Marginal Error of Prayer Directions

[pfssassin]

If you are trying to pray towards Mecca for example (a small point on Earth's, assume spherical, surface) and you are standing on the exact other side your angle doesn't matter since every angle you face at will Cross mecca. Now my question is how do you go about calculating the effect of a 1 degree rotation on a point on a sphere that's not exactly opposite to the point of reference (mecca). So if I am for example praying in NY and face the direction they tell me (exactly) but then rotate by 1 degree how much will I miss mecca by?

I can easily do this on a flat surface using basic trig but it gets weird when you take two points on a sphere.

Note: Although irrelevant to the physics of this, I am an atheist. Just curious.

 

[Dickfore]

the physics of this,

What physics is involved in this?

 

[pfssassin]

What physics is involved in this?

Actually you are right... this is a math question. Seriously never really considered the fact. Just not a member to any math forums so I thought I'd ask here.

 

[Dickfore]

The analogue of straight lines on a spherical surface are great circles.

 

[pfssassin]

Yeah thanks. Just thought about it there can't really be a single forumla for it.

 

[Dickfore]

there is, but depends what you want your input variables to be.

 

[pfssassin]

What I am looking for is consider drawing a circle between New York (single point) and LA but the circle goes around Earth and NY and LA are not necessarily on the opposite sides of the circle which lies on the sphere. If you change the direction of this sphere by 1 degree from NY, how far would the closest distance between LA and the newly formed circle lying on the Earth (sphere) be?

 

[Dickfore]

how are the points on the sphere labeled?

 

[pfssassin]

Not sure what that means.

 

[Dickfore]

Do you want the points input through their latitudes and longitudes?

 

[pfssassin]

That sounds good.

 

[Dickfore]

The distance by which you would miss is:

<br /> s = R \, \varphi \, \sin{\gamma}<br />

where

R - radius of the sphere

\varphi - angular deviation (in radians)

<br /> \cos{\gamma} = \sin{\lambda_{1}} \, \sin{\lambda_{2}} + \cos{\lambda_{1}} \, \cos{\lambda_{2}} \, \cos{(\phi_{2} - \phi_{1})}<br />

where \lambda is the latitude and \phi the longitude of a point on the sphere.

 

[pfssassin]

Thanks a lot man. Couple questions:

1. To get y (looking thing) alone just take the inverse cos of the last equation right?
2. what is angular deviation? (googling returned nothing useful)
3. Whats this formula called? I want to see a sample problem so I can actually learn how it works.
4. Lastly latitude and long are degrees right?
Thanks again!

 

[Dickfore]

EDIT:

Scratch the last formula. The correct one is:

<br /> s = 2 R \, \arcsin{\left[\sin{(\gamma)} \, \sin\left(\frac{\varphi}{2}\right)\right]}<br />

I can't draw a diagram now and show my steps in the derivation, but maybe someone else can check if it's correct.

 

[Dickfore]

angular deviation is the angle that you said is 1^{\circ} in the original statement of your problem.

All the angles that are inside a trigonometric function can be degrees if your calculator is set up in degrees. However, if you find the inverse sine (arcsin), the result will be in degrees again (unless you change it to rad). The last formula (for s) assumes that the result is in radians.

I don't know how this formula is called. I just derived it now.

 

[pfssassin]

Dude. You are ****ing sick. Wish I could pull some **** like that. Working a sample problem will post soon.

 

[Dickfore]

The formula for \cos{\gamma} is actually the dot product formula for two unit vectors in the direction of the respective points on the sphere and expressing their Cartesian coordinates through the latitude and longitude:

<br /> \left\langle \cos{\lambda} \, \cos{\phi}, \cos{\lambda} \, \sin{\phi}, \sin{\lambda} \right\rangle<br />

 

[pfssassin]

The distance by which you would miss is:

<br /> s = R \, \varphi \, \sin{\gamma}<br />

where

R - radius of the sphere

\varphi - angular deviation (in radians)

<br /> \cos{\gamma} = \sin{\lambda_{1}} \, \sin{\lambda_{2}} + \cos{\lambda_{1}} \, \cos{\lambda_{2}} \, \cos{(\phi_{2} - \phi_{1})}<br />

where \lambda is the latitude and \phi the longitude of a point on the sphere.

Random Place in D.C: 38.895112,-77.036366
Mecca: 21.427378,39.814838


cosL = sin(38.895112)sin(21.427378) +cos(38.895112)cos(21.427378)cos(-77.036366-39.814838)
cosL = -0.097855196092127817015497218386228992559245305387623345867

ArcCos(-0.097855196092127817015497218386228992559245305387623345867) = 1.66880837070653510548192373051960254679143170988820787811 Radians

s = 2(6371.0)arcsin[sin(1.66880837070653510548192373051960254679143170988820787811)sin(pi/360)]

s = 110.61 km

**** dude sick! Thanks so much. If you ever get time to show how its derived it would be much appreciated.

 

[pfssassin]

Also I may have hit my questions before becoming annoying quota but I was wondering: does this measure the distance on the surface of the sphere or the direct distance between the two points?

 

[Dickfore]

Example:

New York City: 40^{\circ} \, 43&#039; \, \mathrm{N}, 74^{\circ} 0&#039; \, \mathrm{W}

<br /> \lambda_{1} = +\left(40 + \frac{43}{60}\right)^{\circ} = +40.717^{\circ}<br />

<br /> \phi_{1} = -\left(74 + \frac{0}{60}\right)^{\circ} = -74.000^{\circ}<br />

Mecca: 21^{\circ} \, 25&#039; \, \mathrm{N}, 39^{\circ} 49&#039; \, \mathrm{E}

<br /> \lambda_{2} = +\left(21 + \frac{25}{60} \right)^{\circ} = + 21.416^{\circ}<br />

<br /> \phi_{2} = +\left(39 + \frac{49}{60}\right)^{\circ} = +39.817^{\circ}<br />

<br /> \cos{\gamma} = 0.65232 \times 0.36514 + 0.75794 \times 0.93095 \times (-0.40382)<br />

<br /> \cos{\gamma} = -0.04675 \Rightarrow \sin{\gamma} = \sqrt{1 - \cos^{2}{\gamma}} = 0.99891<br />

<br /> \sin{0.5^{\circ}} = 0.00873<br />

<br /> 2 \times \arcsin{\left(0.99891 \times 0.00873\right)} = 2 \times 0.499 = 0.999^{\circ}<br />

You don't need to know the radius of the Earth. You just need to remember that the circumference of a great circle is very close to 40,000 km. Then, an arc with an angle of 0.999^{\circ} would correspond to:

<br /> 4 \times 10^{4} \, \mathrm{km} \times \frac{0.999^{\circ}}{360^{\circ}} = 111 \, \mathrm{km}<br />

 

[Dickfore]

Also I may have hit my questions before becoming annoying quota but I was wondering: does this measure the distance on the surface of the sphere or the direct distance between the two points?

This measures the shortest distance along the surface. That's why I corrected the first formula.