Yeah, teach me about Cosmology. About milky ways and black holes. Please Register or Log in to view the hidden image! Of course not, that would be .. Please Register or Log in to view the hidden image!
If we assign the radius a value of 1R and the muzzle velocity of the gun 1R/t the tangential velocity of the blue man is R(pi)/2t or ~1.57R/t Given that, the velocity of the bullet with respect to the center of the circle will be \(\sqrt{ \left (\frac{R}{t} \right )^2+\left (\frac{R \pi}{2t} \right )^2}\) or ~1.86R/t At an angle of ~57.5 to the radial. This trajectory will make a chord that intersects the Blue man's path. A little trig gives a length for this chord of ~1.27R divided by the speed of the bullet gives ~0.7t for the bullet to travel this chord. a little more trig shows that the angle formed by the ends of this chord with the center of the circle is ~79 degrees. We've established the angular velocity of the blue man as 90 degrees/t, so the number of degrees he will have traveled in 0.7t is ~63 degrees. This means that the bullet will hit some 16 degrees from the blue man.
hansda is old too? But they are not the same thing. One we can't remember, the other we never forget. Please Register or Log in to view the hidden image! Sigh, I wish I had a sexy physics teacher.
Exactly. In this admittedly peculiar case the gun is actually going sideways faster than it ejects the "bullet" towards the centre. So the motion of the ball or bullet is a combination of the two.
Very nice. So, 15 degrees from the center of the blue man. Eye-balling it says that the blue man would need to be at least twice as wide (to cover 30 degrees) in order to clip himself. Not completely impossible if he started eating a lot Please Register or Log in to view the hidden image!
i started the same thread on another forum. glad it went the same way. what seems a simple exercise at first glance is a bit more involved when looked closely at. someone slowed it down Please Register or Log in to view the hidden image! and another version. Please Register or Log in to view the hidden image! and a "joke". Please Register or Log in to view the hidden image! from KJW
Not rubbish. The only problem with the OP graphic is that the blue guy's gun should be aimed a little to the left. Edit... whoa! A lot of action in this thread since I was here last!
Put me on the edge of a merry go round with my trusty .243 and you in the middle and I betcha I can part your hair with the 1st shot, Coriolis be damned Please Register or Log in to view the hidden image!
Wow. Life and death in practical scientific research isn't it? By the way, my naive understanding is that Coriolis effect depends on friction with the surface of the sphere or disc. On Earth, the cooler polar winds move equatorwards along the ground. As they progress to lower and lower latitudes the Earth surface is heading faster and faster towards East. The polar winds "drag" behind because of friction with the ground at the preceding position they pass through the higher latitudes towards lower latitudes where winds are already moving faster eastwards with the Earth spin friction they had already there. So polar winds swing less towards east than equatorial and other existing air masses between lower latitudes and poles. This makes a vortex pattern of swirling air masses whose lower latitude side is faster towards the East than its higher latitude sides, meanwhile the whole swirling air mass slowly reaches equilibrium with the spin of the Earth and join the Earth air mass as a whole rotating with the Earth. Meanwhile the hotter air that goes towards poles only has the friction with the air masses lower down and not the Earth itself. So the reverse "coriolis" effect is not as strong, and there are many local "horizontal swirls" and lesser-circumference "jet streams". A bullet not in contact with the surface of the Merry go Round would not be much affected by any "coriolis" effect, so its trajectory is simple addition of vectors of muzzle velocity (bullet expulsion velocity) and muzzle-motion sideways during the bullet transit in barrel to exit point (that is the only time it experiences "coriolis" effect because barrel is "swinging" (rotating) and bullet has friction with one side of barrel more than the other, so gains a "rotating" or sideways motion the further along it travels in barrel until expulsion at muzzle. To know "free air" path of bullet one must know swinging effect from in barrel travel/exit at instant of exit. From there on the tajectory is fixed as resultant of tangent velocity (from swinging muzzle tip, like spinning water sprinkler water droplets) and straight along muzzle line velocity due to gunpowder impulse to bullet (like water pressure velocity to water drops from sprinkler). That is all I can think of relevant. Have I forgotten something?
Nope , you can't and you won't because you are totally ignorant about the very effect that alters the shots.
This is beautifully done, except I wish the yellow dot had released the projectile slightly in the direction of his motion of travel; that would be more analogous aiming directly for the center of the circle.
I agree. It shows how the yellow and blue dots are incapable of making an accurate description of the inertial motion of the red dot in the preferred frame. It appears to the yellow dot that the bullet is boomeranging back to him, hence constantly changing direction and constantly accelerating. It appears to the blue dot that the bullet first is accelerating away, then starts decelerating, then starts accelerating towards him, again constantly accelerating and constantly changing direction. Of course, if you ask the yellow or the blue dot what their motion is...their answer is...at rest. (rolls eyes)
.. and go figure! As you will see by my edit, it was a complete error. Wrong person, wrong thread, wrong forum.
LOL How did you make such a mistake? Please Register or Log in to view the hidden image! Anyway, for a mistake, that reply is entirely appropriate.
