Customize

KAP (Kite Aerial Photography)

Discussion in 'Think Tank' started by awesomizer, May 29, 2008.

  1. awesomizer Member

    KAP (Kite Aerial Photography)

    We should take pictures of Anon at the monthly protests using kites because you get cool pictures.

    See KAP for details.
  2. highoverlord Member

    Re: KAP (Kite Aerial Photography)

    i got bad luck my kite would crash in to power lines for me
  3. awesomizer Member

    Re: KAP (Kite Aerial Photography)

    Well, it would only really work when the protest area was in a fairly large open area AND when there was a nice steady wind at the time. Maybe at that DC National Mall thing?
  4. Anon1990.572 Member

    Re: KAP (Kite Aerial Photography)

    Good idea, but again, it would only work in a large open space. I smell Live Raid - it would look fukken awesome to have a few aerial photographs. I will look into it.
  5. vegnej Member

    Re: KAP (Kite Aerial Photography)

    Hey they might be good for taking photos of interesting psilon areas LOL !!!!
  6. Danilov Member

    Re: KAP (Kite Aerial Photography)

    I would think balloons would work better... at least you know they won't suddenly go suiciding into the ground.
  7. Anon1990.572 Member

    Re: KAP (Kite Aerial Photography)

    You'd need a fuck load of balloons to lift a camera unless it were superlight weight. I'm sure you could try it but the kite might be much more viable.
  8. Anonymous Member

    Re: KAP (Kite Aerial Photography)

    Hmmm. It seems to be a fairly popular hobby where I'm from (I've overheard discussions about it a few times).
    Anybody have the equipment for it? I hear it isn't a very inexpensive hobby.
  9. awesomizer Member

    Re: KAP (Kite Aerial Photography)

    Well, I estimate that for about maybe 40$ max you could build a fairly good rig to hoist a normal camera. So, 40$ + cost of kite + (maybe) cost of camera (ie if it crashes).

    If you have a cheap Digic-based Canon camera one can load CHDK and script stuff so as not to have to fuck around with modding the camera.
  10. awesomizer Member

    Re: KAP (Kite Aerial Photography)

    Well, I think my camera weighs about 200g.
    But, we also need to factor in the weight of the string which we use to make sure that the camera doesn't float away.

    Suppose our string weighs 10g/m (and of course is strong enough for our purposes), and we want to have it go max 30m into the air (98 feet for those who are still using the Retarded system of measurement).

    10g*30m=300g worth of string.

    Thus the force pulling it down thru gravity is (0.2 + 0.3) * 9.8 = 4.9 Newton.
    Now, if we want the balloon to go up, we need a force of more than 4.9N so that the net force is going up.

    Knowing that F=ma, put in 0.25 m/s^2 as the acceleration, plus 0.5 as the mass, means in order to get that accel we need to exert 0.125N on our camera (since we already have zero net force going up/down).

    4.9 + 0.125 = 5.025 N

    Therefore we would need to get 5.025 N of lift in order to lift the camera.

    You then need to find the lifting power of a helium balloon from wherever you are, divide 5.025 by that, then get that many balloons.
  11. awesomizer Member

    Re: KAP (Kite Aerial Photography)

    lurk'd moar
    Helium Lift Power
    1g helium lifts 6.125g.

    Converting back from N to g, we see that the "weight" of our payload is 513 g. (including the excess we give to get upwards accel)
    513/6.125=83.75 g of helium

    Suppose that the temperature is a nice 25 deg C (77f) and we have standard pressure of 101.3 KPa.
    83.75g = 20.93 mol of helium
    We know that 1 mol = 22.4L @STP (101.3KPa, 0C) so we need to adjust foir the temp change. The volume of 20.93 mol of helium @ STP is 20.93*22.4 = 469 L
    Convert to Kelvin: 0 = 273.1; 25 = 298.1
    (V_1/T_1)=(V_2/T_2)
    (469/273.1)=(V_2/298.1)
    1.717=V_2/298.1
    1.717*298.1=V_2
    512=V_2
    So @ 101.3 KPa and 25C, the helium needed will occupy 512L. I am, however, too lazy to measure the volume of a balloon right now.

    ULTIMATE FORMULA OF ULTIMATE DESTINY:
    ((((((((((Cw+(Sw*Sl))*9.8)+((Cw+(Sw*Sl))*a))/9.8)/6.125)/4)*22.4)*101.3)/273.1)*(T+273.1))/P)=V
    where V is volume needed to lift [L], Pw is the weight of the payload [g], T is the ambient temp [C], Cw is the camera weight [Kg], Sw is the weight of 1m of string [Kg], Sl is the length of the string [m], a is the desired accel [m/s^2], and P is the ambient pressure [KPa]

    Note that I haven't accounted for the weight of the balloons so leave a little padding.
  12. The Kite Man cometh: Ars does DIY open-source aerial surveillance
    [two page article from arstechnica.com]
    Example: An aerial view of a protest at the University of California-Davis, taken by Stewart Long's balloon from about 200 feet up.
    [IMG]

Share This Page

Customize Theme Colors

Close

Choose a color via Color picker or click the predefined style names!

Primary Color :

Secondary Color :
Predefined Skins