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/sci/ - Science & Math

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>> No.11487442 [View]
File: 67 KB, 1435x890, Screen Shot 2020-03-20 at 12.36.08 PM.jpg [View same] [iqdb] [saucenao] [google]
11487442

My only friend has become a flat earth retard. Normally I'm willing to ignore philosophical differences with friends but he's becoming more and more insufferable with this bullshit. I've tried sending him science videos but he just brushes them off as NASA conspiracies and all this shit. I'm going to attempt to explain it to him myself. Problem is I myself am borderline retarded and haven't taken a math or science class since high school. I need someone to make sure my math isn't shit. He keeps rambling on about how you shouldn't be able to see things from 30 miles away due to the curvature of the earth. So I've set aside an example to illustrate why he's a fucking idiot.

Let's say you have a 900 foot Jesus cross, or 900 feet when you account for the tilt of the earth (so a little over 900 feet) if you were lying prone at sea level 30 miles away and there were no obstructions, without factoring in light refraction or any other weird shit, would you be able to see the statue of our Lord and Savior. Given that the circumference of the earth is 24,901 miles. We can find the angle of this distance.

[math]\frac{30}{24901}=\frac{x}{360}[/math]

[math] x=(360∗30)24901[/math]

[math]x=.43degrees[/math]


We want to find the height of the obstruction caused by the curvature of the earth, which is the middle point of the curve. We can view this as two right trianges, dividing the angle in half. Since we know the radius of the earth is 3959 miles, and we know the angle of the triangle is 0.22 degrees. We can find the adjacent side using the formula
[math]cos(angle / 2) = \frac{x}{3959}[/math]

[math]h=3959−(3959∗cos(0.22))[/math]

convert to feet

[math]h=5280∗(3959−(3959∗cos(0.22)))[/math]

149 feet, which means under ideal conditions for earth curvature obstruction (lying prone at sea level) you can still see a 900 foot object.

>> No.11487433 [DELETED]  [View]
File: 67 KB, 1435x890, Screen Shot 2020-03-20 at 12.36.08 PM.jpg [View same] [iqdb] [saucenao] [google]
11487433

My only friend has become a flat earth retard. Normally I'm willing to ignore philosophical differences with friends but he's becoming more and more insufferable with this bullshit. I've tried sending him science videos but he just brushes them off as NASA conspiracies and all this shit. I'm going to attempt to explain it to him myself. Problem is I myself am borderline retarded and haven't taken a math or science class since high school. I need someone to make sure my math isn't shit. He keeps rambling on about how you shouldn't be able to see things from 30 miles away due to the curvature of the earth. So I've set aside an example to illustrate why he's a fucking idiot.

Let's say you have a 900 foot Jesus cross, or 900 feet when you account for the tilt of the earth (so a little over 900 feet) if you were lying prone at sea level 30 miles away and there were no obstructions, without factoring in light refraction or any other weird shit, would you be able to see the statue of our Lord and Savior. Given that the circumference of the earth is 24,901 miles. We can find the angle of this distance.

[math]\frac{30}{24901}=\frac{x}{360}[/math]

[math] x=(360∗30)24901[/math]

[math]x=.43degrees[/math]


We want to find the height of the obstruction caused by the curvature of the earth, which is the middle point of the curve. We can view this as two right trianges, dividing the angle in half. Since we know the radius of the earth is 3959 miles, and we know the angle of the triangle is 0.22 degrees. We can find the adjacent side using the formula
[math]cos(angle / 2) = \frac{x}{3959}[/math]

[math]h=3959−(3959∗cos(0.22))[/math]

convert to feet

[math]h=5280∗(3959−(3959∗cos(0.22)))[/math]

149 feet, which means under ideal conditions for earth curvature obstruction (lying prone at sea level) you can still see a 900 foot object.

>> No.11487417 [DELETED]  [View]
File: 67 KB, 1435x890, Screen Shot 2020-03-20 at 12.36.08 PM.jpg [View same] [iqdb] [saucenao] [google]
11487417

My only friend has become a flat earth retard. Normally I'm willing to ignore philosophical differences with friends but he's becoming more and more insufferable with this bullshit. I've tried sending him science videos but he just brushes them off as NASA conspiracies and all this shit. I'm going to attempt to explain it to him myself. Problem is I myself am borderline retarded and haven't taken a math or science class since high school. I need someone to make sure my math isn't shit. He keeps rambling on about how you shouldn't be able to see things from 30 miles away due to the curvature of the earth. So I've set aside an example to illustrate why he's a fucking idiot.

Let's say you have a 900 foot Jesus cross, or 900 feet when you account for the tilt of the earth (so a little over 900 feet) if you were lying prone at sea level 30 miles away and there were no obstructions, without factoring in light refraction or any other weird shit, would you be able to see the statue of our Lord and Savior. Given that the circumference of the earth is 24,901 miles. We can find the angle of this distance.

[math] \frac{30}{24901}=\frac{x}{360}[/math]
[math] x=(360∗30)24901[/math]
[math]x=.43degrees[/math]


We want to find the height of the obstruction caused by the curvature of the earth, which is the middle point of the curve. We can view this as two right trianges, dividing the angle in half. Since we know the radius of the earth is 3959 miles, and we know the angle of the triangle is 0.22 degrees. We can find the adjacent side using the formula
[math]cos(angle / 2) = \frac{x}{3959}[/math]

[math]h=3959−(3959∗cos(0.22))[/math]

convert it to feet

[math]h=5280 * (3959−(3959∗cos(0.22)))[/math]

149 feet, which means under ideal conditions for earth curvature obstruction (lying prone at sea level) you can still see a 900 foot object.

>> No.11487387 [DELETED]  [View]
File: 67 KB, 1435x890, Screen Shot 2020-03-20 at 12.36.08 PM.jpg [View same] [iqdb] [saucenao] [google]
11487387

My only friend has become a flat earth retard. Normally I'm willing to ignore philosophical differences with friends but he's becoming more and more insufferable with this bullshit. I've tried sending him science videos but he just brushes them off as NASA conspiracies and all this shit. I'm going to attempt to explain it to him myself. Problem is I myself am borderline retarded and haven't taken a math or science class since high school. I need someone to make sure my math isn't shit. He keeps rambling on about how you shouldn't be able to see things from 30 miles away due to the curvature of the earth. So I've set aside an example to illustrate why he's a fucking idiot.

Let's say you have a 900 foot Jesus cross, or 900 feet when you account for the tilt of the earth (so a little over 900 feet) if you were lying prone at sea level 30 miles away and there were no obstructions, without factoring in light refraction or any other weird shit, would you be able to see the statue of our Lord and Savior. Given that the circumference of the earth is 24,901 miles. We can find the angle of this distance.

[math]\frac{30}{24901} = {x}{360}[/math]
[math]x = \frac{(360 * 30)}{24901}[/math]
[math]x = .43 degrees[math]

We want to find the height of the obstruction caused by the curvature of the earth, so we want to find the height of the curve at its middle point. We can view this as two right triangles, dividing the angle in half. Since we know the radius of the earth is 3959 miles, and we know the one angle of the triangle is 0.22 degrees We can find the adjacent side using the formula. Then subtract that from the radius to get H
[math]cos(angle/2) = \frac {x}{3959}[/math]
[math]h = 3959 - (3959 * cos(angle/2) = x)
[/math]

convert it to feet

[math]h = 5280 * (3959 - (3959 * cos(angle/2) = x))
[/math]

149 feet, which means under ideal conditions for earth curvature obstruction (lying prone at sea level) you can still see a 900 foot object.

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