Help for arithmetically challenged

I'm trying to work out percentage difference in solar radiation
reaching a point on Earth at perihelion and aphelion. (At a basic
level - not allowing for atmospheric absorbtion and reflection etc.)

perihelion distance 147,098,217 kms
aphelion distance 152,097,783 kms
Sun diameter 1,392,520 kms

So far so good - then it starts getting blurry for me.

I think the angular size changes by 3.4 percent. Squaring the angular
size at peri- and aphelion gives a change of 6.9 percent which seems
OTT and therefore wrong.

Help - where am I going wrong?

PS Not - repeat NOT - homework: I left school 44 years ago.
(Correction for lady readers, that should read 3 years ago.)

--
Martin Frey
http://www.hadastro.org.uk
N 51 02 E 0 47

Wasn't it Martin Frey who wrote:
I'm trying to work out percentage difference in solar radiation
reaching a point on Earth at perihelion and aphelion. (At a basic
level - not allowing for atmospheric absorbtion and reflection etc.)

perihelion distance 147,098,217 kms
aphelion distance 152,097,783 kms
Sun diameter 1,392,520 kms

So far so good - then it starts getting blurry for me.

I think the angular size changes by 3.4 percent. Squaring the angular
size at peri- and aphelion gives a change of 6.9 percent which seems
OTT and therefore wrong.

Help - where am I going wrong?

That seems to be an odd way of approaching it, but it is giving you the
correct answer. The more normal method is to observe that the intensity
varies with the inverse square of the distance, so just divide the
squares of the distances.

--
Mike Williams
Gentleman of Leisure

Mike Williams wrote:

That seems to be an odd way of approaching it, but it is giving you the
correct answer. The more normal method is to observe that the intensity
varies with the inverse square of the distance, so just divide the
squares of the distances.

Dohh - thanks.

6.9pc still seems colossal but you can't argue with maths.

--
Martin Frey
http://www.hadastro.org.uk
N 51 02 E 0 47

On Fri, 30 Jul 2004 12:40:44 +0100, Martin Frey
wrote:

I'm trying to work out percentage difference in solar radiation
reaching a point on Earth at perihelion and aphelion. (At a basic
level - not allowing for atmospheric absorbtion and reflection etc.)

Why? Just measure it.

On Fri, 30 Jul 2004 13:09:43 +0100, Mike Williams
wrote:

That seems to be an odd way of approaching it, but it is giving you the
correct answer. The more normal method is to observe that the intensity
varies with the inverse square of the distance, so just divide the
squares of the distances.

An even better answer comes by just measuring the frequency (wave
length) that interests one. It eliminates the errors that math.
estimates. Inverse square is also more important than absorption
barriers in radiation protection - but for the definitive answer,
measurements are always obtained at many points in `protected' area.

On Fri, 30 Jul 2004 14:27:48 +0100, Martin Frey
wrote:

6.9pc still seems colossal but you can't argue with maths.

I can until you show me the actual measurements.

vonroach wrote:

On Fri, 30 Jul 2004 12:40:44 +0100, Martin Frey
wrote:

I'm trying to work out percentage difference in solar radiation
reaching a point on Earth at perihelion and aphelion. (At a basic
level - not allowing for atmospheric absorbtion and reflection etc.)

Why? Just measure it.

See opening post of this thread.

(And I dont want to wait until July 2005 and either Jan or July will
probably be cloudy precluding measurement.)

--
Martin Frey
http://www.hadastro.org.uk
N 51 02 E 0 47

vonroach wrote:

On Fri, 30 Jul 2004 13:09:43 +0100, Mike Williams
wrote:

That seems to be an odd way of approaching it, but it is giving you the
correct answer. The more normal method is to observe that the intensity
varies with the inverse square of the distance, so just divide the
squares of the distances.

An even better answer comes by just measuring the frequency (wave
length) that interests one. It eliminates the errors that math.
estimates. Inverse square is also more important than absorption
barriers in radiation protection - but for the definitive answer,
measurements are always obtained at many points in `protected' area.

See opening post in thread.

--
Martin Frey
http://www.hadastro.org.uk
N 51 02 E 0 47

vonroach wrote:

On Fri, 30 Jul 2004 14:27:48 +0100, Martin Frey
wrote:

6.9pc still seems colossal but you can't argue with maths.

I can until you show me the actual measurements.

See opening post of this thread.

--
Martin Frey
http://www.hadastro.org.uk
N 51 02 E 0 47

vonroach wrote:
On Fri, 30 Jul 2004 13:09:43 +0100, Mike Williams
wrote:

That seems to be an odd way of approaching it, but it is giving you the
correct answer. The more normal method is to observe that the intensity
varies with the inverse square of the distance, so just divide the
squares of the distances.

An even better answer comes by just measuring the frequency (wave
length) that interests one. It eliminates the errors that math.
estimates.

Are you on drugs?


Tim
--
Google is not the only search engine.