Wikipedia:Reference desk/Archives/Science/2014 June 22

= June 22 =

Square light year
Anything whose area is a square light year that you know about?? This is a huge unit of area. If my calculation is correct, it is 3.5 * 10^24 square miles (feel free to give a more accurate value if you can.)

Nothing in our solar system is this big; it's more than 10^12 times as big as the sun. Is this correct?? Georgia guy (talk) 01:21, 22 June 2014 (UTC)
 * it would be a massive understatement to say that nothing in our solar system would have an area of one square light year.  A light year is about 10^16 kilometers, so a square light year would be 10^32 square kilometers.  For comparison, the radius of the solar system is about 100 AU or about 1.5 10^10 kilometers.   Thus, you'd need to lay out 1.3 million so solar systems end to end to make up a light year, and could fit about 1.7 trillion solar systems in an area of one square light year.  -- Jayron  32  02:07, 22 June 2014 (UTC)


 * 1 light-year is $9.461 m$ = $9.461 km$ ≃ $km$, so $1 ly^{2}$ ≃ $km2$. —Quondum 03:05, 22 June 2014 (UTC)


 * Ah yes, 10^16 meters. I messed that up.  Thanks for correcting me.  So, take three decimal places off of all of my length calculations and 6 decimal places off of my area calculations.  In the end, however, it's still a metric fuckton of area, many orders of magnitude larger than the solar system.-- Jayron  32  20:12, 22 June 2014 (UTC)


 * What do you consider to be a "thing" for the purpose of defining the word "anything"? A cross section of the Oort cloud has an area of a few square light-years.  Red Act (talk) 02:59, 22 June 2014 (UTC)


 * A square light year is around 1032 m2. Orders of magnitude (area) ends with the below. PrimeHunter (talk) 20:29, 22 June 2014 (UTC)
 * {| class="wikitable"

!Factor (m2) !Value !Item
 * +List of orders of magnitude for areas 1027 square metres and larger.
 * colspan="3"|...
 * rowspan=2|1032
 * 2×1032 m2
 * Roughly the surface area of an Oort Cloud
 * 3×1032 m2
 * Roughly the surface area of a Bok globule
 * colspan="3"|...
 * 1034
 * 3×1034 m2
 * Roughly the surface area of The Bubble
 * colspan="3"|...
 * 1041
 * 7×1041 m2
 * Roughly the area of Milky Way's galactic disk
 * }
 * When you break up an object into N identical pieces, the total surface area of the system increases by a factor of N1/3. If you were to reduce the Earth into its 1050 individual atoms, and surround each atom by an identical sphere such that the total volume of the spheres equals the original volume of the earth, then the total surface area of the spheres would be about 5x108 km2 * (1050 )1/3 = 2.3x1025 km2, or about one fourth of a square light year.  Do this for the sun and its 1.2x1057  (corrected below) constituent atoms, and you get a total surface area of 6.4x1031km2, or seven hundred thousand square light years.
 * I don't know in what way it makes sense to speak of the total surface area of the constituent atoms of an object, but we could consider the sum total of the atoms' Van der Waals surface area.  If the 2x1030 kg sun is 75% H and 25% He (close enough for these calculations), then it has 9x1056 individual hydrogen atoms with rw 1.2Å and 3x1056 individual helium atoms with rw 1.4Å.  The total, combined surface area is then 9x1056 * 4π(1.2Å)2 + 3x1056 * 4π(1.4Å)2 = 2.4x1032km2, or two and a half million square light years. -- ToE 22:15, 22 June 2014 (UTC)
 * Oops. That should be 75% / 25% by mass, not count.  That will change the results by a small amount. Recalculating ... -- ToE 22:38, 22 June 2014 (UTC)
 * OK, using 75/25 H/He by mass for the sun, we get 9x1056 individual hydrogen atoms, but only 7.5x1055 individual helium atoms.  This brings the first calculation (dividing the sun's volume into this many spheres) down to 6x1031km2, or six hundred seventy thousand square light years, and the second (sum total Van der Waals surface area) down to 1.8x1032km2, or two million square light years. -- ToE 10:33, 23 June 2014 (UTC)
 * 7×1041 m2
 * Roughly the area of Milky Way's galactic disk
 * }
 * When you break up an object into N identical pieces, the total surface area of the system increases by a factor of N1/3. If you were to reduce the Earth into its 1050 individual atoms, and surround each atom by an identical sphere such that the total volume of the spheres equals the original volume of the earth, then the total surface area of the spheres would be about 5x108 km2 * (1050 )1/3 = 2.3x1025 km2, or about one fourth of a square light year.  Do this for the sun and its 1.2x1057  (corrected below) constituent atoms, and you get a total surface area of 6.4x1031km2, or seven hundred thousand square light years.
 * I don't know in what way it makes sense to speak of the total surface area of the constituent atoms of an object, but we could consider the sum total of the atoms' Van der Waals surface area.  If the 2x1030 kg sun is 75% H and 25% He (close enough for these calculations), then it has 9x1056 individual hydrogen atoms with rw 1.2Å and 3x1056 individual helium atoms with rw 1.4Å.  The total, combined surface area is then 9x1056 * 4π(1.2Å)2 + 3x1056 * 4π(1.4Å)2 = 2.4x1032km2, or two and a half million square light years. -- ToE 22:15, 22 June 2014 (UTC)
 * Oops. That should be 75% / 25% by mass, not count.  That will change the results by a small amount. Recalculating ... -- ToE 22:38, 22 June 2014 (UTC)
 * OK, using 75/25 H/He by mass for the sun, we get 9x1056 individual hydrogen atoms, but only 7.5x1055 individual helium atoms.  This brings the first calculation (dividing the sun's volume into this many spheres) down to 6x1031km2, or six hundred seventy thousand square light years, and the second (sum total Van der Waals surface area) down to 1.8x1032km2, or two million square light years. -- ToE 10:33, 23 June 2014 (UTC)

Sexual behavior in animals
Do any other animals (not human beings) engage in: (a) sex with other animals for pleasure, as opposed to procreation; (b) masturbation; (c) sex with different species; and (d) non-intercourse sex such as oral sex or anal sex? Also, if we are talking about scenario "a": how would the scientific researchers know if the animal was doing it for pleasure and not for procreation? Thank you. — Preceding unsigned comment added by 75.44.113.200 (talk) 03:42, 22 June 2014 (UTC)
 * (a) Dolphins and some monkeys such as bonobos; (b) IDK; (c) Deer occasionally try to copulate with cows; (d) Cats, ducks, monkeys and some other species have been known to engage in gay sex. 24.5.122.13 (talk) 03:59, 22 June 2014 (UTC)


 * Yes to all of the above. See Animal sexual behaviour.  Red Act (talk) 04:05, 22 June 2014 (UTC)

Bears having oral sex ScienceApe (talk) 06:59, 22 June 2014 (UTC)


 * I question your position that anal and oral sex are "non-intercourse" sex. --   Jack of Oz   [pleasantries]  09:42, 22 June 2014 (UTC)


 * This is the OP. I had meant to type "non-vaginal intercourse".   75.44.113.200 (talk) 16:28, 23 June 2014 (UTC)


 * That makes sense. Thanks.  --   Jack of Oz   [pleasantries]  07:20, 24 June 2014 (UTC)


 * (a) dogs (b) dogs (c) dogs (d) dogs.--Shantavira|feed me 11:13, 22 June 2014 (UTC)


 * I am not surprised Jack's answers are valid, but they certainly seem overlimited, especially based on Mexican experience. μηδείς (talk) 01:59, 23 June 2014 (UTC)


 * Regarding the question "how would researchers know if the animal was doing it for pleasure and not for procreation?", the answer is simple if the species doesn't have enough mental ability to conceive (no pun intended) the idea that if they have sex then this may cause offspring to be produced later. For the few species that might be smart enough for that sort of reasoning, but which can't use a language that we understand, researchers probably couldn't tell what the animal's intent was. --70.49.171.225 (talk) 04:10, 23 June 2014 (UTC)


 * Of course, non-reproductive sex must be solely for pleasure, as it can't possibly be for reproductive purposes. StuRat (talk) 04:13, 23 June 2014 (UTC)
 * "Individual pleasure" may be a narrow view of non-reproductive sex. Non-reproductive sex may be used for social bonding in some species. 140.254.226.182 (talk) 14:31, 23 June 2014 (UTC)

Bug id
Hi. Today I found an interesting looking bug attached to a pair of my wife's sandals in the garden. Here's the little fella: S/he measures around 4-5cm long (the bug, that is, not my wife), and I'm in the UK. What is it? Thanks :) — sparklism  hey! 17:04, 22 June 2014 (UTC)


 * That's a nice specimen of a beetle. Specifically a scarab beetle. It might be a European chafer, based on the antennae shape. On the other hand, yours would be on the very high end of size for them, and the pronotum might not be quite right. Getting a beetle ID to species level is quite hard unless it's a very common species with few look-alikes. If you want to learn enough about insect morphology, you can use a Key_(biology) to be sure. Here's a nice intro to beetle ID . SemanticMantis (talk) 17:59, 22 June 2014 (UTC)
 * That's great - thanks! :) — sparklism hey! 19:05, 22 June 2014 (UTC)


 * Looks quite a lot like a cockchafer to me. Pretty common in the UK at this time of year. Richard Avery (talk) 07:05, 23 June 2014 (UTC)
 * I thought of that too, but the antennae are different - cockchafers have seven "leaves" on theirs (or six for females). Alansplodge (talk) 13:04, 23 June 2014 (UTC)

Transmission line connection via tapered sections
It is well known that to match two transmission lines of differing impedance using a tapered section introduces a low frequency cutoff that depends upon the type and length of taper. However, I am not clear as to whether connecting 2 TLs of the same impedance but differing dimensions together with a tapered section will introduce such a frequency limitation. Am I correct in thinking that no low frequency cutoff will be apparent in the latter case? --86.180.143.223 (talk) 20:40, 22 June 2014 (UTC)


 * It should not introduce a cutoff, provided that the taper maintains the same impedance throughout. Intuitively, as the impedance of the two transmissions lines is changed from mismatch to match while keeping the length of the tapered region constant, the cutoff does not change in frequency, but the reflection coefficient approaches zero for frequencies in the cutoff band, just as one would expect without the taper section. The geometry of the taper naturally is important to get a perfect match with differing dimensions. —Quondum 21:17, 22 June 2014 (UTC)
 * (ec) You are correct. For example, coaxial TLs of the same impedance are routinely joined by adapters like this with no low frequency loss. However such adapters need careful design and testing to achieve accurate impedance matching (shown by low VSWR) at high frequencies. 84.209.89.214 (talk) 21:21, 22 June 2014 (UTC)