Author Topic: Math Question  (Read 7397 times)

Offline Tree

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Re: Math Question
« Reply #60 on: July 20, 2017, 12:22:59 pm »
Ok, you said coconuts...

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« Last Edit: July 20, 2017, 01:47:44 pm by Tree »
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Offline VirginiaJim

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Re: Math Question
« Reply #61 on: July 20, 2017, 12:49:20 pm »
Yes, that one came to mind as well...excellent choice.  By the way, when you post in a YouTube vid, before you save the post, remove the s on https.   That way the vid will show up instead of just the link.
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Offline B.D.F.

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Re: Math Question
« Reply #62 on: July 20, 2017, 01:43:41 pm »
Post altered 21 July by BDF to remove my initial reply to Rhino, which I now think was wrong. See next post for why I now think Rhino was right.

Brian

Interesting. I will have to study the Eötvös Effect more. This is why NASA launches to the east the vast majority of the time. Extra 1000 mph is a lot of rocket fuel.

But I still maintain that the Coriolis Effect is near zero at the equator and more pronounced at the poles due the the difference in the rate of angular movement. I did find a website that agrees with me:

https://www.quora.com/Why-is-the-Coriolis-force-zero-at-the-equator-and-maximum-at-the-poles

The Coriolis force acts in a direction perpendicular to the rotation axis and to the velocity of the body in the rotating frame and is proportional to the object's speed in the rotating frame.
The Coriolis effect is caused by the rotation of the Earth and the inertia of the mass experiencing the effect. Because the Earth completes only one rotation per day, the Coriolis force is quite small, and its effects generally become noticeable only for motions occurring over large distances and long periods of time, such as large-scale movement of air in the atmosphere or water in the ocean. Such motions are constrained by the surface of the earth, so only the horizontal component of the Coriolis force is generally important.
 The horizontal deflection effect is greater near the poles and smallest at the equator, since the rate of change in the diameter of the circles of latitude when travelling north or south, increases the closer the object is to the poles.

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« Last Edit: July 21, 2017, 04:36:19 am by B.D.F. »
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Offline B.D.F.

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Re: Math Question
« Reply #63 on: July 20, 2017, 05:28:04 pm »
So this stuck with me and I kept coming back to it..... and I think you are right and I was wrong (but may be getting 'righter' after shaking off my  earlier, wrong, thinking).

First: launching into orbit, yep, the velocity is a 'freebie' and absolute regarding orbital velocity. I missed it because I was thinking about the entire flight (including landing) but that is irrelevant. All that is required is a fixed amount of acceleration to an altitude and velocity, and starting off going with the Earth's rotation would just add or subtract.

The second part is harder for me to see because I am stuck on the fact that 'force' would be lost at launch. And that is fine and well but does not mean anything if there is nothing trying to alter the projectile's initial velocity in the first place. This is what I think I missed or rather kept injecting but does not belong: there is a velocity added or subtracted, so what? Nothing external will exert any force on the projectile to reduce or increase that velocity! The entire path becomes  So again, I think you are and were right and I was wrong- the Coriolis effect does reduce to zero as the angle approaches and reaches a direction parallel with the Earth's rotation, any line of latitude. This only applies at the equator but that is what we have been talking about.

Moving away from the equator makes it more complex because I do not think there are any 'straight' lines of latitude anymore that will not cross a curve. Still thinking about that one though.

So thanks for the input and correction- it is always best to be correct as much as can be. Your insomnia turned into my.... afternoonstuckinthethought onmia.  ;) ;D

I did go looking for calculating 'accessories', formulas, a spreadsheet or similar but everything I found is commercial and has to be purchased. So much for the easy way out of 'plug 'n chug' calculating. Going to have to do this the hard way and work with the rules and physics.

Thanks again!

Brian

Interesting. I will have to study the Eötvös Effect more. This is why NASA launches to the east the vast majority of the time. Extra 1000 mph is a lot of rocket fuel.

But I still maintain that the Coriolis Effect is near zero at the equator and more pronounced at the poles due the the difference in the rate of angular movement. I did find a website that agrees with me:

https://www.quora.com/Why-is-the-Coriolis-force-zero-at-the-equator-and-maximum-at-the-poles

The Coriolis force acts in a direction perpendicular to the rotation axis and to the velocity of the body in the rotating frame and is proportional to the object's speed in the rotating frame.
The Coriolis effect is caused by the rotation of the Earth and the inertia of the mass experiencing the effect. Because the Earth completes only one rotation per day, the Coriolis force is quite small, and its effects generally become noticeable only for motions occurring over large distances and long periods of time, such as large-scale movement of air in the atmosphere or water in the ocean. Such motions are constrained by the surface of the earth, so only the horizontal component of the Coriolis force is generally important.
 The horizontal deflection effect is greater near the poles and smallest at the equator, since the rate of change in the diameter of the circles of latitude when travelling north or south, increases the closer the object is to the poles.

Hey! My insomnia fueled brain farts can't be wrong!
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Offline Rhino

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Re: Math Question
« Reply #64 on: July 21, 2017, 08:01:57 am »
Hey Brian, more insomnia last night (I don't think I'm getting enough alcohol). Thought some more about the Eötvös effect last night and how much it would actually effect ballistics of a gun like a 1 mile sniper shot. Due to this effect you would weigh more standing at one of the poles then you do at the equator. I googled it this morning and found this NASA response. 0.35%! You weigh 0.35% less at the equator than you do at the north pole. Way more then I would have thought. I don't know the equation to calculate centrifugal force but if I use a linear extrapolation (I realize this may not be a linear function), a 338 Lapua muzzle velocity of about 3000 ft/sec is about 2000 mph. Therefor at the equator a sniper shooting east the bullet will be traveling 3000 mph relative to the center of the earth, but shooting west only 1000 mph. The felt gravitational pull on the bullet will be about 1% different. I don't know how to calculate the ballistics based on this but if we are talking 1 shot 1 kill, you would definitely need to calculate that into the shot. And 1% on a 100 mile shot from the Paris gun could be a mile? If this is the case, then it has as much or more of an effect as the Coriolis effect. I think you were correct in that you need to calculate both effects weather you are at the equator or a pole and everywhere in between. Now I wonder if the Shooter app I have on my iPhone calculates the Eötvös effect as well as the Coriolis effect (not that it really matters when I'm at the trigger, my skills do not come close to having to worry about this).

https://image.gsfc.nasa.gov/poetry/ask/a11511.htm