Equatorial mount declination and right ascencion ring problem?
924362
The answer to this depends on the specific equatorial mount you havve.  Different manufacturers will have different ways to do this.  The rings should be able to move on their stem, but not freely -- they should (if you don't touch them) follow the scope's motion.\n\nIn general, you want the RA and Decl rings to be adjustable so that you can calibrate then (or align, which might be a better word.).\n\nAfter aligning your polar axis closely, select a recognized star near the equator (+/- 30 degrees is fine), point the scope to it, and then adjust the RA ring to agree with the known right ascension of that star.  Similarly set the declination ring, and \nyou are all set.  \n\nIf the rings are too loose, then you need to check the scope directions (don't you hate that?) and see if there isn't a collet\nring or nut or setscrew or something to tighten it a bit.  It still should be moveable when you need to re-adjust the rings (each time you set up the scope you need to set the rings.
810893
Oscar. You have written too much. This 125m tower stuff does not matter compared to r=6.38*10^6. Why is this talk about 1.00 km and angle theta? See what “zipper” above says. Besides I answer your last question << Why is the trajectory of the cannonball not parabolic?>>. Why should it be if both you and Newton demand it to be circular? Assume that V1 of cannonball is the 1st cosmic speed corresponding to a circular track. Let V2 be > than V1, then you receive an oval track. Take V3 > than V2, then the track may become parabolic, so that cannonball would never return back to the cannon to linger out among planets and stars. If you are intrigued calculate a set of velocities and corresponding orbits of the cannonball. If cannon is not available consider a football! I like your inventive wits – to be honest!
721540
Common problem you need to redirect the Focal Light Projector to triangulate the correct quadrant, for the Philippines it is 63 degrees left and Negative 64.5678 East and 657 south south due west minus 645.45 that should fix the problem
535952
Assuming a big ring that rotates, we need to fix the radius and the velocity out on the edge.  The acceleration toward the middle (which causes the feeling of gravity) is given by\n\na = v^2/r\n\nTo match gravity on earth, assume 9.80 m/s^2\n\nYou can make the radius what ever you want.  Assume 100 meters.\n\nThen 9.80 = v^2/100.  So, v^2 = 980.  v = 31.30495.\n\nSo, how fast is it spinning?\n\nThe circumference is 2*pi*r = 2*100*3.14159 = 200*pi.\n\nThe time it takes for it to do one rotation is:\n\n200*pi (meters)\n--------------------\n31.30495 (meters/sec)\n\n= or 20.07088 seconds.\n\nThe angular velocity is 360 deg/20.07088 seconds = 17.9364 deg/sec\n\nor, 2pi radians/20.07088 = .3130495 rads/sec.\n\nNotice, this is exactly what you would have gotten by taking v/r and getting the angular velocity.  I don't know if you are studying that part of rotational dynamics yet.\n\nSo, the bottom line is, as long as v^2/r = 9.80, you can then get the rotational velocity in radians from v/r.
25445
The earth is APPROXIMATELY round due to gravitational forces.
791771
Equatorial diameter = 12,756.2 km\nPolar diameter = 12,713.6 km\nVolumetric mean diameter = 12,742.0 km\n\n(BTW there is no such thing as centrifugal force, it's an engineers' myth; there is only inertia.)  :)
737740
Follow ditz's advice (make a diagram) as you read these directions for solving the 4 problems:\n\n1. assuming that the sun is a point source and the earth's surface is flat:\nangle = arctan (6.25/10.1)\n\n2. assuming that the string forms a straight line, and that the end of the string is on the ground:\nsqrt(50^2 - 30^2)\n\n3. arccos (1.75/4)\n\n4. 30 cos (75 degrees)
154590
I bought a field guide that helps you find things with your telescope and basic telescope knowledge. i know you have to know where north/south (which ever is closer to you) and what your latitude is in degrees (how far you are from the equator).
27065
I've been studying this stuff recently, and I'm pretty sure this is how it goes:\n\nK is now (-2,1) and D is now (-3,-2).\n\nFor future reference on determining rotations, simply turn the graph counter-clockwise one time for 90 degrees, upside down for 180 degrees, and for 270 degrees, just turn the graph clockwise once to see what the new cooridnates are.
935099
Kepler had something to say about this type of problem.\nGMT^2=4*pi^2*a^3
38995
In this situation a compass needle will not point to anywhere on the dial very strongly - it will just wander around weakly with no real direction. Explorers now need GPS to determine the exact pole location; it shifts slightly every year.\n\nDC :-)
681007
If you mean the latitude, then it's 41.873712° N (+41.873712°).\n\nIf you mean the magnetic declination, then it's currently\n14°15'W changing by 0°3'E/year.
1035386
Astronomers generally use either the equatorial or ecliptical reference frames. The standard reference plane for the equatorial reference frame is the earth's equator. The standard reference plane for the ecliptical reference frame is the plane of the earth's orbit around the Sun. The intersection of these two planes is a line that points in one direction towards the vernal equinox, and in the other direction towards the autumal equinox.\n\nIn both systems, the zero point of longitude (for ecliptic) or Right Ascension (for equatorial) is defined as the direction of the vernal equinox, also called the First point of Aries.
337819
The direction that a clock goes is entirely dependent on how it is made.  You can easily put the numbers in the opposite direction above the equator and have it work, so which side of the equator it's made on doesn't really matter.
93783
I think one has to recalibrate your barometer for every different location (including altitude), according to what I read on the internet and my intuition. \n\nI am working on a more quantitative analysis which I will try to complete to post here later.
709479
eclipze Big Big news for you!\n★http://www.osoq.com/funstuff/extra/extra02.asp?strName=eclipze
105883
First we note that the two bases of the trapezoid are parallel and the segment joining the midpoints of the altitudes is parallel and equidistant from the bases. Therefore the vertical displacement of the midpoints of the bases from the midpoints of the altitudes is equal.\n\nIf we also contruct the segment joining the midpoints of the bases we see this intersects our other segment at its midpoint. We have now cut our quadrilateral into four triangles and it is an easy task to show that opposite triangles are congruent using SAS.