limited brain capacity. The brain is in fact far more powerful than the fastest supercomputers. Mental arithmetic is difficult because the brain doesn't have the correct software for doing computations. You have to work around that by hijacking other things your brain can do well. This is necessary to be able to do a computation at all; attempting to do computations efficiently is irrelevant.

Good mental arithmetic stategies work by breaking the problem down in many easy to compute substeps. This in such a way that you can mentally keep track of all the substeps. Count Iblis (talk) 16:50, 28 April 2010 (UTC)[reply]

A trick I use is similar to that. I memorized all squares up to about 50 and wanted to do more but didn't do much, although 2-digit squares are pretty easy to calculate any way. That takes time but if you want to be good at mental math you have to take time to memorize some things I assume. With the knowledge of all squares up to 50, I can do calculations like 38 * 46 easily. Note these are both 4 from the number 42, so we have (42 - 4) * (42 + 4) = 42^2 - 4^2 = 1764 - 16 = 1748. This uses the formula (x - y)(x + y) = x^2 - y^2. It works as long as you have the squares memorized and the difference between the numbers is even. If it's not an even difference, there's one little extra step. For example, 38 * 47 = 38 + 38 * 46 and then the difference is even so you can use the same method. StatisticsMan (talk) 14:55, 30 April 2010 (UTC)[reply]