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Hi;

I need help with this one if anybody is versed in astronomy.

Duffy's idea is that when he gets a certain bit of information, the position of the sun and moon at that time are significant.

If within 30 days a mag 5+ quake happens, the sunrise/sunset curve will be passing thru one of those points.

Now I can compute the subsolar and sublunar points and I can compute sunrise/sunset times for the quakes but I can't compute the curve representing all the places on earth where they're rising or setting.

I can compute the distance between two points on earth so can I say if sunrise is the right time and it's 90 degrees from a certain location that the curve falls thru that location?

I think so but I'm not sure.

Roger
Roger,

I was pretty good at astronomy but that question is too difficult for me. Can't help you.

Chris
(02-02-2017, 09:40 PM)Island Chris Wrote: [ -> ]Roger,

I was pretty good at astronomy but that question is too difficult for me. Can't help you.

Chris

Chris;

Ok, thanks for replying.

I'm pretty sure it's correct for a sphere but the earth isn't a sphere.

Roger
(01-31-2017, 02:55 AM)Roger Hunter Wrote: [ -> ]Hi;

I need help with this one if anybody is versed in astronomy.

Duffy's idea is that when he gets a certain bit of information, the position of the sun and moon at that time are significant.

If within 30 days a mag 5+ quake happens, the sunrise/sunset curve will be passing thru one of those points.

Now I can compute the subsolar and sublunar points and I can compute sunrise/sunset times for the quakes but I can't compute the curve representing all the places on earth where they're rising or setting.

I can compute the distance between two points on earth so can I say if sunrise is the right time and it's 90 degrees from a certain location that the curve falls thru that location?

I think so but I'm not sure.

Roger

Sorry everyone for being away so much. Hopefully that will change within a couple of months.

Am I understanding?

  1. we have lat/lon points (crosses) [how determined isn't relevant at the moment]
  2. we have quake lat/lon
  3. the correlation is true if the cross is in sunrise/sunset at the time of the quake
I anticipate being able to be online again this Tuesday.

Brian
(02-06-2017, 03:03 AM)Skywise Wrote: [ -> ]
(01-31-2017, 02:55 AM)Roger Hunter Wrote: [ -> ]Hi;

I need help with this one if anybody is versed in astronomy.

Duffy's idea is that when he gets a certain bit of information, the position of the sun and moon at that time are significant.

If within 30 days a mag 5+ quake happens, the sunrise/sunset curve will be passing thru one of those points.

Now I can compute the subsolar and sublunar points and I can compute sunrise/sunset times for the quakes but I can't compute the curve representing all the places on earth where they're rising or setting.

I can compute the distance between two points on earth so can I say if sunrise is the right time and it's 90 degrees from a certain location that the curve falls thru that location?

I think so but I'm not sure.

Roger

Sorry everyone for being away so much. Hopefully that will change within a couple of months.

Am I understanding?

  1. we have lat/lon points (crosses) [how determined isn't relevant at the moment]
  2. we have quake lat/lon
  3. the correlation is true if the cross is in sunrise/sunset at the time of the quake
I anticipate being able to be online again this Tuesday.

Brian

Brian;

Yes, if I understand Duffy, that's right.

But check with him to be sure. 

My question was if the quake is 90 degrees from the cross is that sunrise/set by definition.

The earth isn't a perfect sphere so there's bound to be some difference but is it significant?

Roger
(02-06-2017, 03:34 AM)Roger Hunter Wrote: [ -> ]My question was if the quake is 90 degrees from the cross is that sunrise/set by definition.


No. Let's look at a simplified case.

Let's put the cross on the North pole.

On the Winter Solstice, we already know that the North Pole is in perpetual darkness. So no matter the time on this day, it will never be in sunrise or sunset on that day.

Now imagine a quake on the equator. Doesn't matter where as long as the latitude is zero.

This quake is exactly 90 degrees from the North Pole - exactly 90 degrees from the cross.

Yet, no sunrise/sunset.

This is an extreme case, but unfortunately it shows that you cannot use your assumption.

Just calculate the sun angle of the cross locations at the time of the quake. This should give you an angle you can use to determine angle to the horizon, and from that decide if the location is in sunrise/set.

Reversing this, the prediction would be, any time a cross is in sunrise/sunset, there may be a quake somewhere on the planet. This is TOO LOSE of a specification to have any statistical significance.

Am I missing something?

Brian
(02-06-2017, 04:39 AM)Skywise Wrote: [ -> ]
(02-06-2017, 03:34 AM)Roger Hunter Wrote: [ -> ]My question was if the quake is 90 degrees from the cross is that sunrise/set by definition.

Brian;

Quote:No. Let's look at a simplified case.

Let's put the cross on the North pole.

On the Winter Solstice, we already know that the North Pole is in perpetual darkness. So no matter the time on this day, it will never be in sunrise or sunset on that day.

Now imagine a quake on the equator. Doesn't matter where as long as the latitude is zero.

This quake is exactly 90 degrees from the North Pole - exactly 90 degrees from the cross.

Yet, no sunrise/sunset.

This is an extreme case, but unfortunately it shows that you cannot use your assumption.

Obviously which shows I'm too ignorant of astronomy. I should have seen that myself.

Quote:Just calculate the sun angle of the cross locations at the time of the quake. This should give you an angle you can use to determine angle to the horizon, and from that decide if the location is in sunrise/set.

What angle? Sun to equator?

Quote:Reversing this, the prediction would be, any time a cross is in sunrise/sunset, there may be a quake somewhere on the planet. This is TOO LOSE of a specification to have any statistical significance.

Agreed but it happens so often I need to prove it's chance.

Quote:Am I missing something?

No, you nailed it. I'm embarrassed, thanks.

Roger
(02-06-2017, 02:56 PM)Roger Hunter Wrote: [ -> ]
(02-06-2017, 04:39 AM)Skywise Wrote: [ -> ]
Quote:Just calculate the sun angle of the cross locations at the time of the quake. This should give you an angle you can use to determine angle to the horizon, and from that decide if the location is in sunrise/set.

What angle? Sun to equator?


OK. I think I messed that up.

Compute the subsolar lat/lon at the time of the quake. Determine the distance from the subsolar point to the cross in degrees. If the cross is in sunrise/sunset this angle will be 90 degrees.

Brian
(02-07-2017, 06:52 AM)Skywise Wrote: [ -> ]
(02-06-2017, 02:56 PM)Roger Hunter Wrote: [ -> ]
(02-06-2017, 04:39 AM)Skywise Wrote: [ -> ]
Quote:Just calculate the sun angle of the cross locations at the time of the quake. This should give you an angle you can use to determine angle to the horizon, and from that decide if the location is in sunrise/set.

What angle? Sun to equator?


OK. I think I messed that up.

Compute the subsolar lat/lon at the time of the quake. Determine the distance from the subsolar point to the cross in degrees. If the cross is in sunrise/sunset this angle will be 90 degrees.

Brian

But - but that's what I've been doing!

That's not an angle anyway, it's a distance.

It's also a geocentric angle, if that's what you mean.

Roger
(02-07-2017, 02:16 PM)Roger Hunter Wrote: [ -> ]It's also a geocentric angle, if that's what you mean.


Yes. The 'cross' must be in sunrise/sunset if it is 90 degrees away from the subsolar point.

But this is an unusual definition of sunrise/sunset, as some locations, particularly those towards the poles, have very long sunrise/sunset times. For example, locations on the arctic circle around the time of winter solstice have the sun going around the sky never getting off the horizon.

But for the test here, it works.

BTW, if I've done my math right, a band +-1 degree wide 90 degrees from a point on a sphere covers about 1.75% of that sphere. In the case of a spherical earth of 6371 km radius, that's a surface area of 510,064,472 km^2.

That's equivalent to a circular area with a radius of 12,742 kilometers. Basically a bulls eye that big. From past experience this tells us that the odds of a hit are very high. Recall my Earthquake Dart Board experiment only used 1000km radii.

In fact, continuing the comparison to my EQDB, my 11 daily 1000km radius circles had a total area 34,557,519km. So Duffy's prediction area is 14.76 times larger than mine, and his windows are 30 days compared to my 7 days (4.28 times larger).

A significant difference is that my locations targeted high seismic regions by design, whereas Duffy's do not. But I think this may be offset by the much larger prediction area and time windows. Although his windows aren't truly a full 30 days since they are only valid for sunrise/sunset.

Brian
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