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Physics, Astronomy & Cosmology / What are the forces on a car travelling in a circle?

« on: 15/07/2010 18:35:50 »

A vehicle is traveling in a straight line, say a space vehicle, or an efficient rolling one on a huge flat surface, like Libya only smoothly paved.

The one in space requires no power to keep going.  The one on the desert flat needs some power to overcome the air and mechanical friction, however little or much that might be.

I have mounted on the right side either vehicle a jet thruster, pointed at a 90 degree angle away from the body.  When I fire it gently off, each of the vehicles begins a turn to the left.  By definition each is now "registering" lateral g force since the formula for that only includes time and radius.  I am inputting energy, power, or whatever we call it, the vehicle is turning.  When I stop, the vehicles return to traveling in a straight line.   If the jet left in a continuous burn, both keep circling.  The more I ramp up the power the shorter the radius becomes and the more lateral g involved.  Which leads me to the (premature?) conclusion that it takes energy to have lateral g.

I am not far from imagining that work is being done by a circling vehicle, in that it is always accelerating away from the straight and natural course.

What am I overlooking?

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Physics, Astronomy & Cosmology / What are the forces on a car travelling in a circle?

« on: 11/07/2010 15:27:51 »

As Geezer points out the question is:

"Would there be even more energy needed just to offset the lateral g force generated by the circling car?"

All the talk of tires and friction, and motorcycles being able (or not) to negotiate a turn of given radius is interesting.  The key word in the question is "just" or maybe better put should be "only".  And "even more" gives credit to all who imply that the screeching tires are somehow involved in the need for more energy.   But it all may figure in to an answer, somehow.

There is a formula for lateral g, and I think it goes:   1.22 times radius in feet divided by the square of the lap time (one revolution) in seconds.

We know g force would "cost" if in a straight line, as in an accelerating car needing more gas (even without or beyond any friction of air or mechanical/tires).,,

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Technology / Is it really efficient to generate electricity from speed bumps?

« on: 11/01/2010 16:43:09 »

Here is my bottom line.  The energy recovery, inefficient at best, is not the answer.  Don't spend it in the first place.  A modern car could have its computer receive a signal in certain neighborhoods limiting the top speed in that area.  Wait a minute, we already have that, a computer (human brain), a speed limit sign (visual signal), and ability of the computer (human brain) to do the correct thing.  The energy used to create the speed bumps and recovery systems could be put to direct use elsewhere.

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General Science / The molecules in a single drop of water diluted evenly throughout the Earth's oceans would result in

« on: 16/09/2008 12:05:37 »

I have seen the number 20.000 which would result in even more molecules per drop.  However I have also seen the 24,000 number.  I think drops come in different sizes and therefore not a standard precise volume for one.

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General Science / The molecules in a single drop of water diluted evenly throughout the Earth's oceans would result in a density of one molecule per litre of sea water!

« on: 12/09/2008 16:18:56 »

Erik  asked the Naked Scientists:

Dear Dr. Chris:   My brother and I have come to a "conclusion", and
have a high degree of  confidence in the following statement:

"The molecules in a single drop of water, if diluted evenly throughout the "seven seas" (the oceans of the world), would result in a density of one molecule (from the drop) per litre of sea water."

This is astounding!  Please find below our assumptions and maths.  I
hope you can confirm the correctness of the assumptions and the accuracy of the maths!

This is useful in illustrating the mind-boggling tiny size of molecules.

Erik Moeser
Menomonee Falls, WI

Robert Moeser
Boston, MA

One litre = 24,000 drops

The volume of the sea is 1.4 billion cubic kilometres

18g (or cc) of water = one mole

Number of molecules per drop:  1000/18 (6.022X10 to the 23rd)/24000 = 1.39 X 10 to the 21st

Number of litres in the oceans:  1.4  X  10 to the 9th X  10 to the
ninth  =  cubic metres of sea water, X 1000 = litres of sea water = 1.4 X 10 to the 21st

What do you think?