Janus

Contraction is a squeeze to both observers.

Show me' date=' how the outside observer is going to measure contraction, by looking to reference books? If contraction occurs it should be real, not surreal. How can a squeeze be not a squeeze? Actual experiment is when we have actual squeeze, real to all observers inside and outside the spaceship.[/quote']

 

Time and space are relative and not absolute and thus depend on who is making the measurement.

 

Here we have an example of what I was talking about. You have made the assumption that length is absolute (that observers in and out of the space ship will measure the same length.) Relativity says this is not the case. Thus your assumption and Relativity contradict each other. But you have not provided any actual evidence that your assumption is correct (no matter how much you believe it to be so) and Relativity wrong.

 

On the other hand, Relativity has made numerous predictions that have been confirmed by actual physical experiments.

In your image you did not mention the reflector and misapplied direction of vector in vt2: the arrow should point to the left' date=' not to the right. Yellow distance is misleading distance, nobody sees along this path anything, and this distance shown at angle BETA, not ALPHA: or else in your calculations BETA angle should be mentioned. Anyway when the light pulse is at the end of the yellow distance, reflector moves together with the sphere S to a position of the sphere S'. So spheres S and S' is the same spaceship!

 

Distances Ct1 and Ct2 and Y are seen together at one common angle ALPHA. To say, that they are different, means fun or mischief. So, if gamma-contraction is assigned to distances ct1 and ct2, then assign it to the distance Y: contraction is a squeeze, and a squeeze is a squeeze from the point of view of anybody... or God.

 

Ha-ha: the same spaceship! Einstein was not good at school, he could allow himself any misapplication of vectors and mundane|vain calculations. [Your picture showed "a spider got entangled in his own web". It is not a shock, Swansont, it's this spider dizziness of exhaustion'].

 

Of course it is the same space ship! The oval to the left is the spaceship at the instant when the light hits the mirror. The oval to the right is the Same spaceship at the instant when the light returns to the center. This is as seen by an observer outside of the ship, watching it go by at v. (to keep the image uncluttered, I omitted the spaceship at the instant the light is emitted from the center from the image)

To someone in the ship, the light simply travels out at angle alpha (wrt the relative motion) for a distance Y, hits the mirror and returns along the same path to the center. Since the light returns to the center of the ship in S' (the frame of the ship), it must also return to the center in frame S (the frame in which the ship is moving at v) Thus in S' the light follows the path shown in my images.

Alpha is the angle the light is launched as measured by someone traveing with the ship. Y is the radius as measured by someone traveling with the ship The length of the ship is only contracted along the direction of motion when measured by someone to which the ship has a relative motion. The ship has zero relative motion wrt to someone traveling with the ship, thus it has no length contraction for that person.

To say otherwise shows a complete and utter lack of understanding of what Special Relativity is about. To refute Relativity you must first understand it.

The simple fact is that Relativity is a completely self-consistant theory. It contains no mathematical or logical errors. It makes numerous predictions which have held up under every real experimental test put to it.

No thought experiment you can ever dream up will ever refute it. If you think you've come up with one, all it means is that you have made an assumption that is contradictory to one of the postulates of Relativity. All you've done in this case is show that Relativity is not compatible with your assumption. This in of itself provides nothing in the way of proving which is correct, your assumption or Relatiivity.

The only way to refute Relativity is to produce results from a Actual Physical Experiment that gives results that differ from that which Relativity predicts.

In order to examine this claim that light emmited at different angles somehow changes the results let's try the the folowing exercise.

Let's call the frame of the sphere frame S' and the frame in which the sphere moves at v, S

In S the light travels a distance of ct1 while our sphere travels a distance of vt1.

Now considering image 1:

Assuming the light is emitted at at an angle of alpha from the center of the sphere as measured in S' , then the point at which the light hits the sphere is as measured in S is:

[math] ( Y \sin \alpha , vt_1+ \gamma Y \cos \alpha ) [/math]

where Y is the radius of the sphere as measured in S'.

and that the center of the sphere at the instant of impact is at (0,0)

[math]\gamma = \sqrt{1- \frac{v^2}{c^2}}[/math]

The gamma factor is needed because the sphere is length contracted in S

Thus:

[math]c^2t_1^2 = Y^2 \sin^2 \alpha + ( vt_1 + \gamma Y \cos \alpha)^2 [/math]

[math]c^2t_1^2 = Y^2 \sin^2 \alpha +v^2t_1^2 + 2 \gamma Y v t_1 \cos \alpha + \gamma^2 Y^2 \cos^2 \alpha

[/math]

[math]c^2t_1^2 = Y^2 \sin^2 \alpha +v^2t_1^2 + 2 \gamma Y v t_1 \cos \alpha + \left( 1- \frac{v^2}{c^2} \right)Y^2 \cos^2 \alpha

[/math]

[math]c^2t_1^2 = Y^2 \sin^2 \alpha +v^2t_1^2 + 2 \gamma Y v t_1 \cos \alpha + \left( \cos^2 - \frac{ \cos^2 v^2}{c^2} \right)Y^2\alpha

[/math]

re-arranging:

[math]c^2t_1^2 = v^2t_1^2 + 2 \gamma Y v t_1 \cos \alpha +Y^2 \left( \sin^2 \alpha + \cos^2 - \frac{ \cos^2\alpha v^2}{c^2} \right)

[/math]

[math]c^2t_1^2 = v^2t_1^2 + 2 \gamma Y v t_1 \cos \alpha +Y^2 \left( 1- \frac{ \cos^2\alpha v^2}{c^2} \right)

[/math]

Gathering the t1^2 terms together:

[math]c^2t_1^2-v^2t_1^2 = 2 \gamma Y v t_1 \cos \alpha +Y^2 \left( 1- \frac{ \cos^2\alpha v^2}{c^2} \right)

[/math]

[math]t_1^2 (c^2-v^2) = 2 \gamma Y v t_1 \cos \alpha +Y^2 \left( 1- \frac{ \cos^2\alpha v^2}{c^2} \right)

[/math]

Moving all the terms to the left side:

[math]t_1^2 (c^2-v^2) - 2 \gamma Y v t_1 \cos \alpha -Y^2 \left( 1- \frac{ \cos^2\alpha v^2}{c^2} \right) =0

[/math]

We get an equation that can be solved for t1 using the quadriatic equation thusly:

[math]t_1 = \frac{ 2 \gamma Y v \cos \alpha \pm \sqrt{ 4 \gamma^2 Y^2 v^2 \cos^2 \alpha + 4 (c^2-v^2) Y^2 \left( 1- \frac{ \cos^2\alpha v^2}{c^2} \right) } }{2(c^2-v^2) }[/math]

[math]t_1 = \frac{ 2 \gamma Y v \cos \alpha \pm \sqrt{ 4\left( 1- \frac{ v^2}{c^2} \right)Y^2 v^2 \cos^2 \alpha -+4 (c^2-v^2) Y^2 \left( 1- \frac{ \cos^2\alpha v^2}{c^2} \right) } }{2(c^2-v^2) }[/math]

[math]t_1 = \frac{ 2 \gamma Y v \cos \alpha \pm 2 Y \sqrt{ \left( 1- \frac{ v^2}{c^2} \right) v^2 \cos^2 \alpha + (c^2-v^2) \left( 1- \frac{ \cos^2\alpha v^2}{c^2} \right) } }{2(c^2-v^2) }[/math]

[math]t_1 = \frac{ 2 \gamma Y v \cos \alpha \pm 2 Y \sqrt{ \left( 1- \frac{ v^2}{c^2} \right) v^2 \cos^2 \alpha + \frac{(c^2-v^2)}{c^2} c^2 \left( 1- \frac{ \cos^2\alpha v^2}{c^2} \right) } }{2(c^2-v^2) }[/math]

[math]t_1 = \frac{ 2 \gamma Y v \cos \alpha \pm 2 Y \sqrt{ \left( 1- \frac{ v^2}{c^2} \right) v^2 \cos^2 \alpha + \left( 1- \frac{ v^2}{c^2} \right) \left( c^2- \cos^2\alpha v^2 \right) } }{2(c^2-v^2) }[/math]

[math]t_1 = \frac{ \gamma v Y \cos \alpha \pm Y \sqrt {1- \frac{ v^2}{c^2}} \sqrt{ v^2 \cos^2 \alpha + \left( c^2- v^2 \cos^2\alpha \right) } }{(c^2-v^2) }[/math]

[math]t_1 = \frac{ \gamma v Y \cos \alpha \pm Y \sqrt {1- \frac{ v^2}{c^2}} \sqrt{ v^2 \cos^2 \alpha + c^2- v^2 \cos^2\alpha } }{(c^2-v^2) }[/math]

[math]t_1 = \frac{ \gamma v Y \cos \alpha \pm Y \sqrt {1- \frac{ v^2}{c^2}} \sqrt{ c^2 }} {(c^2-v^2) }[/math]

[math]t_1 = \frac{ \gamma v Y \cos \alpha \pm cY \sqrt {1- \frac{ v^2}{c^2}}} {(c^2-v^2) } [/math]

Now consider image 2 which shows the path the light takes on its return to the center of the sphere:

Here we can see the light travels a distance of ct2 in frame S, while the sphere moves a distance of vt2.

As a result we can see that:

[math]c^2 t_2^2 = Y^2 \sin^2 \alpha + ( \gamma Y \cos \alpha - vt_2 )^2 [/math]

and following the same steps as above we find that solving for t2 we get:

[math]t_2 = \frac{- \gamma v Y \cos \alpha \pm cY \sqrt {1- \frac{ v^2}{c^2}}} {c^2-v^2 }[/math]

The total time T it takes for the light to leave the emitter and return is

[math]T = t_1 + t_2[/math]

Thus

[math]T=\frac{ \gamma v Y \cos \alpha \pm cY \sqrt {1- \frac{ v^2}{c^2}}} {(c^2-v^2) } + \frac{- \gamma v Y \cos \alpha \pm cY \sqrt {1- \frac{ v^2}{c^2}}} {c^2-v^2 } [/math]

[math]T=\frac{ \gamma v Y \cos \alpha \pm cY \sqrt {1- \frac{ v^2}{c^2}} - \gamma v Y \cos \alpha \pm cY \sqrt {1- \frac{ v^2}{c^2}}} {c^2-v^2 } [/math]

[math]T=\frac{ \pm cY \sqrt {1- \frac{ v^2}{c^2}} \pm cY \sqrt {1- \frac{ v^2}{c^2}}} {c^2-v^2 } [/math]

atr this point we have a choice to make. There is only one possible answer for T, but there are four ways to solve this equation due to the "plus or minus" signs. Luckily we can automatically elliminate 3.

If both are minus we end up with a negative time for an answer, so that doesn't work. And if one is minus and the other plus we end up with a answer of 0 for T and that makes no sense. this leaves us with both being plus giving us:

[math]T=\frac{ cY \sqrt {1- \frac{ v^2}{c^2}} + cY \sqrt {1- \frac{ v^2}{c^2}}} {c^2-v^2 } [/math]

[math]T=\frac{ 2cY \sqrt {1- \frac{ v^2}{c^2}} } {c^2-v^2 } [/math]

Now let's relate T to T' the total round trip time in S'

In S' the light travels a distance of Y to the sphere and then travels a distance of Y back again.

this gives us:

[math]T' =\frac{ 2Y}{c}[/math]

and

[math]\frac{T'}{T}= \frac{\frac{ 2Y}{c}}{\frac{ 2cY \sqrt {1- \frac{ v^2}{c^2}} } {c^2-v^2 } }[/math]

[math]\frac{T'}{T}= \frac{ 2Y}{c}} \frac {c^2-v^2 }{ 2cY \sqrt {1- \frac{ v^2}{c^2}} } [/math]

[math]\frac{T'}{T}= \frac {c^2-v^2 }{ c^2 \sqrt {1- \frac{ v^2}{c^2}} } [/math]

divide top and bottom by c²:

[math]\frac{T'}{T}= \frac {1- \frac{v^2}{c^2} }{ \sqrt {1- \frac{ v^2}{c^2}} } [/math]

simplify and re-arrange:

[math]T = \frac{T'}{\sqrt{1- \frac{v^2}{c^2}}}[/math]

And we see that no matter what angle alpha the light is aimed, the stardard time dilation holds, thus falsifying the claim.

[math]c^2 t_2^2 = Y^2 \sin^2 \alpha + ( \gamma Y \cos \alpha - vt_2 )^2 [/math]

[math]t_1 = \frac{ \gamma v Y \cos \alpha \pm Y \sqrt {1- \frac{ v^2}{c^2}} \sqrt{ c^2 }}}{(c^2-v^2)}[/math]

[math]t_1 = \frac{\gamma v Y \cos \alpha \pm cY\sqrt{1-\frac{v^2}{c^2}}}{(c^2-v^2)}[/math]

Is the book pretty good? Is it worth reading after I have already seen both movies?

Definitely read the book, if for no other reason than a sense of history. Written in 1898 by H.G. Wells, a man ahead of his time who more or less invented many of the "themes" of modern SF. (Alien Invasion, time travel, antigravity, invisibility)

ANy way, Do you people know why all the tripod machines are defeated? I mean, Tom Cruise only blew up only one tripod machine, and simultaneously, all the others were either weakened or froze, vulnerable by human attacks. For example, there was a scene that the US air force bombers and helicopters were inefective to the shell of the machine. But once one was blown up by Tom, even the infantry could blow up the machine with a bezooka. Btw, what were those birds doing on the machine??

As explained the ivaders were destroyed by microbes. This is explained in the voise over at the end of the movie, which, with your weak French, probably why you missed that part.

The destruction of one of the machines by Cruise had nothing to do with the rest of the machines malfunctioning. It did not happen simultaneously. (You have no idea how much time passes between the destruction of the first machine and the scene where they show the rest inactive.

I think the machine destroyed by cruise and the one done in later were nods to the novel again. In it, the Machines were not invunerable(they had no shields) and the humans were able to take a few out; it was just that it wasn't enough to stop or slow the advance of the invaders. Also, once we found an effective tactic against them, the Martians would come up with something to circumvent it.

The shields were added in the first movie made in the 50's and retained in this movie to give the invaders the same degree of "edge" over modern weapons as the original novel gave the Martian's over late 1800's military technology.

The birds were merely an indication that the shield was not functioning for that machine(most likely due to the ailing invaders inside), and thus it was vunerable to weapons now. It is also a nod to the original novel, where the protagonist sees birds circling the hood of a machine and picking at the remains of the dead Martians inside.

Secondly, why do they need human blood?

In the original novel, the martians "fed" on the blood of others, so they used humans as a food source. In order to keep the aspect of the story where the invaders captured live humans, the director just changed this a little so that they used the blood as fertilizer instead. The Red weed is taken from the novel also, so he incorporated the blood into its cultivation.

Agreed. The relative velocity on its own[/i'] can't make the age difference.

Yes it can. According to brother B, it is only the relative velocity that causes A to age less.

The value of the relative velocity determines the age difference. There will be an age difference only if there is a difference in the type of motion (acceleration) of the twins.

 

Therefore' date=' essentially, the twin's relative velocity generates the possibility of an age difference and a one-sided acceleration makes this an actuality. Is this basically correct?[/quote']

 

I'd say that what "makes" the time difference depends on which twin you ask.

If that is true how come can succesfuly calculate/measure

the amount of space we have traveled through?

 

We can't. What we can do is choose some convenient reference point and calculate our relative motion with respect to it.

But that reference point is can only be consider "stationary" for convenience's sake we can know nothing about our absolute motion through space

 

A speed through that distance is absolute.

Since there is no absolute reference to judge that speed by, speed can never be absolute.

Sorry, but you are mistaken. This is easily demonstrated every day with particle accelerators. The Earth itself accelerates as it orbits the Sun, thus the Earth's orbital velocity of 30km/sec would be added and substracted to the particle's final velocity as it leaves the accelerator during different points of its orbit. Today's accelerators can reach velocities close enough to c that this 60 km/sec variation would cause a noticeable difference in relativistic effects seen in those particles over the course of a year. No such variations are ever noticed.

If thats true than the moon contributes 2.2 times as much force to the tides as compared to the sun.

 

So with no moon' date=' tides will crest 2.2 times lower, right?[/quote']

 

Well, you have to remember that the tides vary. When the Sun and moon are aligned we get the highest high tides and the lowest low tides (spring tides). When they are at right angles to each other, we get a much smaller variation between low and high tides(neap tides). Neap tides have so little variation because the Sun and Moon actually work against each other at this time. The shifting positions of Sun and Moon in the sky also cause the times of the high and low tides to shift from day to day.

 

So if you remove the Moon, you get tides that are smaller than spring tides, but larger than neap tides. The height of tides will remain fairly constant, and occur at the same time of day every day. (high tides around noon and midnight, low tides near sunrise and sunset)

I worked the whole day' date=' now here's the reference to the article.

Anyway, it looks as if it is published.

http://www.rainbow-calendar.hotmail.ru/Einstein Relativity Refutation (2).htm

shortly:

If you move after light impulse with the speed c/2, you approach it with the speed c/2, and relative speed is c/2 also. In the same way, if someone moves after the light impulse at an angle alpha with the speed [c*cos alpha'], you receive the relative speed [c*sin alpha]. If someone denies relative speed, than v should be zero.

 

By the way I expect a good friendly guy to join soon my thread, I expect to appear between his reviews, not between attacks.

I am not a scientist but have a general interest in relativity.

I am annoyed by the ever growing need to introduce exotic matter and additional dimensions into theories to account for effects in our universe.

 

I have what maybe a completely silly idea (but would really like that to be confirmed! ) that may have a germ of interest ....

 

1. Suppose that the maximum velocity posssible in our universe was 0.0000005% higher than c. Call this c++.

 

2. light still has a measurable velocity c. But c is less than c++. Possibly from this a photon has mass.

 

3. If a photon has mass' date=' why have we not detected it? Maybe because it is so so minimal. So minimal that even when traveling at 99.9999995% of c++ when it had maybe 10,000 times the mass of a photon at rest, it is still a minimal and undetectable mass (todate).

 

4. What would be the consequencies of a photon having minimal mass, rather than zero mass?

 

A. Possibly this is the allusive Dark Matter?

B. photons colliding with anything (planets/eyeballs) inflict a force. (However, assuming we receive countless photons from all directions all the time. these forces cancel out).

C. Several formula to deal with singularities and expansion would be 'slightly' out and annoying ambiguities accounted for? (eg possible variance in value of c over time, possible variance of gravitational effects over distance)

 

Anyhow, my question is, is this idea a dead duck non-starter? or could it possibly be just very unlikely?[/quote']

 

 

I'd have to say dead duck non starter for a couple of reasons:

 

1. Observations of galaxies and their rotation show that dark matter concentrates in halos surrounding these galaxies, light does not do this.

 

2. There simply isn't enough light to account for dark matter. Dark matter is estimated to make up over 90% of the matter of the universe. The upper limit for the mass of a photon(if it were to have one) due to our present sensitivity of equipment is 4 x 10^-48 g. Thats 2 * 10^20 times smaller than the mass of the electron. The night sky would have to be literally glowing.

Hmmm.... this seems to imply that the light emitted by the particle going left will never reach the particle going right' date=' in my frame of reference.

 

[/quote'] No, because light travels at c relative to you in your frame. Therefore it will catch the particle going to the right at 3/4c.

This implies that the light from the first particle will always reach the other particle, in the frame of reference of the first particle. Now, reality cannot change (at least, I don't think it can ) just because we change frames of reference. Genuine paradox or pure ignorance on my part?

 

Yes it does reach the other particle in this frame, just as it does in every frame.

I thought it could only approach the speed that the particles are being accelerated at' date=' which was rouhgly 200 thousand miles an hour.

 

[/quote'] The only thing that limits the final velocity of any rocket is how much fuel it carries.

It was when you said That equation was not developed to provide answers (or predictions) for FTL speeds. Period. To better illustrate this point you need to look no further than the equation itself. It uses c as it's upper limit for providing real answers. If it had been developed for FST speeds' date=' it would not use c as it's upper limit.[/i']

 

This implies that you think c as a speed limit was part of the development, and it wasn't.

Saint also goes on to say

"The backbone of SR is not simply the constancy of c, but also that it is the universal "speed limit". Everything was developed based on that idea. Since it was clearly developed with the assumption that FTL speeds are impossible to achieve, it cannot provide useable answers to inputs beyond c."

Where he quite literally says that SR was based on the Speed of light being a universal speed limit.

Yes

Relative to the Earth it travels the following distance:

In one lunar month it moves Orbital diameter x pi = 440' date='000 miles x 3.14= 1,382,000

In one year there are 12.38 lunar months.

So distance travelled in one year = 1,382,000 x 12.38 = 17,104,000

 

Relative to the sun it is also moving, on average, an additional distance equivalent to the earth's orbit.

This is 3.14 * 186,000,000 = 584,040,000

So, the total distance travelled, relative to the sun, is 600,000,000 miles[/quote']

 

If you are looking for the distance relative to the Earth you should use the sidereal month rather than the synodic month. There are 13.38 sidereal months in a year. this increases the distance traveled to 18,491,000 miles.

I know when something is accelerated it will be moving slower.

and I know when something is going faster than the speed of light - in theory - time would go backwards.

 

 

so when something IS going the speed of light, does time stop, or just move very slowly?

 

thanks.

It would be undefined. Look at the time dilation equation:

[math]T = \frac{T'}{\sqrt{1- \frac{v^2}{c^2}}}[/math]

Note that as v equals c, the equation becomes

[math]T = \frac{T'}{0}[/math]

And division by zero is undefined.

But since nothing with mass can travel at c, it'll never come up.

If absolute zero is when particles come to a complete hault, then would the hottest temperature be when those particles are zooming around at light speed? If so, what temperature is that?

As close to infinite as you want to get. Temperature is related to the Average kinetic energy of the particles involved. Now while you can't get particles up to light speed you can get as close as you want.

Kinetic energy, taking Relativity into account is

[math]KE = mc^2 \left( \frac{1}{1- \frac{v^2}{c^2}}-1 \right)[/math]

note that as v approaches c, the kinetic energy approaches infinity and so does the temp.

maybe' date=' i'll look it up.

 

What im talking about, is like you looking at a miniature train set.

 

Except the train set is like a solar system. and you see things moving all around and such.

 

So, if something is moving at .5c - will the person that is 1 light year away, and you (the view of the entire "train set") see the same time dialation?[/quote']

 

As long as that something is moving at 0.5c relative to him, yes. And just to make it clear, "moving relative to" does not have to mean "moving directly towards or away from". Thus in the person's view the something could be moving from left to right at .5c and it would be moving at .5c relative to him.

 

If you were already aware of this, I apologise, I'm just trying to figure out where you are coming from with this question.

Your third POV would see things depending on its relative velocity to the ship and Earth. It is no different than either the POV of the Earth or the POV of the ship.

There is no POV that can say that it sees what "actually" happens. All POV's are equally valid, and what each POV measures as happening is what "actually" happens in that frame of reference.

 

Now these calculations involved an estimate of time' date=' but it is unclear to me whether they are talking Sun time, or Earth time.[/quote']

 

It doesn't really matter. Over a period of, say, four biilion years, Earth time and Sun time would differ by less than 8000 yrs. This is this is much smaller than any reasonable margin of error in that estimate.

So' date=' what would it take to blow up the moon?

 

 

[/quote']

A release of energy equal to about 1.25 x 10^29 joules.

 

That is equal to the conversion of 1.39 x 10^9 tonnes( about 1.9 X 10^-11 the mass of the moon itself) of matter to energy.

 

In terms of an asteroid collision this is the the equivalent of a body the size of Deimos (6 km) hitting at a velocity of 4% of the speed of light.

 

Or at velocities you are more likely to encounter: It would take a body of about the size of Saturn's moon Mimas (200 km radius) falling from the outer solar system and striking the moon head on (such that the objects velocity and Moon's orbital velocity are added together.)

does that mean light is always constant relative to an absolute space or constant relative to a third stationary observer?? i thought there was no absolute space, that everything was relative. doesnt that mean we can figure out where the centre of the expanding universe is?

To add to what Swansont has already stated. The speed of light in a vacuum is always c, where c is 299,792,458 meters/sec as measured by any observer.

Thus say you have three observes: A; standing next to the light source, B; moving away from the light source, and C; moving towards the light source. Each has a device capable of measuring the speed of the light from the source relative to themselves.

They will each measure that the light has a velocity of c to relative to themselves.

It depends on your orientation of the ship's gravity to the impact.

If the impact is at a right angle to the "up" of your ship's gravity and strong enough, then, yes, you will be thrown off your feet.

While the gravity will pull you down to the deck plates, it has no sidewise component. If the ship is hit hard enough to shift it sideways, your feet, due to the friction between them and the deck plate, will go with it. your upper body, however will try and stay in place (law of inertia). Your feet come out from under your center of gravity and you lose balance.

If you were hit from the "bottom", you would feel heavier for a moment. If the impact is large enough and you weren't prepared, you could also end up falling.

If you were hit from the top, you would feel lighter, and again, depending on the severity of the impact and what you were doing at the time, it could throw off your balance.