Will a cylindrical shaped kite fly?
481512
I believe a cylindrical kite can fly. If you think about it there is space for the air to move through and yet there is a structure to create resistance. The air would move the the open cylindrical design, but at the same time, the walls of it would allow resistance for the air to hit and allow resistance in order to keep it in the air......that's what i think, i dont know about the no wind part though.
672455
A kite is a four-sided planar figure with two equal-length sides "on top" and two equal length sides "on bottom", so that the lines connecting opposite vertices intersect at a right angle.\n\nThere are two triangles formed (both isosceles): one with edge A and two sides of length 6.  The other shares edge A but has sides length 14.\n\nI wish I could help you further, but apparently my geometry is too rusty.
733269
h = height\nd = distance from kite at start\n\nsin(47) = h/d\nd = h/sin(47)\nd = 0.73h\n\nsin(39) = h/(d+30)\n(0.73h+30)sin(39) = h\nh = 34.898 ft
434974
not at all - just most wakeboards are slightly lower volume than beginners kiteboards so can be a bit harder to start with.\ngood for higher winds though.
762124
I wouldn't use the sine function to solve this problem.\n\nYou know the angle of elevation, and something about both legs of two right triangles. This looks like a job for the tangent function.\n\nCall the initial distance to the kite x, and the elevation of the kite y.\n\ntan 47 = y / x\n\nWhen you increase x by 30, the angle of elevation becomes 39 degrees.\n\ntan 39 = y / (x + 30)\n\nSolve both relations for y and equate the two:\n\ny = x tan 47\ny = (x + 30) tan 39\n\nx tan 47 = (x + 30) tan 39\n\nx tan 47 = x tan 30 + 30 tan 39\nx tan 47 - x tan 30 = 30 tan 39\nx = 30 tan 39 / (tan 47 - tan 39)\nx = 92.516 feet.\n\nNow you can find y:\n\ny = x tan 47\ny = 92.516 tan 47\ny = 99.21 feet.\n\nSince we didn't work with the hypotenuse of either right triangle, we didn't use the sine function, but since you did ask about it:\n\nh^2 = y^2 + x^2\nh = sqrt(x^2 + y^2)\n\nWhere x and y are defined as before.\n\nsin 47 = y / h\nsin 47 = y / sqrt(x^2 + y^2).\n\nSimilarly:\n\nsin 39 = y / sqrt[(x+30)^2 + y^2]\n\nI would take issue with the fact that we don't know x. As it turns out, we can figure it out without using the sine function, as shown previously.
236075
You might try http://www.intothewind.com .  I think they sell spinning box kites (not exactly sure what it is you're looking for).
617842
it certainly wont (provided it is not run over when it reaches the ground.)because it is very light it will be affected by the wind and will manage to make a smooth landing,however if the wind velocity is high it will die.(i am answering takin into consideration that the ant does not get a heart attack!)
705060
If I understand the question, you are giving a velocity to the cube and don't want it to touch the hemisphere at any time in its flight. I think you want the minimum, not the maximum hor. velocity to do this, right? Because any arbitrarily high velocity will do it. I'll use a planar analysis where the hemisphere is a semicircle.\nAssuming the semicircle is centered at x=y=0, the trajectory will be a parabola y=r-kx^2 and at y=0, |x| must be >r. Another constraint arises when we consider the region of the semicircle near the top. Here the semicircle approximates a parabola, since 1-cos(small angle) approaches small angle^2/2. Thus deltay = r(1-cos(small angle)) = r-y approaches r*small angle^2/2 = r(x/r)^2/2, or x approaches sqrt(2/r*deltay*r^2) = sqrt(2r*deltay).\nDownward acceleration is g and horizontal velocity is v. When y=0, 0.5gt^2=r, so t=sqrt(2r/g), and x=vt must be >r. Thus the 1st condition (for clearance at the bottom) is v > r/sqrt(2r/g), or\nv > sqrt(rg/2).\nThe 2nd condition requires that when r-y=deltay is small, then and x=vt must be >sqrt(2r*deltay) where t=sqrt(2deltay/g), or  \nv > sqrt(2r*deltay) / sqrt(2deltay/g) or \nv > sqrt(rg).\nThis latter is the more stringent of the two requirements and thus is the minimum clearance velocity.
627979
i've heard of him.\n\ndo you kite then ?
243350
Yes.  A wheel in the shape of a circle is going to go significantly farther than a wheel in the shape of a square.
341855
I used to make them as well..if you hold up three of your fingers.. this is the shape to put the sticks to start...Then insert the other two sticks horizontaly between the other ones. There is also a kite version..but its very hard to describe how to make it.
1069494
It will be more. Mathematically:\n\nno wind case:\n\nLet the distance from A to B = d\nAnd let the speed of the plane without wind = v\nSince distance/speed = time, in the case with no wind the time for the return trip =  d/v + d/v = 2d/v\n\nWind case:\n\nLet d,v be as above, and let the wind speed blowing from A to B = w. The planes wind-aided speed = v+w and it's wind reduced speed is v-w. The time for the return trip is now:\n\nd /(v+w) + d /(v-w) = (d(v-w)+d(v+w))/(v^2 - w^2)\n\nThis simplifies to:\n\n2dv /(v^2 - w^2) compared to 2d /v for the no wind case.\n\nThe ratio of the no wind case to the wind case is:\n\nv^2 / (v^2 - w^2)\n\nWe see that if w > 0, this ratio will be greater than 1, which means the return trip will take more time if there is a wind.
400216
It depends on windspeed and direction, and the planes direction in relavence to that speed and direction. It depends on the speed of the plane. It depends on the density of the object, because of wind resistance. EX. feather would be close, lead ball would be further. Your question is like me saying, "I am in a car, how long will it take me to get to McDonalds?" Not enough info
783086
Gravity is pulling down on the object, air is resisting it (pushing up, if you will) and causing friction, and if there's any wind it's pushing sideways on the object. I doubt you'd have wind in your problem, though. Wind is complicated :)
948754
we juss need to take vector sum of the two forces. assuming the wind is directly in horizontal direction....... sqrt(9.25sq+1.05sq)=9.3N
316237
light line or use a flyrodand throw with the wind
917959
My suggestion is don't be cheap and buy a real one.\nYou don't understand the positive and negative forces\non the frame and wing spar. It's not like building a kite\nand take it out for it's first flight. But If you think that you can\nbuild one and fly it with no experience with building or even\nflying one go right a head, I love it when I see some knuckle\nhead make the news, I just laugh my azz off.