Just finishing up reviewing some notes on predicting the sun's
position, but I'm stuck on an apparent inconsistency in an equation:

You can predict the sun's elevation angle (EA) using the well-known
formula:

cos EA = (sin L * sin D) / (cos L * cos D * cos H)

where L is latitude, D sun's declination and H the hour angle. This
works fine when the sun's above the horizon.

But if you turn this round to try to solve for H when the sun is below
the horizon with EA0, eg to predict sunrise/sunset times, then the
algorithm bombs out. For SRSS predictions you seemingly use the zenith
angle rather than the altitude angle, ie with zero dead overhead and
the SRSS angle at eg 90.833°, but in exactly the same equation!

I've always got this work fine with the appropriate adjustment but
never been quite sure as to why the difference and I'd really like to
find out. There must be some trigonometric sublety I guess to do with
which quadrant the angle ends up in as to exactly how the equation is
implemented. Assuming this makes sense to anyone at all, can anyone
shed any light on why the change from elevation to zenith angle as the
sun dips below the horizon?

JGD

On Mon, 28 Mar 2005 16:58:44 +0100, John Dann wrote in


Just finishing up reviewing some notes on predicting the sun's
position, but I'm stuck on an apparent inconsistency in an equation:

You can predict the sun's elevation angle (EA) using the well-known
formula:

cos EA = (sin L * sin D) / (cos L * cos D * cos H)

where L is latitude, D sun's declination and H the hour angle. This
works fine when the sun's above the horizon.

I cannot find my old notes, but have a memory that divison sign "/" should
in fact be a postive sign "+". As it stands, that would lead to a
numerator of zero at the equinoxes.

--
Mike 55.13°N 6.69°W Coleraine posted to uk.sci.weather 28/03/2005 16:54:33 UTC

On Mon, 28 Mar 2005 17:54:33 +0100, Mike Tullett
wrote:


I cannot find my old notes, but have a memory that divison sign "/" should
in fact be a postive sign "+". As it stands, that would lead to a
numerator of zero at the equinoxes.

Thanks Mike, you're right about the plus sign - it should be

sin EA = (cos L * cos D * cos HA) + (sin L * sin D)

to solve for EA, but that was unfortunately just a typo in my writing
the post. The original I've used is indeed as per the corrected
version above.

It's in solving for HA that the division is used, ie

cos HA = (sin EA - sin L * sin D) / (cos L * cos D)

But the inconsistency remains. (Well, I'm sure there's a good and
obvious explanation - it just escapes me ATM!)

JGD

On Mon, 28 Mar 2005 18:05:05 +0100, John Dann wrote in


It's in solving for HA that the division is used, ie

cos HA = (sin EA - sin L * sin D) / (cos L * cos D)

But the inconsistency remains. (Well, I'm sure there's a good and
obvious explanation - it just escapes me ATM!)

Ok John - ignore my last post as I was typing it as you posted the correct
version. Let's take a situation again at the equinox to make life simple.
Then

cos HA = (sin EA) / (cos L)

Let's assume L = 55 and EA is 10 degress below horizon. Then

cos HA = (sin -10) / (cos L) = -0.174 / 0.574 = -0.303

So HA = +108 degrees or -108 degrees and that is just over 7 hours from
noon, either before or after. That would be about right I think.

Or am I working along the wrong lines?:-)

--
Mike 55.13°N 6.69°W Coleraine posted to uk.sci.weather 28/03/2005 17:55:15 UTC

On Mon, 28 Mar 2005 18:05:05 +0100, John Dann wrote in


It's in solving for HA that the division is used, ie

cos HA = (sin EA - sin L * sin D) / (cos L * cos D)

But the inconsistency remains. (Well, I'm sure there's a good and
obvious explanation - it just escapes me ATM!)

Ok John - ignore my last post as I was typing it as you posted the correct
version. Let's take a situation again at the equinox to make life simple.
Then

cos HA = (sin EA) / (cos L)

Let's assume L = 55 and EA is 10 degress below horizon. Then

cos HA = (sin -10) / (cos L) = -0.174 / 0.574 = -0.303

So HA = +108 degrees or -108 degrees and that is just over 7 hours from
noon, either before or after. That would be about right I think.

Or am I working along the wrong lines?:-)

--
Mike 55.13°N 6.69°W Coleraine posted to uk.sci.weather 28/03/2005 17:55:15 UTC

On Mon, 28 Mar 2005 18:05:05 +0100, John Dann wrote in


It's in solving for HA that the division is used, ie

cos HA = (sin EA - sin L * sin D) / (cos L * cos D)

But the inconsistency remains. (Well, I'm sure there's a good and
obvious explanation - it just escapes me ATM!)

Ok John - ignore my last post as I was typing it as you posted the correct
version. Let's take a situation again at the equinox to make life simple.
Then

cos HA = (sin EA) / (cos L)

Let's assume L = 55 and EA is 10 degress below horizon. Then

cos HA = (sin -10) / (cos L) = -0.174 / 0.574 = -0.303

So HA = +108 degrees or -108 degrees and that is just over 7 hours from
noon, either before or after. That would be about right I think.

Or am I working along the wrong lines?:-)

--
Mike 55.13°N 6.69°W Coleraine posted to uk.sci.weather 28/03/2005 17:55:15 UTC

On Mon, 28 Mar 2005 18:05:05 +0100, John Dann wrote in


It's in solving for HA that the division is used, ie

cos HA = (sin EA - sin L * sin D) / (cos L * cos D)

But the inconsistency remains. (Well, I'm sure there's a good and
obvious explanation - it just escapes me ATM!)

Ok John - ignore my last post as I was typing it as you posted the correct
version. Let's take a situation again at the equinox to make life simple.
Then

cos HA = (sin EA) / (cos L)

Let's assume L = 55 and EA is 10 degress below horizon. Then

cos HA = (sin -10) / (cos L) = -0.174 / 0.574 = -0.303

So HA = +108 degrees or -108 degrees and that is just over 7 hours from
noon, either before or after. That would be about right I think.

Or am I working along the wrong lines?:-)

--
Mike 55.13°N 6.69°W Coleraine posted to uk.sci.weather 28/03/2005 17:55:15 UTC

On Mon, 28 Mar 2005 17:54:33 +0100, Mike Tullett
wrote:


I cannot find my old notes, but have a memory that divison sign "/" should
in fact be a postive sign "+". As it stands, that would lead to a
numerator of zero at the equinoxes.

Thanks Mike, you're right about the plus sign - it should be

sin EA = (cos L * cos D * cos HA) + (sin L * sin D)

to solve for EA, but that was unfortunately just a typo in my writing
the post. The original I've used is indeed as per the corrected
version above.

It's in solving for HA that the division is used, ie

cos HA = (sin EA - sin L * sin D) / (cos L * cos D)

But the inconsistency remains. (Well, I'm sure there's a good and
obvious explanation - it just escapes me ATM!)

JGD

On Mon, 28 Mar 2005 17:54:33 +0100, Mike Tullett
wrote:


I cannot find my old notes, but have a memory that divison sign "/" should
in fact be a postive sign "+". As it stands, that would lead to a
numerator of zero at the equinoxes.

Thanks Mike, you're right about the plus sign - it should be

sin EA = (cos L * cos D * cos HA) + (sin L * sin D)

to solve for EA, but that was unfortunately just a typo in my writing
the post. The original I've used is indeed as per the corrected
version above.

It's in solving for HA that the division is used, ie

cos HA = (sin EA - sin L * sin D) / (cos L * cos D)

But the inconsistency remains. (Well, I'm sure there's a good and
obvious explanation - it just escapes me ATM!)

JGD

On Mon, 28 Mar 2005 17:54:33 +0100, Mike Tullett
wrote:


I cannot find my old notes, but have a memory that divison sign "/" should
in fact be a postive sign "+". As it stands, that would lead to a
numerator of zero at the equinoxes.

Thanks Mike, you're right about the plus sign - it should be

sin EA = (cos L * cos D * cos HA) + (sin L * sin D)

to solve for EA, but that was unfortunately just a typo in my writing
the post. The original I've used is indeed as per the corrected
version above.

It's in solving for HA that the division is used, ie

cos HA = (sin EA - sin L * sin D) / (cos L * cos D)

But the inconsistency remains. (Well, I'm sure there's a good and
obvious explanation - it just escapes me ATM!)

JGD