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June 10

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evidence of time travels

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Baseball Bugs It's called Cruise to Japan 1932 --Thegooduser Life Begins With a Smile :) 🍁 02:08, 10 June 2020 (UTC)[reply]
Do you have a link? ←Baseball Bugs What's up, Doc? carrots→ 02:18, 10 June 2020 (UTC)[reply]

here --Thegooduser Life Begins With a Smile :) 🍁 02:20, 10 June 2020 (UTC)[reply]

Just someone jokingly commenting about the fact that Japanese school uniforms look the same now as they did in the 1930s. Watch at 2:49. --Guy Macon (talk) 03:31, 10 June 2020 (UTC)[reply]
Early silent films, which moved through the camera during recording by the operator hand-cranking the reel, usually had slower frame rates than is common today. When a movie recorded at 18 fps is played back at 24 fps as is usual for modern movies, everything – including the speed at which people walk – is sped up by 33 percent. BTW, if you discover real evidence of time travel, it is best to keep it to yourself, or the Time Police will come and erase your memory.  --Lambiam 06:44, 10 June 2020 (UTC)[reply]
As noted below, the frame rate can be compensated for, IF someone bothers to take the time to do it. ←Baseball Bugs What's up, Doc? carrots→ 20:37, 11 June 2020 (UTC)[reply]
In fact the real problem is not the different frame rates, you can always do something like what telecine does to get the proper frame rate. The problem is that humans cranking the handle sometimes turn it faster, sometimes slower. Sometimes they're helped along by gravity, sometimes they fight it, and soemtimes they can get tired towards the end of the scene. This makes for an irregular frame rate which makes for the characteristic jerky motion of silent movies. 93.136.79.38 (talk) 15:25, 11 June 2020 (UTC)[reply]
Genuine time travel is not possible backwards as time will always be linear for the traveler...Man A steps into the time travel device, then he travels back, watches the sporting event then has a beer. This is linear for him. On a second note, I can travel in time, I move forward continuously. In fact I can't stop. 109.151.74.33 (talk) 09:19, 10 June 2020 (UTC)[reply]
Yes. Time is a vector. ←Baseball Bugs What's up, Doc? carrots→ 09:22, 10 June 2020 (UTC)[reply]
No, it's not. --Wrongfilter (talk) 09:43, 10 June 2020 (UTC)[reply]

Lambiam Is it a joke that they will erase my memory? I find it hard to tell between a joke and serious message when online, as explained here. Also when will time travel become a reality? --Thegooduser Life Begins With a Smile :) 🍁 16:22, 10 June 2020 (UTC)[reply]

@Thegooduser: – It was a good-natured joke, referring to the overused trope of Time Police. It is not known if time travel in any form will ever be possible (see our article Time travel), but the scientific consensus appears to be that even if theoretically possible, we will not be able to make it work on a macroscopic scale. For further discussion of the possibility, see this Scientific American article.

Heritability of Parkinson's disease

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Have there been any studies quantifying heritability of Parkinson's disease? For example what's the best estimate for the averaged extra risk factor of getting Parkinson's disease when one parent was affected? 95.168.116.34 (talk) 17:29, 10 June 2020 (UTC)[reply]

In what is called "the largest genome-wide association study of PD to date"[1] (caveat: not peer-reviewed preprint on bioRχiv – but the authors have established academic credentials) the "liability-scale narrow-sense" heritability of PD is estimated as 22% (with a 95% confidence interval of 18–26%), based on a meta-analysis of 11 datasets. The modifiers I put between scare quotes are not explained in the article. Also, this specific item of information is reported in an ill-coordinated sentence that is ungrammatical (while understandable but lacks a subject).  --Lambiam 08:35, 11 June 2020 (UTC)[reply]
If the baseline incidence of PD is ~1% does this mean people with (that particular amount of) family history are ~22x more likely to get it? 95.168.118.175 (talk) 14:16, 11 June 2020 (UTC)[reply]
No. It means that genetic variation accounts for 22% of the chance a person will get Parkinson's. Abductive (reasoning) 17:57, 11 June 2020 (UTC)[reply]
So if the incidence of PD was exactly 1.00% then the probability of someone with no family history of PD whatsoever getting PD would be in the neighborhood of 0.78%? Hmm this alone doesn't really help estimate the probability of someone with family history getting it. --OP 31.45.225.163 (talk) 13:23, 12 June 2020 (UTC)[reply]
The heritability estimate doesn't look at family histories. It looks at correlations between shared alleles and people getting the disease. I doubt very much that the 22% estimate will hold up as larger genome-wide association studies are conducted. If one were to do all 7.7 billion human genomes, there would be no alleles and zero heritability correlated to Parkinson's (and many other diseases). Abductive (reasoning) 00:01, 13 June 2020 (UTC)[reply]
I fail to comprehend your final sentence. Surely if all people were tested there would be more alleles, not zero? Rmhermen (talk) 14:55, 15 June 2020 (UTC)[reply]
More alleles in general, but not necessarily alleles correlated with Parkinson's. That wide a study may find alleles for other things. That said, I'm not sure what basis Abductive has for claiming that a wider study would find that there were no alleles correlated with Parkinson's disease. --OuroborosCobra (talk) 15:07, 15 June 2020 (UTC)[reply]
Every time a genome-wide association study is conducted on a wider sample of the population, the correlations decrease because the effect is spread over more genes/alleles, each of which is found to have a smaller effect. Alleles discovered in earlier studies lose their statistical significance. It is a statistical certainty that if a study of a few tens of thousands is at 22%, the actual heritability would be 0% if the whole human species were checked. Abductive (reasoning) 21:58, 16 June 2020 (UTC)[reply]

Looking for scientific research on Solunar theory

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Wikipedia's article on Solunar theory does not cite any scientific sources. The credited inventor of ST, John Alden Knight, is described as an accomplished angler and published at least one book on bass fishing, but does not appear to have had any academic credentials. Solunar tables are widely published in American media and have been for decades. Yet I cannot find any scientific studies via Google. Can anyone unearth some rigorous research? Or is this pseudoscience that was persisted unchallenged for almost a century? DrewHeath (talk) 17:36, 10 June 2020 (UTC)[reply]

Google scholar yields 106 results. 107.15.157.44 (talk) 18:08, 10 June 2020 (UTC)[reply]
There are quite a few unrelated returns in that search, but also some articles that look useful. If I can access some of them I'll see about improving our own article here. Thank you. DrewHeath (talk) 18:18, 10 June 2020 (UTC)[reply]
User:DrewHeath Google scholar 2 brings up results including Sieh 1950, Pulver 2017, Theroux 1998, Milardi 2018 etc. Amousey (they/then pronouns) (talk) 13:36, 11 June 2020 (UTC)[reply]

Transformers: relation of the primary current's frequency or rate of change, and the output voltage of the secondary

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I've asked this question (worded a bit differently) before, but I'd like a second opinion. More than a few people/sources give answers that seem to contradict the conclusion of the previous Q&A session, and I'm not sure what to make of it. Are the "rules different" for a non-sinusoidal wave, or is "rate of change" not effectively the same as "frequency"?

ZFT (talk) 18:46, 10 June 2020 (UTC)[reply]

There are high-end 1:1 audio transformers where the output is the same as the input (not perfectly -- no component is perfect -- but close) over a frequency range from 20 Hz to 20,000 Hz. There are 50 Hz power transformers that perform poorly at 60Hz, 60 Hz power transformers that perform poorly at 50Hz, and power transformers that work well at 50 Hz or 60 Hz. Some aircraft use 400 Hz power, and use transformers that perform poorly at other frequencies. And there are transformers that are designed for radio that operate at far higher frequencies. In addition to being designed for different frequencies, there is wide variation on how much power, how much current, and how much voltage they can handle.
Fun story: One time I was working on a prototype, and someone from accounting came over and said "they sent me here because they say you know a lot about how transformers work.". I started asking what the application was so I could give the correct advice. They looked confused and said "I just wanted to know how they change from cars to giant robots"... --Guy Macon (talk) 00:14, 11 June 2020 (UTC)[reply]
That doesn't answer the question. I'm getting rather tired of glib responses. Please read the question and the links carefully. ZFT (talk) 01:51, 11 June 2020 (UTC)[reply]
Congratulations on your ability to compose a snarky response when faced with your own inability to draw a reasonable conclusion from an answer. You certainly put me in my place! If multiple people on multiple websites -- people who are able to successfully answer questions asked by others -- all respond with answers that are unacceptable to you, you might want to ponder what the common factor is.
Let me dumb it down for you. Audio transformers work the same with any signal (within the audio frequency range) from a pure sine wave to Death Metal. The type of signal is irrelevant for audio transformers. Power transformers only work properly with sine waves in a narrow range of frequencies. Frequency is the number of occurrences of a repeating signal per unit of time. Rate of change, also known as Slew rate, is the change of a signal per unit of time. They are not the same thing. --Guy Macon (talk) 02:29, 11 June 2020 (UTC)[reply]
ZFT wrote "please read the question and the links carefully." I'm willing to read the question carefully. The links are to quora, which is garbage, so I won't pay any attention to the links.
'Are the "rules different" for a non-sinusoidal wave...'? For an ideal transformer, no. For a real transformer, yes. Depending on the design of the transformer, applying a non-sinusoidal voltage will result in an output voltage that is approximately in proportion to the turns ratio, but there will be some distortion.
'or is "rate of change" not effectively the same as "frequency"?' Frequency is the number of times a certain wave shape is repeated. In the case of a power transformer, the wave shape is a sinusoid, and the frequency is usually 50 or 60 cycles per second (Hz). In the case of an audio transformer, the input is a voltage that represents speech or music, and there is no repetition, so the frequency is undefined. However, a prototype of the audio transformer was probably tested in a laboratory, and during the tests a range of sinusoidal inputs was applied.
"Rate of change" is the instantaneous rate of change of a waveform at a certain instant. For a sine wave, the rate of change is 0 at the top or bottom of a crest, and a maximum as the wave crosses 0 volts while moving from the negative half-plane to the positive half-plane. Jc3s5h (talk) 14:12, 11 June 2020 (UTC)[reply]
The first Quora link is actually what prompted me to (re)start this Q&A session, because it seems (in my obviously and admittedly incomplete understanding) to contradict the conclusion reached from last time. Am I wrong in assuming that, in this context (ideal transformer laws), a higher/lower frequency entails a higher/lower rate of change in current or voltage? If I am, does that mean that the "rate of change" (but not "frequency") of the input does have an effect on the output voltage and current? ZFT (talk) 20:25, 11 June 2020 (UTC)[reply]

If we're talking about frequencies, then we're most likely talking about sinusoidal waveforms. The same waveshape is repeated over and over; ideally it began infinitely long ago and will continue forever. Here is a diagram:

 

One cycle is what is shown during the period labeled with λ. If we consider the common voltage from a wall outlet in the US, the period of one complete cycle is 1/60 second. The root mean square voltage is 120 volts, so the peak voltage is about 170 volts.

The mathematical waveform, if we choose to have the wave cross from positive to negative when time, t, is zero, is

V ≅ 170 sin (60 × 2πt)

(The 60 in this equation comes from the frequency, 60 Hz.)

The rate of change is the derivative of this function. One notation for the derivative is to add a prime to the variable, V'. Those familiar with calculus will know

V' ≅ 170 × 60 × 2π cos (60 × 2πt)

or

V' ≅ 64088 cos (60 × 2πt)

One of the times a maximum rate of change will occur is t = 0. 120π × 0 = 0, and cosine of 0 is 1, so the maximum rate of change is 64088 volts per second. We see that the maximum rate of change is proportional to the frequency, if the maximum amplitude of the input voltage is kept constant.

So you are correct, a higher/lower frequency does entail a higher/lower rate of change of voltage and current if the if the maximum amplitude of the input voltage is kept constant. Jc3s5h (talk) 22:54, 11 June 2020 (UTC)[reply]

So does the rate of change of the primary coil's current have an effect on the output coil's voltage? ZFT (talk) 03:11, 12 June 2020 (UTC)[reply]
If we model a transformer circuit with an ideal voltage source as the input, an ideal transformer, and an ideal resistor connected across the secondary coil, then cause & effect can be thought of as follows:
  • The voltage source causes a current to flow in the primary.
  • The amount of the primary current is determined by the amount of voltage, and the transformed resistance, in accord with Ohm's law.
  • The transformed resistance is the resistance attached to the secondary multipled by the turns ratio (Np/Ns
Once these quantities have been computed, any other desired quantities can be derived from them. Jc3s5h (talk) 12:21, 12 June 2020 (UTC)[reply]
I'm surely OT, but what about the relation between frequency and impedance? Increasing the frequency increases the impedance in an inductor like the primary of a transformer, so the current in the primary should decrease. As long as the transformer is not loaded, probably nothing will change, but it seems to me that under load the voltage (or the current?) in the secundary could decrease with increasing frequency. 2003:F5:6F06:DF00:41D2:1EBA:1630:84C5 (talk) 20:55, 12 June 2020 (UTC) Marco PB[reply]
The primary of a loaded transformer is not an inductor, although it does have inductance. An ideal inductor draws no power, only apparent power, because the magnetic field puts back as much energy as was put into it. A loaded transformer draws actual power, because the magnetic field sends energy to the load. --Guy Macon (talk) 19:40, 15 June 2020 (UTC)[reply]

Measurement of a cut cylindrical object, by a primitive way

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If I take a hose and cut it at the middle along its width (=vertical cross-section), but I don't cut it completely into to parts, but around half of it remain connected, in a way it's difficult to determine by eyes sight if it's exactly half or not while I need it exactly in the half.

The issue is that I need a primitive way to check it while I'm outside in nature (not by electronic equipment etc.), but maybe other available means such as string / thread / yarn / wire (as I heard in past but couldn't find it again). Is there an accurate way to do so? (How did people do it in the ancient times?) ThePupil (talk) 21:15, 10 June 2020 (UTC)[reply]

Depends how long it is, and how long a piece of string you have. If the string is as long as the hose, cut it to length (or tie a knot to mark the length), then double it over to give half the length and put this against the hose. If the string is shorter, cut it into a piece (or mark it with a knot) much shorter, then count how many such lengths there are in the hose, then use the short piece of string half that number of times along the hose. 2A00:23C6:AA08:E500:89D5:DEA2:A805:722C (talk) 21:49, 10 June 2020 (UTC)[reply]
Thank you. I made a mistake and wrote horizontal while I needed to write vertical. I edited it. Please if you can adjust your answer to the update. Thank you.ThePupil (talk) 22:13, 10 June 2020 (UTC)[reply]
Is there any particular reason you cannot arrange your hose in a U shape and then cut at the bottom of the U ?DrewHeath (talk) 22:18, 10 June 2020 (UTC)[reply]
I'm asking about the size of the vertical cross-section. It means that even if you make U shape and cut it at the bottom, you're still not sure you cut it only half, and I want to check it was really cut by half, accurately, not more or less.ThePupil (talk) 23:05, 10 June 2020 (UTC)[reply]
It might help to understand the question if we knew why you want to do such a thing. My current understanding is that if you cut a horizontal cylindrical tube vertically in two, the fresh cut forms a circle (or rather two circles, one for each of the two severed parts). But you want to cut only halfway through, so that the cut forms a semicircle and the tube remains connected along the remaining other half of the cross-section circle. OK? Take a piece of (non-elastic) string and draw it tightly in a loop all the way around the tube. Using a marker, make a mark across the ends of the string where they meet, completing the loop. Taking the string off the tube, it should now be marked in two spots that are apart by a distance equal to the circumference of the tube. Check whether that distance is about three times the diameter of the tube; if not, something went wrong. Now fold the string so that the two marked spots come together, keep them tight together and pull the loop out till it is straight. Mark the farthest spot on the string, which should be halfway between the other two marks. Draw the string again tightly around the tube so that one of the old marks and the new halfway mark are at the same height on opposite sides. Draw a line segment on the tube from one mark to the other. Its length is half the tube's circumference, so this line segment is a semicircle. Now cut through that semicircle, being careful not to cut farther than its endpoints.  --Lambiam 23:40, 10 June 2020 (UTC)[reply]
Assuming Lambiam has correctly understood the goal, this boils down to a practical way of constructing the diameter of a circle because a diameter exactly cuts a circle in half. One way is to note that the tangents to a circle are parallel where the diameter crosses. If you lay two pieces of the hose on a flat surface (call it "table") and place another straight-edge (call it "ruler") to it is balanced across both of them, the half-way point is to where the hose touches the table and the hose touches the ruler. If you place one piece of hose projecting off the edge of the table at the distance you want to cut it, it's easy to see and cut to those two contact points. DMacks (talk) 04:42, 11 June 2020 (UTC)[reply]
Thank you very much for the answers. I asked it for purposes of understanding the history of the measurement.ThePupil (talk) 20:32, 13 June 2020 (UTC)[reply]
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