c.f question

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johannesbender
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c.f question

Post by johannesbender »

hi . why dont object on earth show c.f
but as soon as you attach it to something else c.f is shown
is this because everything moves with the atmosphere and earth and gravity also has a hand in it ?

imagine a long rod with a sliding weight , placed into the
ground so it stands upright ...why doesnt earth rotation
impart c.f on this weight ?

im not good at physics , is this some kind of reference frame situation ?
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re: c.f question

Post by oldNick »

Hi J.

The answer is : It does not spin fast enough.

C.F effect depends on R.P.M revs per min.

The second hand on a watch does 1 rpm slow no c.f effect

The earth spins at 1 rev per 24 hours. very slow no c.f.

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Post by jim_mich »

oldNick wrote:The second hand on a watch does 1 rpm slow no c.f effect
Whenever rotation exists, there exists CF. Even the second hand on a watch has CF, though it is very small.

Any object co-rotating with the Earth at the equator has its measured weight reduced by 0.34 percent, because of CF caused by the Earth's rotation.

All objects moving East (same direction of Earth rotation) weigh less because of increased CF. If they move fast enough, then they get launched into outer space. Conversely, objects moving West (opposite direction of Earth rotation) usually weigh more because of decreased CF, but it depends upon latitude and speed, since you can only counter-act existing CF.

This difference of weight when moving East versus West is known as the Eötvös effect.

The Eötvös effect is the change in perceived gravitational force caused by the change in centrifugal acceleration resulting from eastbound or westbound velocity. When moving eastbound, the object's angular velocity is increased (in addition to the earth's rotation), and thus the centrifugal force also increases, causing a perceived reduction in gravitational force.

The Earth's speed at the equator is about 1040.42 MPH. An object at the equator would need to move West at 1040.42 MPH to become stationary and not cause any CF. An object at the equator moving East at 1040 MPH (twice as fast as the Earth) would weigh 1.36% less than an equal object moving West at 1040 MPH.

CF increases exponentially relative to rotational speed. Riding on the Earth's equator as it spins at 1040.42 MPH produces CF equal to 0.34% of weight. Doubling the speed produces 1.36% of weight. Doubling again produces 5.44% of weight. Doubling again produces 21.76% of weight. Doubling again produces 87.04% of weight. Just a little more speed, at slightly over 17 times normal earth rotation (about 16,700 MPH East or 18,780 MPH West) and an object at the equator flies off into space as its CF equals its weight.

Now you know why rockets are launched East over the Atlantic Ocean from Florida and not West over the Pacific Ocean from California.


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re: c.f question

Post by Oystein »

Actually it is not the RPM alone that dictate the force, but just as much the radius.

Centrifugal force increases in direct proportion to the radius. (if RPM stays the same)

At equator you will weigh less because of CF..
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Post by johannesbender »

a new question , so if i dont misunderstand (with my extremely limited knowledge) if you rotate something slow enough cf wont be that big ?
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Post by jim_mich »

I've posted these formulas any number of times before...

CF = k × w × r × rpm × rpm

k = constant, which depends upon the type of measurement units of the radius.
w = weight, any units
r = radius
rpm = revolution per minute

The constant value of 'k' must be according to the radius units.
For inches and rpm, the value of k is 0.000028403397
For meters and rpm, the value of k is 0.001118243976

Weight units used does not effect the formula. If you measure the weight in pounds then the CF is in pounds. If you measure the weight in Kilograms then the CF is in Kilograms.

If you double the radius then the CF is doubled.
If you double the weight then the CF is doubled.
If you double the rpm then the CF increases by a factor of four.

Pick some realistic values and plug them into the formula. Suppose you have a 6 inch kitchen wall clock. The radius is 3 inches. The second hand rotates at 1 RPM. Assume a weight of 1 fly on the tip of the second hand.
The CF = 0.000028403397 × 1_wt_of_fly × 3_inch_radius × 1_rpm × 1_rpm.
CF = 0.000085210191_wt_of_fly.
This is probably about the weight of one hair on the little fly's leg. Not very much. But also not zero either. My point is that CF is never zero when an object is being rotated.

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Post by oldNick »

johannesbender wrote:a new question , so if i dont misunderstand (with my extremely limited knowledge) if you rotate something slow enough cf wont be that big ?
That is correct.
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re: c.f question

Post by johannesbender »

thank you jim , now i understand much better .

any weight shifted inside a wheel will affect cf too.
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Re: re: c.f question

Post by oldNick »

Oystein wrote:Actually it is not the RPM alone that dictate the force, but just as much the radius.

Centrifugal force increases in direct proportion to the radius. (if RPM stays the same)

At equator you will weigh less because of CF..
only if the RPM of the smaller radius is enough to cause the CF effect to start with.

Jim_mich.

Shrink the earth to the size of your clock. The hour hand on the clock does two revolutions per 24 hours. The Earth only does one revolution per 24 hours. Please do your cf calculations all you want but what is the point?
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Post by jim_mich »

oldNick wrote:Please do your cf calculations all you want but what is the point?
oldNick wrote:The second hand on a watch does 1 rpm slow no c.f effect

The earth spins at 1 rev per 24 hours. very slow no c.f.
Jim_Mich wrote:My point is that CF is never zero when an object is being rotated.
My point is that you were wrong when you said there is "no c.f. effect" at slow rotation.

My point is that there is ALWAYS a CF effect whenever there is rotation. The CF might be very small, but it still exists whenever an object follows any curved path.

The original question by johannesbender was, "why dont object on earth show c.f" and "imagine a long rod with a sliding weight , placed into the ground so it stands upright ...why doesnt earth rotation impart c.f on this weight ?

OldNicks wrote: "The answer is : It does not spin fast enough."

This answer is not correct, because the Earth's spin does indeed impart CF to the weight, as demonstrated by the Eötvös effect.
oldNick wrote:Shrink the earth to the size of your clock.
I'm not a wizard. I cannot shrink the Earth. :)}

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re: c.f question

Post by oldNick »

Hi Jim_mich

So your point was to make sure everyone new I was a fool, I don't need you to do that I can do it myself thank you.

Believe it or not other people know of all the formulae you keep spouting as if you thought them up.

I think your green dots have gone to your head, and you have to teach us nobodies a lesson.

I use to believe you had a motion wheel! but not now because you have no imagination!.

This is supposed to be a discussion board!.
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re: c.f question

Post by justsomeone »

Jim, are you also Josh- Mich?
. I can assure the reader that there is something special behind the stork's bills.
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Post by jim_mich »

oldNick wrote:This is supposed to be a discussion board!.
And the discussion was about CF caused by Earth rotation. And in the discussion you stated that at slow RPM there would be no CF. And that little tidbit of information was obviously false.

I've seen it said many times here on the forum that slow rotation produces no CF. And every time I read such statements, I think, "What foolishness! Rotation ALWAYS causes CF."

Why get upset with me for pointing out a simple truthful fact? There is no need to make a mountain out of a mole hill.

If, as you suggested, you figure the CF at 3 inch radius rotating 2 rotation per day (1/720 RPM), the CF is 0.0000000001637151 times the weight of the object. Granted, that is an exceedingly small amount of CF, but it is still not zero CF.

If there was on the end of that clock hand a black hole that weighed as much as the QE2 ship (about 92,000 tonnes) then its CF while rotating at 2 rotations per day would be about 1/2 ounce. Not zero. My point is that whenever there is rotation, then there is CF. You never have rotation with no CF. Whether the CF is strong enough to overcome friction and cause outward motion, is a totally other subject.

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re: c.f question

Post by rlortie »

"Whether the CF is strong enough to overcome friction and cause outward motion, is a totally other subject."

Overcome friction? how about giving Centripetal a little credit, if it was not for centripetal you would not have rotation or Cf, seems to me it wins hands down! What is the primary property or force that creates rotation making Cf? I believe its called "centripetal"...

Hey Jim! how are the vacuum formed plastic models doing? I have received three shipments of plastic, yet no working model has shown up yet.

Ralph
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Post by johannesbender »

okay , now i need to know something .

does the formulae take into consideration the amount of cf
that comes from the earth also ? weight i think would include that to a degree ?
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