Math/geometry -- area calculation

I think these are triangle and trapezoids, not cones.

The final shape for calculation 2b is a trapezoid.
What is the distance from the point of intersection to right side of the figure?
What determines it?  Is it a fixed range. or something else?

You may be able to find all the distances (and all the areas) from similar triangles.

Is the boat always pointed directly at the target?

If this really is a 3D problem, the the shape is properly called a frustum of a cone.

If the area you want in Q2 is one the surface of the sea, then you have a 1D problem with triangles and trapezoids.

If the area you want is a giant, expanding circular tunnel (half above the water and half below), then you have cones and frustums.

Correcting an earlier error:
>>  If the area you want in Q2 is one the surface of the sea, then you have a 2D problem

Here is the similar triangle solution.
This drawing is to scale, with  x =  2 boxes/foot    and   y =  4 boxes/mile
 There are three similar triangles:
The smallest includes the target and the intersection point.
The next largest includes the intersection point and the box.

Since the triangles are similar, the distances to the intersection point will be in the ratio 3:1

I don't know how you want to define the largest triangle, but if you know either the range or the width, you can solve for the other data since this triangle is similar to the first two.

This will let you find the area of the trapezoid.

Assuming you mean triangles when you say cones,
What are the two triangles in question?
Are the three sided of one triangle formed by the two red lines and the yellow box?
Are the three sided of the other triangle formed by the two red lines and some other line?
Are the two red lines supposed to be tangent to the green circle?
Do you need the area to greater precision than the approximation cosine(1 ft/1 mile)=1 ?
Where are the grey triangles?

d-glitch:

You're spot on with your questions... here's a response to your comments:

1. Yes... it's a trapezoid.

2. Distance from intersection to figure (green circle)... I'm not sure.  I only know the distance from boat to green circle as well as width of the yellow rectange.  I hope this can be computed.

3. 3D vs. 2D.  Green circle may be on either surface (as illustrated) or airborne.   This would make it a 3D; however, I am not concerned about sub-surface.  Thus, it'll be only a half of the 3D, right?

How can this be computed in Excel?   Thank you in advance!!

EEH

With the accuracy of cosine(1 ft/1 mile)=1, and assuming that the lines are tangent to the circle, the distance from the intersection to the green circle would be distance from boat to green circle * diameter of green circle / (diameter of green circle+ width of the yellow rectangle)

A trapezoid is not a 3D figure, are you talking about the area bounded by the dotted lines and the lines labeled 3FT and 6FT in d-glitch's http:#a40686044 diagram?  I don't see a 6FT label in your http://filedb.experts-exchange.com/incoming/2015/03_w13/905001/Area-calculation.xlsx diagram, so how is the right side defined?

If the green circle is actually a 3D sphere, is the yellow box actually a disk? and do you want the surface area of the frustrum formed by rotating d-glitch's trapezoid around the axis formed by the bisector of the two red lines passing through the yellow box and green circle?
If so, and if the base of the trapezoid is actually 6FT, then the height of the trapezoid would be distance from boat to green circle * diameter of green circle / (diameter of green circle+width of the yellow rectangle) * (6FT+3FT)/3FT
The surface area of the side of frustrum, not counting the circular ends, would be pi*(6FT+3FT)*sqrt((6FT-3FT)^2+height^2)
The area of the 3FT diameter circle would be pi*(3FT)^2/4 and the area of the 6FT diameter circle would be pi*(6FT)^2/4

>>   I only know the distance from boat to green circle as well as width of the yellow rectangle.  

That is good, but we still need one more piece of information:

>>  What is the distance from the point of intersection to right side of the figure?
     
From the circle to the right, there is no information.
This is also what ozo was asking.
I arbitrarily set it to four miles in my drawing.  The lines go on to infinity if you don't limit them.

What are you trying to do?  Why does the area matter?

The geometry can be perfect, but it may not be practical.  Do you really think you can see a 1 FT target at a range of 2 miles?

d-glitch:

Yes, I imagine that more information is needed.   Right now, the assumption is that all required values are known to an operator.   So, plugging in any required is fair game right now.

Right now, the purpose of this "red trapezoid" is to determine the a "danger zone" (as part of a fundamental military application).   Things are very conceptual right now so I may not have all the answers at this time.

With respect to seeing a "1 FT target at a range of 2 miles"... the answer is "yes".... however, I can't go into details here as to why/what.

Again, right now this is more conceptual... I'm just trying to gauge whether or not this could be calculated in Excel (i.e., instead of using a modeling tool).    Thank you in advance!

EEH

ozo:

"If the green circle is actually a 3D sphere,"
- Not it could be any surface... no matter what shape, the left/right boundaries will determine the angle (from boat).

"Is the yellow box actually a disk"
- View it as a surface...

Not sure if I track the proposed math.    Any chance this could be put into Excel to compute the area (2D... surface to surface is fine for right now).

EEH

The trapezoid is the danger zone that exists on the surface of the flat ocean if you aim precisely at some part of the target from the yellow box, but miss because the target moves after you fire.

If you miss because you didn't aim properly, then the danger zone will be bigger.

So the missing piece of information is the range of your weapon.  The drawing I posted earlier solves the problem approximately for a range of 4.5 miles.

The midpoint of the 6 ft line is 4.5 mi from boat.  So the ends of the line are a little further away.

We can agree on 3 ft box, 1 ft target, 2 mi distance.
Tell us what do you want to use for a range.

The danger zone would be more like the volume of the frustrum than the surface area of the frustrum,
or the more relevant area may be the area of the sea surface within the frustrum
Aiming at a target 2 miles away, I would think there would be danger from a red line missing the target by a few inches as well as danger from a red line tangent to the circle.
And if the target can move, then the circle should cover all places the target could be, or where the operator of the weapon might think it could be.

First of all my apologies for not having responded to the question in a while...

Both solutions work great... I appreciate the insights and spreadsheets for calculations!!!

EEH