Read the book, then you'll get it.

Clarke wrote the book while he was working on the script with Kubrick. I think they are both better when viewed as companions to eachother.

One of my favorite stories, but I too find it hard to stay awake during the symphony portion.

chiefthe

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# chiefthe

Member Since 08 Nov 2004Offline Last Active Jul 01 2009 03:50 PM

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- Member Title Hasbro Engineer
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### In Topic: 2001: A Space Oddysey

30 August 2006 - 08:10 AM

### In Topic: What Is Nerf To You?

15 June 2006 - 12:37 PM

My job.

chiefthe

chiefthe

### In Topic: Fall '06 Blasters

24 February 2006 - 10:44 AM

It's awesome that you guys are looking forward to the blasters.

Sorry I couldn't be there to see your reactions. I am sure they were priceless.

chiefthe

Sorry I couldn't be there to see your reactions. I am sure they were priceless.

chiefthe

### In Topic: Micro Darts

03 October 2005 - 10:15 PM

Unfortunately, there is really no way to tell. It will be whatever they happen to have in stock in the warehouse, so it may be older darts or newer ones.

Sorry I can't be of more help.

chiefthe

Sorry I can't be of more help.

chiefthe

### In Topic: Stefan Theory

03 October 2005 - 08:15 PM

Ah! But you accelleration is known--it is actually a decelleration due to gravity. That said:

The time it takes something to fall from a given height (h) when its initial angle is 0 degrees can be derived from the y equation (you can derive it yourself for other angles, zero is just the one I don't have to look up):

y=(v_0 sin (theta))t -0.5(g)(t)^2+h

y=0-0.5(g)(t^2)+h

set y=0 to see when it hits the floor:

t_floor=sqrt(h/(0.5g))

Using a little calculus magic, the velocity equations are:

v_x=v_0

(remember, no drag, so there is no opposing force slowing the x-velocity, so it has the same velocity it started out with---this equation is only for the zero degree case, for the general case: v_x=(v_0 cos(theta)) )

v_y=-(g)(t)

(negitive because the vector is pointing down---this equation is only for the zero degree case, for the general case: v_y=(v_0 sin(theta))-(g)(t) )

So, going back to vector algebra:

v=sqrt(v_x^2 + v_y^2)

If you want to solve for v_f, then plug t_floor from above into the v_y equation, then use the resulting v_x and v_y to put into the v equation above.

LOTR434, I hope I'm not doing your homework... ;)

chiefthe

The time it takes something to fall from a given height (h) when its initial angle is 0 degrees can be derived from the y equation (you can derive it yourself for other angles, zero is just the one I don't have to look up):

y=(v_0 sin (theta))t -0.5(g)(t)^2+h

y=0-0.5(g)(t^2)+h

set y=0 to see when it hits the floor:

t_floor=sqrt(h/(0.5g))

Using a little calculus magic, the velocity equations are:

v_x=v_0

(remember, no drag, so there is no opposing force slowing the x-velocity, so it has the same velocity it started out with---this equation is only for the zero degree case, for the general case: v_x=(v_0 cos(theta)) )

v_y=-(g)(t)

(negitive because the vector is pointing down---this equation is only for the zero degree case, for the general case: v_y=(v_0 sin(theta))-(g)(t) )

So, going back to vector algebra:

v=sqrt(v_x^2 + v_y^2)

If you want to solve for v_f, then plug t_floor from above into the v_y equation, then use the resulting v_x and v_y to put into the v equation above.

LOTR434, I hope I'm not doing your homework... ;)

chiefthe

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