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longgone's avatar

Do you remember any particularly good lessons?

Asked by longgone (19724points) January 28th, 2015

In your school days, were there any lessons that stood out – “different” enough for you to still remember today?

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12 Answers

elbanditoroso's avatar

A couple, but they are so ingrained I am not sure I could even describe them.

We were one of the first classes to be taught the “new math” using some educational theories in place in the late 60s and early 70s.. Anyway, they taught us about a different (better?) method of figuring out equations in your head,. To this day, I can do more math in my head than the average person.

janbb's avatar

My seventh grade science teacher brought in the special section from the NY Times when the Watson and Crick DNA model story broke. That was exciting to learn about and he explained it really clearly. He also taught us how an automobile engine works. It’s funny because science is not my field but he was a great teacher.

Dutchess_III's avatar

Um…dissecting frogs and making their dead legs move via electrical impulses.

My creative writing class stands out in my mind. Teacher once gave me an F for a story I wrote, along with a note to come see her. Turns out she thought I had plagiarized it! I hadn’t. It was all me. So I wound up with an A++. :D

kritiper's avatar

Generally speaking, never assume anything.
There are two sides to every story, so don’t jump to conclusions before hearing what the other person has to say.
Junior high school is a social hell.

LuckyGuy's avatar

I learned in 7th grade how to determine the distance to the horizon. The teacher said “It’s as easy as 1,2,3.”

The distance to the horizon in miles is 1.23 x the square root of your height above the water in feet.
Over the decades I’ve used that to estimate how far ships are in the distance, curvature of the Earth, height of radio towers, line of sight communication for microwave dishes. Very useful.
It’s as easy as 1,2,3.

There are others. Compound interest and the rule of 72.

elbanditoroso's avatar

@LuckyGuy – I didn’t know 1–2-3 trick but I’m going to try it this afternoon on my drive home. Cool

janbb's avatar

@LuckyGuy Butthat is predicated on always know how far above sea level you are. Is that very useful?

LuckyGuy's avatar

@janbb It only depends upon how far above the water you are. I live near lake Ontario, elevation about 250 ft. If I am on the shore of Hamlin Beach and am about 10 ft above the water the horizon is ~ 4 miles away. If I am on the shore in Miami about 10 ft above the water the horizon is still about 4 miles away. What is really neat is you can see and understand why the bottom of ships are cut off when they are in the distance. If you are 10 ft above the water the horizon is 4 miles away and if the lower 20 of the ship is missing that means it is about 1.23 x Sqrt(20) = 5 miles past that or 9 miles away. With that info you can dial in your artillery rounds. :-)

On a clear day, for an experiment try to watch a ship slowly sink out of sight before it fades away.

Here’s a practical, real world example. If I have a tower on a 2000 ft high mountain, say in the Southern Tier, and want to talk with a city with an elevation of 500 ft, Rochester NY. I figure a difference of 1500ft. The horizon from the mountain top is 1.23 x sqrt (1500) = ~50 miles away Since Rochester is 60 miles away I need to add a tower at that end to guarantee line of sight. How high should it be? I want 10 miles so solving for 10 miles I get 10/1.23 => 66 ft tall tower. I’ll use Rohn 25G with 6 – 10 ft sections and add a 6 ft aluminum mast. . Perfect communication 24/7 direct line of sight. Or I can mount the small receiving dish on the 7th floor of a building with an unobstructed view of the antenna. Stealthy and easy.
And it was from 7th grade science class!

LuckyGuy's avatar

That rule works for SatCom too. Example: Ignoring refraction, what is the farthest distance the satellite can cover using only ground based communication.
Example: Assume a LEO communication satellite at 400 miles up. 400×5280 = 2112000 ft up. Horizon distance is 1787 mile radius.
Corollary. What is the minimum number of LEO satellites needed to cover the area that includes 1000 miles above and below the Equator 24/7. 23000 miles / (2×1787) = ~ 7 satellites. Better put up several trailing spares just in case. Call it 10 or 11. Of course this is an approximation but the 1, 2, 3 answer is very close.

I’ve used it many times in my life.

janbb's avatar

Ok – now you lost me but I’ll take your word for it that is is useful.

LuckyGuy's avatar

Come on. Don’t give up on me. You can do this. Next time you are at the shore watch the ships. You’ll see. Here is a Wiki article on the subject: Distance to horizon .

longgone's avatar

@elbanditoroso How does it work, can you teach us?

@LuckyGuy Love that!

To all: Thanks for replying! I had forgotten all about this thread, had a nice surprise just now.

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