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# Why Humes Thought the Back Missile Hit at a Sharp Angle -- a Hypothesis

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Robert,

I've been paying quite a bit of attention to that for some time now. In fact, on your thread I just commented on how such a low sternum (the top of which the clavicle bones are attached) could result in many tracheal rings being located above the sternal notch.

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It is rather odd, isn't it. As I pointed out on your thread a few minutes ago, I thought at first it was an alteration of the chest x-ray, or the x-ray was taken from a perspective way above JFK's head. However, after looking at photos of a shirtless JFK, it appears his clavicles really did descend rather steeply, and that he did have a very low suprasternal notch. He must have been the very opposite of you!

I wonder if this anatomical oddity contributed to JFK's perpetual slouch; the one grossly exaggerated by Dale Myers in his Dealey Plaza cartoon.

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• 6 months later...
On 8/25/2016 at 7:18 PM, Michael Walton said:

According to Sandy you have to aim the ball "150 feet" in order to make it go where you want. Think about that for a minute. That's equal to a 15-story building. Imagine a baseball field with a 15-story building sitting in the middle of it, a guy goes way out in the outfield and aims for home but projects the ball the same trajectory as the 15-story building. Somewhere along the line, his math analysis is way, way off.

There is no freaking way that you have to aim the ball the equivalent of a 15-story building to throw the ball 110 to 120 yards. It seems like Sandy doesn't care that I've actually done this numerous times before my arm fell apart.

You need to throw the ball high, aiming it way over the catcher's head. Otherwise the ball will fall far short of the catcher... because of gravity. This diagram illustrates this:

I used physics to determine how far over the catcher's head the thrower needs to aim the baseball in order for it to come down in the catcher's vicinity. I first calculated 150 feet. A few days later I realized I had used a formula suitable for bullets (suitable because a bullet's velocity doesn't change much from beginning to end). But for a ball, that formula would give a poor approximation. So I re-did the calculation using the correct formula. This gave me the correct answer, 197 feet.

Mike refuses to believe me to this day, even though I got the same answer using an online ballistics calculator. So I decided to get some corroboration. I asked members of Quora to formulate and post their own answers.

In response, I got one guy answering who has both a baseball background (he even makes wood bats) and some knowledge of physics. He made the same mistake I originally did, and got the same poor approximation, 150 feet.

Quote
James Canale, Mechanical Engineer and wood bat turner

Nobody “aims” above the target. They just aim at the target, and experience has them put the right arc on it.

But, 80mph is something like 120 ft/s, 40 yards/s, so the ball is in the air for 3 seconds.

Distance dropped is s=1/2 at^2. So 1/2*32 3^2 = 16*9 = 144ft above the catcher. That seems high to me. But it's probably right. An outfielder might bounce it well in front of the catcher too - which might shorten the overall arc length.

The reason he got 144 ft instead of my 150 ft is because he estimated 80 mph to be 120 ft/s. Had he used the correct number (117 ft/s) he would have gotten 150 feet.

Another Quora member got the correct answer. I can't show his calculations in this editor, so I have to show screen shots:

This guy got an answer of 60 meters, which is equal to 197 feet. The very same answer I got.

(Here is the forum post where I calculated 150 ft. Here is the forum post where I calculated 197 ft.)

Will Mike believe me now??

EDIT: Corrected the phrase "needs to throw" to "needs to aim.)

Edited by Sandy Larsen
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Sandy,

Your quote "I used physics to determine how far over the catcher's head the thrower needs to throw the baseball in order for it to come down in the catcher's vicinity."

The distance was approx 320ft in 3.2 sec.

The top blue line is the top of the baseball arc. (Red arrow points to baseball).

If you replay the video and time the travel of the ball, in relationship to the runner getting ready to slide, (straight line distance between 3rd base and home plate = 90ft), the runner has approx 7 strides, is in an arc outside the straight base line distance and slides, all from when the ball in flight reaches it's top height which is 1/3 of the remaining time of 3.2 sec = 1.07sec.

7 strides @ 3.5ft per stride = 24.5ft

extra baseline running arc = 5ft approx

slide before ball hits glove = 3.5ft

Total approx 33ft

Fastest runner in the world Usain Bolt at 27MPH Olympic Record 100meters

Base runner 33ft in 1.07 sec = 21mph

Ball in flight, caught by catcher approx 5ft before runner slides.

At 1.07 seconds (travel time from top height of ball arc to catcher's mit), how fast/far does the ball have to travel to beat the runner by approx 5ft?

chris

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8 hours ago, Chris Davidson said:

Sandy,

Your quote "I used physics to determine how far over the catcher's head the thrower needs to throw the baseball in order for it to come down in the catcher's vicinity."

The distance was approx 320ft in 3.2 sec.

The top blue line is the top of the baseball arc. (Red arrow points to baseball).

If you replay the video and time the travel of the ball, in relationship to the runner getting ready to slide, (straight line distance between 3rd base and home plate = 90ft), the runner has approx 7 strides, is in an arc outside the straight base line distance and slides, all from when the ball in flight reaches it's top height which is 1/3 of the remaining time of 3.2 sec = 1.07sec.

7 strides @ 3.5ft per stride = 24.5ft

extra baseline running arc = 5ft approx

slide before ball hits glove = 3.5ft

Total approx 33ft

Fastest runner in the world Usain Bolt at 27MPH Olympic Record 100meters

Base runner 33ft in 1.07 sec = 21mph

Ball in flight, caught by catcher approx 5ft before runner slides.

At 1.07 seconds (travel time from top height of ball arc to catcher's mit), how fast/far does the ball have to travel to beat the runner by approx 5ft?

chris

Chris,

Is the story problem you present here somehow related to the simple ballistics principle I and Robert Prudhomme have been trying to impart on Mike Walton?

BTW, you quoted me as saying, "I used physics to determine how far over the catcher's head the thrower needs to throw the baseball." I misspoke when I wrote that. The word "throw" should be "aim."

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5 hours ago, Sandy Larsen said:

Chris,

" I misspoke when I wrote that. The word "throw" should be "aim."

Thanks for correcting that.

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