Math Is Fun Forum

harpazo65

Member

From: Brooklyn, NY

Registered: 2024-09-07

Posts: 52

How Far Can You See?

Use the facts that the radius of Earth is 3960 miles and 1 mile = 5280 feet.

A person who is 6 feet tall is standing on the beach in Florida, and looks out onto the Atlantic Ocean. Suddenly, a ship appears on the horizon. How far is the ship from shore?

A friend suggested the following solution.

Let d = distance of ship from shore

Let r = radius of Earth

Let h = height of person

Let m = a mile in terms of feet

Here is the equation my friend developed:

d^2 + r^2 = [(r + (h/m)]^2

The formula above is basically applythe Pythagorean Theorem.

Questions

1. Is my friend right?

2. Is there an easier way to solve this problem even if it means trigonometry or graphing?

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Focus on the journey for there will be those who would love to see you fall.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,627

Re: How Far Can You See?

Not quite sure about the m conversion.  You just need everything in the same units.

That formula works.  Pythag is a part of trigonometry.   I think it's the easiest and quickest method.  d comes out as the straight line distance, not the curvature distance.

Bob

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

KerimF

Member

From: Aleppo-Syria

Registered: 2018-08-10

Posts: 248

Re: How Far Can You See?

Which part of the ship appeared?
I mean, isn't its height important too?

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harpazo65

Member

From: Brooklyn, NY

Registered: 2024-09-07

Posts: 52

Re: How Far Can You See?

Not quite sure about the m conversion.  You just need everything in the same units.

That formula works.  Pythag is a part of trigonometry.   I think it's the easiest and quickest method.  d comes out as the straight line distance, not the curvature distance.

Bob

I will work it out later and show my effort here.

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Focus on the journey for there will be those who would love to see you fall.

harpazo65

Member

From: Brooklyn, NY

Registered: 2024-09-07

Posts: 52

Re: How Far Can You See?

Which part of the ship appeared?
I mean, isn't its height important too?

The question does not give the height of the ship. It is not needed to find the distance it is from shore. I will work it out later on today.

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Focus on the journey for there will be those who would love to see you fall.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,627

Re: How Far Can You See?

That's a fair comment. If the ship was a mile high, you'd be able to see the top long after the bottom had disappeared below the horizon.  The method also disregards the curvature of the Earth and where the person's eyes are in that height figure.  But it gives a good approximation for most practical purposes.

Bob

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

harpazo65

Member

From: Brooklyn, NY

Registered: 2024-09-07

Posts: 52

Re: How Far Can You See?

That's a fair comment. If the ship was a mile high, you'd be able to see the top long after the bottom had disappeared below the horizon.  The method also disregards the curvature of the Earth and where the person's eyes are in that height figure.  But it gives a good approximation for most practical purposes.

Bob

I am not criticizing KerimF. The question does not give the ship's height.
By the way, this problem does involve the curvature of Earth.

d^2 + (3960)^2 = [(3960) + (6/5280)]2

Plugging that into my calculator, I get approximately 3 miles.

What do you say?

d ≈ 3 miles.

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Focus on the journey for there will be those who would love to see you fall.

harpazo65

Member

From: Brooklyn, NY

Registered: 2024-09-07

Posts: 52

Re: How Far Can You See?

Which part of the ship appeared?
I mean, isn't its height important too?

The height of the ship is important but perhaps not for this particular question. The problem does not give the ship's height. By the way, this problem does involve the curvature of Earth.

I will plug the data given into the formula.

d^2 + (3960)^2 = [(3960) + (6/5280)]2

Plugging that into my calculator, I get approximately 3 miles.

What do you say?

d ≈ 3 miles.

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Focus on the journey for there will be those who would love to see you fall.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,627

Re: How Far Can You See?

I get 3 as well.

Bob

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

Oculus8596

Banned

From: Great Lakes,Illinois

Registered: 2024-09-18

Posts: 126

Re: How Far Can You See?

Which part of the ship appeared?
I mean, isn't its height important too?

The height of the ship is important but perhaps not for this particular question. The problem does not give the ship's height. By the way, this problem does involve the curvature of Earth.

I will plug the data given into the formula.

d^2 + (3960)^2 = [(3960) + (6/5280)]2

Plugging that into my calculator, I get approximately 3 miles.

What do you say?

d ≈ 3 miles.

I have seen this problem many times in various forums.
I also got about 3 miles in my calculation.

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