The “supermoon bloodmoon lunar eclipse” is coming to North and South America, as well and the UK and parts of Eastern Europe this month. The total lunar eclipse will start late on Sunday, Jan. 20, and finish early on Monday, Jan. 21, and because it occurs during a “supermoon”, it will appear about 14% bigger than normal.
It will also be one of the few times when you’ll be able to photograph the moon and stars simultaneously. While it will be a deep blood red, the contrast difference will be low enough that cameras will be able to pick up stars in the background.
Unlike a solar eclipse, which can be viewed only from a certain relatively small area of the world, a lunar eclipse may be viewed from anywhere on the night side of Earth. A total lunar eclipse lasts a few hours (this one will be about 5.5 hours long), whereas a total solar eclipse lasts only a few minutes as viewed from any given place, due to the smaller size of the Moon’s shadow. Also unlike solar eclipses, lunar eclipses are safe to view without any eye protection or special precautions, as they are dimmer than the full Moon.
Total lunar eclipse: A total lunar eclipse occurs when the moon and the sun are on exact opposite sides of Earth, according to NASA. When this happens, Earth blocks the sunlight that normally reaches the moon. Instead of that sunlight hitting the moon’s surface, Earth’s shadow falls on it. Starting at 9:36 p.m. EST Jan. 20. At 10:34 p.m., it moves into a partial eclipse, and starting at 11:41 p.m., the full eclipse begins; a maximum eclipse occurs at 12:12 a.m. Jan. 21. The total eclipse ends at 12:44 a.m.
Supermoon: A supermoon occurs when the full moon is at the closest point of its orbit to the Earth (perigee). That makes the moon look extra-close and extra bright — up to 14% bigger and 30% brighter than a full moon at its farthest point from Earth, known as the apogee, NASA said. This is the first of three supermoons in 2019. The others will be on Feb. 19 and March 21. Of these, the Feb. 19 full moon will be the closest and largest full supermoon of 2019.
“Blood” moon: That is just the reddish color the moon will appear during the total lunar eclipse.
We’ll be seeing half the moon right? But it will look like a flat disk with an area 1/4 that of the entire surface area of the moon, so does that mean that we’ll actually only see 1/4 of the moon?
No, we’ll see half the moon since half the moon is illuminated.
How do we reconcile these two items:
“Without the greenhouse effect, the average surface temperature would be 255 degrees kelvin (-18 degrees Celsius or 0 degrees Fahrenheit)” Tufts University professor
“There is no significant atmosphere on the moon, so it cannot trap heat or insulate the surface. When sunlight hits the moon’s surface, the temperature can reach 260 degrees Fahrenheit (127 degrees Celsius).” space.com
We’re told that the Earth would be 0F without an atmosphere, almost 60F colder that it actually is. But the moon reaches 260F without an atmosphere.
I encourage any and all to check my math, but when I modify the typical application of the Stephen-Boltzmann law to divide by two when averaging two halves, I arrive at the expected Earth temperature of:
303.24K 85F 30C
Not the widely accepted value of 255K 0F -18C when two halves are added together and divided by four for an ‘average’.
So, when correctly applying the Stephen-Boltzmann law we find that the Earth is cooler than expected and there is no indication that ‘greenhouse gases’ are back-radiating 60F of warmth at all times. This seems to mesh with reality better. In many locations on Earth the temperature range in a single day has a 25F difference between high and low. So the sun itself heats up the Earth’s surface by 20 degrees on any given day, but less than 2% of our atmosphere is constantly back-radiating 60F and we can’t figure out a way to measure it?
Thomas, you are still confusing total energy with flux. Therefore, your calculations are incorrect. You must divide solar flux by 4 to obtain the average flux over the surface of a sphere – not 2. On the other hand, if you first calculate the AVERAGE flux experienced by the entire surface of the day side (integrating the different flux values seen by each point on the surface and dividing by the area of the hemisphere), it then would be correct to divide that value by 2 to obtain the average flux over the entire surface of the sphere. I posted a very detailed FINAL attempt to set you straight under our previous discussion at:
https://wattsupwiththat.com/2019/01/07/pbs-washing-clothes-in-cold-water-can-help-prevent-global-warming-if-we-overcome-the-dragons-of-inaction/
Evidently, you didn’t read it, because you’re still confused. I included remedial geometry in that post in response to yours, so I encourage you to return to that thread and read my entire post. I have copied only a portion of that post below, so others might more easily see what we were discussing:
Your argument for dividing by 2 is correct if we were considering TOTAL energy in watts (or average flux in watts per square meter) over the entire day side hemisphere. But we’re not. We’re talking about actual flux experienced by each surface point of the day side hemisphere, which is NOT constant. You’re confusing total, or average energy with flux, or energy distribution.
Please follow this simple explanation: Let’s have a weak sun illuminating a sphere of radius R meters. Let’s also say that a surface at a right angle to the direction of the flux would experience 100 watts per square meter. First replace the sphere with a disk of radius R meters. To state the obvious, that disk intercepts EXACTLY the same amount of energy as the original day side hemisphere oriented towards the sun. Each point on the disk experiences precisely 100 watts per square meter because the surface of the disk at each point is perpendicular to the radiant energy. The TOTAL energy illuminating the disk is (100*PI*R^2) watts. Now make that disk into a stretchable membrane, and stretch it into a hemisphere with the convex side towards the sun. The edges of the hemisphere and original disk are coincident. The new membrane hemisphere also experiences a total of (100*PI*R^2) watts because the cross sectional area is the same as the disk. Isn’t that obvious? Now the total (100*PI*R^2) watts illuminating the day side hemisphere must be distributed over the entire sphere. So divide the total (100*PI*R^2) watts by the area of the sphere, which is (4*PI*R^2), to obtain the average flux over the entire sphere, which is 25 watts per square meter. Hence the factor of 4.
Look at it another way. The membrane hemisphere experiences the same total energy as the disk, namely (100*PI*R^2) watts. But the area of the hemisphere is twice that of the disk. So the AVERAGE flux per square meter of hemisphere surface must be half that of the disk, or 50 watts per square meter. So now you can do your divide by 2 thingy to get the average flux over the entire sphere, because the flux we just calculated for the day side hemisphere is already an average.
What’s twisting you off is that you assume the radiant energy received by every square meter of the hemisphere is the same, and equal to (in the example above) 100 watts per square meter. It isn’t. If it were, then a hemisphere would receive a total of (100*2*PI*R^2) watts, but a disk with the same cross sectional area would only receive (100*PI*R^2) watts. That would mean that a cylinder of parallel (nearly) rays from the sun with radius R and a specific flux density would impart twice as much total energy to a hemisphere than it would to a disk with the same area as the cylinder of rays. That’s insane, and reminds me of the kind of nutty arguments made by alarmists.
Floyd Doughty – Thanks for your response.
When averaging two halves, I add them together and divide by 2. Would you like to dispute that?
“What’s twisting you off is that you assume the radiant energy received by every square meter of the hemisphere is the same” – actually, that’s what your divide-by-4 problem is assuming. I assume no such thing.
Break down the Earth into two hemispheres, day and night. We use that hypothetical flat disk to help calculate the Day.TSI using a monotonic radiant energy distribution. Now we have the Day.TSI that is received by half the Earth. The energy isn’t spread out evenly across half a sphere in the same way as the hypothetical disk, but that doesn’t matter, it’s still distributed across half the Earth. We have a value for Day.TSI representing half the Earth. Averaging two halves requires dividing by two. You’ve said it yourself: “The membrane hemisphere experiences the same total energy as the disk”
This is not difficult.
If I were to drive 40 miles in an hour and then be at rest for an hour, and I want to average both those halves together I add them and divide by two. It doesn’t matter how that speed was distributed over the first hour, it could actually graph to a semi-circle, starting slowly, accelerating to the half way mark and then decelerating to the end. It’s still just:
[(total from first half) plus (total from second half)] / 2
I know you’re invested in believing the divide by 4 but that doesn’t make it right.
Who is Stephen?
Tell him (or her) that the Stefan–Boltzmann law surface for the Earth is *INSIDE* the Earth – and it’s a sphere – which a much better approximation of the surface than a disk.
So whatever result you obtain from your calculation should be referred to as the “dirt” effect .
That is provided the Stephen drilled a hole to center of the Earth in order to approximate the electromagnetic radiation at the core – which is the only place which can approximated as having a perfect reflecting surface – which is unlikely.
Otherwise, the Stefan-Boltzmann calculation is similar to all other calculation in the sense that garbage in leads to garbage out.
But hey, who needs a thermodynamics book – the name of game in climatology is to do whatever you need to do in order to obtain the result you need.
https://www.space.com/14725-moon-temperature-lunar-days-night.html
The average of 100 and -173 is -37.
If the Earth had no atmosphere, the surface temperature would be very hot during the days and very cold at night. The average would be very cold, similar to the Moon.
If the Earth’s atmosphere lacked a so-called greenhouse effect, the average surface temperature would be about -18 °C instead of 14-15 °C.
David Middleton:
“If the Earth’s atmosphere lacked a so-called greenhouse effect, the average surface temperature would be about -18 °C instead of 14-15 °C.”
How did you arrive at the -18 °C temperature value?
“If the Earth had no atmosphere, the surface temperature would be very hot during the days and very cold at night.”
Remove just the water vapour and nights would be much colder over land. But the far greater thermal reservoir is the oceans, their surfaces barely cool at night.
Hi, David. I was thinking about this latter comment, and wondered your opinion on some questions.
Often we read of the ‘If the Earth did not have an atmosphere…’ sorts of descriptions. You mention the temp ‘…if the Earth lacked a so-called greenhouse effect…’. For purposes of discussion, let’s assume that the atmosphere has only Nitrogen and Oxygen and trace Inert gasses, that the surface has no water, and the albedo is the same as the moon.
My questions:
What would be the diurnal min and max temps at the equatorial surface at perihelion and apohelion?
What would the difference be between the daytime and nighttime lapse rates at that location and times of year.
By what physical mechanism would atmospheric sensible heat leave the earth?
The -18 C calculation still assumes the “Earth average” Albedo of 0.3, but the Albedo of an “airless” planet earth would be similar to that of the moon, the main difference being the missing 60% cloud cover at an albedo of anywhere between 0.6 and 0.95…..Clouds being the main factor affecting how much incoming sunshine is reflected back into space.
DMacKenzie: [ The -18 C calculation still assumes the “Earth average” Albedo of 0.3, ]
Right, the Earth’s Albedo is dynamic, not a constant. But we’ll address that after we resolve the main issue.
If we were to calculate the Total Solar Irradiance (TSI) for each of the Earth’s day hemisphere and the Earth’s night hemisphere we would have something like:
Day.TSI = Day.TSI
Night.TSI = 0
And to average these values together we’d have:
(Day.TSI + 0) / 2
The widely accepted application of Stephen-Boltzmann law does this instead:
(Day.TSI + 0) / 4
That is incorrect.
Thomas, you are getting into more detailed calcs that are better explained here:
http://www.drroyspencer.com/2016/09/errors-in-estimating-earths-no-atmosphere-average-temperature/
This is a much more complicated heat transfer problem, even for the Moon, than the simple Stefan-Boltzmann equation. If you look at thge problem mathematically, you cannot average the two sides, since the function is T^4. In some heat transfer problems (radial through the insulation around a pipe, for example) the log-mean temperature difference is used instead of a simple arithmetic mean. If you set up the entire problem, the surface temperature of the moon is a function of heat transfer from below the surface and radiative heat transfer from the surface. These two heat flows are controlled by the thermal conductivity of the rock, its heat capacity, etc. At a minimum, the daylit side and the shaded side need to be calculated independently, or perhaps at a much smaller grid to simulate the physical moon. On earth, it gets even more complicated due to the heat generated by the Earth itself, the oceans, clouds versus no clouds, etc.
Wrong. Refer to my response to your post above. DMacKenzie’s primer link below is correct.
Loren,
thanks, keep it up
Loren Wilson – Thanks for your response.
“you cannot average the two sides, since the function is T^4.”
It doesn’t matter that the function is T^4, that was only a proxy function to determine how much TSI Earth’s dayside hemisphere is receiving. We apply that function to arrive at an answer for Day.TSI.
Now that we have a total value for Day.TSI, we can average it with Night.TSI, representing the two hemispheres of Earth.
(Day.TSI + Night.TSI) / 2
… assume Night.TSI = 0
(Day.TSI + 0)/2
If we were to calculate the Total Solar Irradiance (TSI) for each of the Earth’s day hemisphere and the Earth’s night hemisphere we would have something like:
Day.TSI = Day.TSI
Night.TSI = 0
No, TSI is the power received per unit area perpendicular to the incoming sunlight. Therefore the power incident on the earth’s surface is TSI x the cross sectional area, which is πr^2: TSI *πr^2
So the average power received at the surface on the illuminated side is shared over the surface area of the hemisphere: 2πr^2
So the average on the lit hemisphere is TSI *πr^2/2πr^2 = TSI/2
and the average over the whole sphere is TSI/4
And to average these values together we’d have:
(Day.TSI + 0) / 2
The widely accepted application of Stephen-Boltzmann law does this instead:
(Day.TSI + 0) / 4
That is incorrect.
A primer on the -18 calculation
http://www.instesre.org/Solar/SimpleGreenhouseModel.html
After reviewing the referenced URL, I’m okay with the -18C number as calculated. I’m also okay (but with less assurance) with the 15C measured average earth surface temperature, which for the purpose of this comment, I’ll accept. However, I am not okay with saying either (a) the greenhouse effect is the reason the earth’s surface is warmed by approximately 33C, or (b) the earth would be 33C cooler in the absence of atmospheric greenhouse gases. The reason being that as I understand it, the albedo of 0.29 used by the referenced URL to compute the -18C number is primarily dependent on the presence of clouds, which in turn is dependent on the existence of water vapor in the atmosphere. Water vapor is the primary earth atmospheric greenhouse gas. If greenhouse gases are removed from the earth’s atmosphere, water vapor will also be removed. If water vapor is removed from the earth’s atmosphere, the clouds that reflect the incoming solar radiation will not be present. If the clouds are absent, the earth’s albedo won’t be 0.29, but will likely be closer to 0. Using the algorithm in the referenced URL, as the earth’s albedo approaches 0, the earth’s energy-rate-equilibrium temperature increases. For an albedo of 0, the value of Te (equation 4 in the referenced URL—note, however, that equation 4 is incorrect in that the right side of equation 4 should be the fourth root of the expression) is 278.3K (not 255.4K) which corresponds to a temperature of +4.9C. Thus, using the algorithm in the referenced URL and an albedo of 0, the temperature difference between the measured value of 15C and the sans-greenhouse-gas atmospheric model value is approximately 10C, not 33C. Now if someone can argue that if no clouds existed in the earth’s atmosphere the albedo of the earth would be 0.29, then I’m okay with the 33C difference. Until someone makes that argument, the claim that in the absence of atmospheric greenhouse gases the earth’s surface would be 33C colder than it is today is nonsense.
At a more fundamental level, to (a) compute an average earth surface temperature using a model that represents an earth atmosphere with greenhouse gases (which is the case if an albedo of 0.29 is used), and then (b) claim the output of that model represents the earth temperature devoid of greenhouse gases is illogical; but in my opinion, entirely consistent with much of the logic used by the AGW community.
Great, another wildly off-topic food fight over “who’s righter about thermodynamics”.
Please don’t post off-topic comments.
Thanks Ant. That kind of guff really gets on my nerves anyway. Shame that Oz won’t be seeing anything of the supermoon stuff but knowing our luck it’d be cloudy anyway.
If the moon is between the earth and the sun, don’t you think than the sunny side will be facing the sun, and we will see the dark side?
No, the rotation speed of the moon keeps one face constantly towards the earth. You’re apparently mixing “sunny side” and “dark side” of the moon (as it rotates around earth, all parts of the moon are illuminated equally by the sun) with the “visible side” and “far side” of the moon.
@Thomas Homer – Not sure where your location is, but as I understand it, it is a full eclipse, thus the whole Moon will be in the shade of the Sun for some time, during the event.
Will the moon still be illuminated such that we can see it?
If we wanted to average the amount of light that still illuminates half the moon during this eclipse, with the other half in the dark, we would calculate how much light is still illuminating the hemisphere we can see, and then divide that value by two. Averaging two halves together, requires dividing by two to get the average. Even though the moon appears as a disk to us with 1/4 the area of the entire moon, to average two halves we would divide by two.
My experience having photographed full lunar eclipses is that the moon never becomes completely dark. But close to it.
Will the moon still be illuminated such that we can see it?
Yes. It will be a full moon in lit area and dimension (and a bit bigger than usual due to being at orbital perigee). The only difference is that it will be much, much dimmer than usual and probably have a coppery tint. This is because it is in the shadow of the Earth and illuminated only by sunlight refracted through our atmosphere.
Thomas, I’m invested in nothing but mathematical accuracy. Unfortunately, you’re hopelessly un-teachable. I can explain it to you, but I can’t learn it for you. Read DMacKenzie’s primer link, if you can understand it. The analyses there are correct. Also, the post by Phil is correct. This is absolutely the last time I will have anything to say on the subject, unless you try to hijack another thread with mathematical malfeasance.
I bet if I did a search I’d find 50 guys telling us the “blood moon” is a sign that Jesus return is imminent.
I dislike the “supermoon” hype. It sounds great, but the change is barely perceptible to the naked eye. People expect something spectacular, but it’s just a slightly larger full moon. If they do perceive a larger or brighter moon, it’s likely just the power of suggestion.
I find the full moon bright and gorgeous regardless of whether it’s “super”…
Given the ratio between Earth and Moon size we might be more accurately described as a double planet.
Greg: Well watch this one. During the 5.5 hours, it will not be bright, but it will appear larger than average, and quite larger than when at apogee.
Check it out at:
http://astropixels.com/ephemeris/moon/moonperap2001.html
It also causes king tides – about 20 cm higher than usual in my area. Where it coincides with a low pressure system, onshore winds and large waves you can expect increased erosion and some coastal inundation. Those effects are very noticeable.
Oh noes.. Gorebull Warbling causes blood supermoons forcing king tides, greater coastal erosion, inundation et al.. Toyota and Nissan have a lot to answer for, especially leaving their tyre tracks all over the Moon’s maria.
/sarc as if it were necessary.
I agree with Greg. The name “supermoon” can only confuse people. There is nothing special to see at such a moon, that is slightly larger than normally.
I agree!
Otherwise you end up suffering from Super Moon fatigue – which is just artifact of the Julius Caesar’s calendar .
The Super Moon is the Full Moon or the New Moon which is closest to the perigee.
I also love the impact on the tides by the moon which can be visually stunning,
Especially during the winter eclipse season – or roughly a month after the NH Winter Solstice – which lasts roughly a month.
No one is gonna mention that it will be redder than “normal” due to human atmospheric pollution?
I have a prediction, once again we will be clouded over during the whole thing! 😉 Hoping not, got a nice digital camera for Christmas and would love to put it on the big tripod and get some decent shoots.
The “supermoon bloodmoon lunar eclipse” is coming to North and South America, as well and the UK and parts of Eastern Europe this month
It’s more than that it’s a ‘super blood wolf moon’ lunar eclipse.
SuperMoon because it wears red and blue tights. Oh, wait that’s Superman….
Supermoon because it is a full moon that occurs while the moon is at or near perigee (closest to Earth)
Bloodmoon because it’s a total lunar eclipse which results in a moon with a reddish hue (like blood)
Wolfmoon because that’s a nickname given to full moons in January dating back to colonial times.
Shame there’s only 1 full moon in January this year otherwise this could have been a
“Super blue blood wolf moon”
John, the ‘blue moon’ and ‘wolf moon’ terms are mutually exclusive. A ‘blue moon’ is the 2nd full moon in a month. The ‘wolf’ moon is the first full moon of the year.
It’s not so clear cut. the wolf moon is the Jan. moon (Wolf moon means “January moon”, not as you insist “first moon of the year” – it just so happens that since Jan is the first month of the year and the length of the moon cycle being shorter than the month results in the first moon of the year always being a Jan./wolf moon). snow moon is the Feb. moon (unless there is no full moon in Feb., the only month of the year that is possible, then it’s called the black moon), Worm moon is March and so on – each month has his own name for full moons. Some insist only the 1st full moon in the month gets the name others would have any full moon in the month getting the name. It really depends on whom you ask as there is no universal agreement (though someone somewhere will probably insist on a 97% consensus 😉 ).
blue refers to the 2nd full moon of the month (doesn’t matter which month, even Feb can have two during leap years).
So the second full moon in Jan is a blue (for 2nd) wolf (for Jan) according to some. and just blue according to others.
All I know is the Blue Moon is a dangerous moon.
About two decades ago on the last Sunday in August, we were cruising in my 1991 Crown Victoria along interstate 90 just west of Tomah, Wi, when I came across a heat-wave induced buckle in the concrete pavement. As I drove across the buckle, one of my front tires kicked up a loose piece of concrete that bounced and rattled around under the car. The engine cut off a few seconds later with an annoying loss of power in the steering and braking systems, which are important in 4,400 lb vehicle. Fortunately I-90 is dead straight there, so I just put the car in neutral and tried to restart the engine. Nothing. The engine turned over, but would not start. Hmmm. So I put on the flashers and let the car slow down as was its wont, and eventually eased onto the shoulder. After a few minutes of poking around under the car I discovered that the road debris had managed to strike the fuel filter and knock it from its mounting bracket on the inside of the frame just in front of the passenger-side rear tire, disconnecting the fuel line and starving the engine in the process. Fortunately the fuel line retaining clip was still attached, and it all went back together with a bit of duck tape, and we were on our way pretty quickly.
But seriously, what are the odds that a rock could be kicked up from the pavement on exactly the trajectory required to hit a two inch diameter target mounted along the frame and tucked up against the body pan?
I would calculate those odds as “Once in a Blue Moon.”
As it turns out, IT WAS a Blue Moon that day!
Postscript: Turns out there were enough Crown Vics out there getting their fuel filters knocked off during Blue Moons that later models (including my 1996 and 2001 Mercury Marquis) had a right angle shield incorporated into the front of the mounting bracket to protect the fuel filter from bouncy road debris.
This is an “existential opportunity” to see something that we don’t see very often.
What am I missing here? Didn’t we have something like this last January? Nothing apocalyptic happened then. Nothing disastrous will happen this time unless you count having your predictions proven wrong disastrous.
People get excited about new years and Christmas and they happen every year without fail. While super moons and blood moons each happen, on average, a few times a year, they don’t always coincide with each other (and they aren’t always wolf moons, IE January moons, either) Next year for example there is no super moon in January. And if I’m reading the timeline for each correctly, the next time a Super and blood moon coincides is December 2021 (in other words there won’t be any super blood moons at all next year)
Always critically evaluate what you read concerning science and things human social related if it is portrayed as “non-fiction.” Keep the BS detector working, ‘cuz on the internet there is so much BS out there.
Take for example the WeatherPicture of the Day for today January 9, 2019 that Anthony has on the right side of the web page.
https://weatherpictureoftheday.com/2019/01/09/venus-rising-at-sunset
It claims to show Venus at sunset on November 9, 2018. Well, it can’t be. Venus crossed in front of the sun relative to Earth on October 26th, 2018 and became a brilliant early morning planet since then. I know, I watched it disappear from the evening sky in late October, and reappear just before sunrise in November. Venus is currently magnificent and bright just before sunrise. It was even more so in December as it is now pulling ahead of the Earth in its 225 Earth day orbit around the Sun.
For our friends in the UK and Europe (weather permitting and you can get out of bed in time) get to see the lunar eclipse end at moonset, Venus will be brilliant and rising in the eastern sky behind you.
Well, the photographer admitted as much – saying it was a typo in the post. Which I am perfectly willing to allow. I lived for four years near to Mt. Monadnock, and if you are outside under a clear morning sky in early November, the fingers don’t work all that well for a while!
Keep the BS detector working, ‘cuz on the internet there is so much BS out there.
wait a minute, you mean “it’s on the internet so it must be true” isn’t true. But I read it on the internet, so it must have been true 😉
Below the Lunar Eclipse Chart,
“Total lunar eclipse: A total lunar eclipse occurs when the moon and the sun are on exact opposite sides of Earth, according to NASA.”
Well yeah, I knew that, but, ” according to NASA.” ?
Suddenly, I have my doubts!
Is it just me?
The Gavin and the boys and girls at GISS will get right on it… as soon as their shut-down furlough ends.
Until then, non-essential NASA operations aren’t happening and this lunar eclipse thus may* not happen until Trump and Congress agree to some more money for NASA.
*note 1: I used “may”
Note 2: the thought of GISS on furlough has me all choked up… in tears of joy — Climate scam hustlers on hold. Film at 11.
Yes indeed. That “according to NASA” is ridiculous. It sounds as if NASA discovered it, or as if it were not true according to other persons. Will we say that, “according to Einstein”, 1 + 1 = 2?
Mick Walker: Well yeah, I knew that, but, ” according to NASA.” ?
Suddenly, I have my doubts!
Is it just me?
yeah, Hansen, Gavin and the gang have given NASA such a bad name that the phrase “according to NASA” does immediately cause doubt of the veracity of whatever that phrase is attached to.
Jean Meeus: That “according to NASA” is ridiculous. It sounds as if NASA discovered it,
I agree that it is an unneeded tag. But I don’t think it’s there to imply “NASA discovered it”, but more to indicate that the it was someone at NASA who was the source of the definition that the caption writer was using. It doesn’t mean no one else could have given a similar definition if asked.
Anthony Watts gone made.
Singing “blue moon” in 2nd LA Niña year.
After https://www.google.com/search?q=paradise+burning+2018&oq=paradise+burning+2018&aqs=chrome.
post traumatic stress disorder.
Paradise lost + 2nd La Niña year
heavy precipitations come down.
__________________________________________________
calm down.
Your children won’t know how chicken ice cream tastes.
Children won’t know what a Super blood wolf moon is. they’ll be things of the past. 😉
According to Sky and Telescope: “The average angular diameter of the full Moon is 31 arcminutes, but during the eclipse it will be 2.2 arcminutes wider.”
That’s about 7.5% larger than the mean size of the moon. For us on the west coast of the US it peaks at 9:12pm and should be a beautiful coppery red just like last January.
Blood moon is the same as a lunar eclipse.
Lunar eclipses occur only when the moon is full.
“Supermoon’s” only occur when the moon is full.
This is not a rare occurrence.
the rarity is in how frequently you get both a Super moon and a blood moon at the same time. it will be a couple of years before both happen at the same time again where as it’ll be only a matter of months for each individually.