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#1 2024-05-15 11:17:08

mathxyz
Member
From: Brooklyn, NY
Registered: 2024-02-24
Posts: 1,053

Gibb's Hill Lighthouse

The Gibb’s Hill Lighthouse, Southampton, Bermuda, in operation since 1846, stands 117 feet high on a hill 245 feet high, so its beam of light is 362 feet above sea level. A brochure states that the light itself can be seen on the horizon about 26 miles distant. Verify the correctness of this information. The brochure further states that ships 40 miles away can see the light and planes flying at 10,000 feet can see it 120 miles away. Verify the accuracy of these statements. What assumption did the brochure make about the height of the ship?


Too much information here threw me into a loop.


You say?

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#2 2024-05-15 20:41:27

Bob
Administrator
Registered: 2010-06-20
Posts: 10,627

Re: Gibb's Hill Lighthouse

Yes, it does get complicated.  I think this is the diagram:

EMO0lg0.gif

LB represents the (hill + lighthouse) and H is a point on the horizon.

You'll have to convert the feet into miles.

Angle LHO is 90 so you can use Pythag to get LH.

I'll leave the ship for a moment. Once we have a method for the plane it will be easy to adapt it for the ship so I'll attempt the plane next.

For this part you'll still only be able to see to the horizon but the plane P is above the Earth's surface, so we'll just be able to see it if LHP is a straight line.

The difficulty this time is that triangle LPO does not have a right angle.

LOP looks a bit like 90 in my diagram. It isn't!

I need to think about this part so I'll come back to this later.

LATER EDIT: Ok I have it, I think.

We know these distances:  LO, OH, AP and OP.  We can use trig on triangle LHO to calculate angle OLH = OLP as LHP is a straight
line

So in triangle LOP we have two sides and an angle. That's enough to use the cosine rule to find LP.

Do you know the formula for this?

Now replace P by S, the ship. Assume the ship's distance is as given and work backwards to get SA.


Bob


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 smile

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#3 2024-05-16 01:48:04

mathxyz
Member
From: Brooklyn, NY
Registered: 2024-02-24
Posts: 1,053

Re: Gibb's Hill Lighthouse

Bob wrote:

Yes, it does get complicated.  I think this is the diagram:

https://i.imgur.com/EMO0lg0.gif

LB represents the (hill + lighthouse) and H is a point on the horizon.

You'll have to convert the feet into miles.

Angle LHO is 90 so you can use Pythag to get LH.

I'll leave the ship for a moment. Once we have a method for the plane it will be easy to adapt it for the ship so I'll attempt the plane next.

For this part you'll still only be able to see to the horizon but the plane P is above the Earth's surface, so we'll just be able to see it if LHP is a straight line.

The difficulty this time is that triangle LPO does not have a right angle.

LOP looks a bit like 90 in my diagram. It isn't!

I need to think about this part so I'll come back to this later.

LATER EDIT: Ok I have it, I think.

We know these distances:  LO, OH, AP and OP.  We can use trig on triangle LHO to calculate angle OLH = OLP as LHP is a straight
line

So in triangle LOP we have two sides and an angle. That's enough to use the cosine rule to find LP.

Do you know the formula for this?

Now replace P by S, the ship. Assume the ship's distance is as given and work backwards to get SA.


Bob

1. Nice picture.


2. I will work on this on paper and return here if needed.

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