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I really feel like Father Dougal for asking this ...

  • 19-08-2006 11:56am
    #1
    Closed Accounts Posts: 10,921 ✭✭✭✭


    but I'll ask anyway.

    Say you see an object in the distance that you know for a fact to be size 'x' (eg your friend on the other side of street is 6ft tall) but from this distance they appears to be just size 'y' (eg just 4 inches from however many feet away)) is there any simple ratio rule for determine approxmiately how far away the object is?


Comments

  • Closed Accounts Posts: 7,794 ✭✭✭JC 2K3


    Get a pencil or similar object(a ruler would work better actually) and hold it in front of your eye and adjust it until it appears to be the same height of your friend in the distance and measure the distance of the pencil to your eye. Then divide the height of the pencil into your friend's actual height and multiply it be the distance of the pencil to your eye.

    You can probably derive a simple ratio from a simple method like that, I'm too tired to think though...

    EDIT: I think you'd need to know about the lens in the human eye to get a ratio actually...


  • Registered Users, Registered Users 2 Posts: 5,166 ✭✭✭enda1


    Well just doing some back of the handkerchief maths there but it seems that if you know a person to be six foot tall, at what distance from you do they seem like six foot tall. Say your "real height" distance is six foot from you. Anyway. The angle this person makes (from thwe top of their head to your eye to the bottom of their feet, assuming your eyelline is at their midheight) at six feet from you is 53.13degrees.

    If a three foot high person stood at the same point from you the angle would be 28.072degrees. So since this is not quite half the height but related to tan, it is quite hard to estimated (from i the head calculations) distances. however if you restrict yourself to long distances form you, more than about 20 feet only, tan of a small angle becomes approximately proportional to the angle. So if the person appears twice as small if they are distance away than another distance, you could say they are twich the distance way.

    Now to put a base distance in to do your estimation with.
    Set someone up who is 6 foot tall, at a known distance form you, then figure out what height they appear to you, then you'll be easily able work out ther distance form you.

    If you couldnt be bothered doing this ill come up with a simple (fits most situatuions) rule.


  • Registered Users, Registered Users 2 Posts: 5,166 ✭✭✭enda1


    Assuming your index finger is 5cm long and the distance from your finger to your eye is 80cm, then if you place your finger in front of your eye, and it appears the same height as a person who is known to be 2m tall, then they are 32m away.

    their distance is just... [(distance of finger to eye)*(person's actual height)] / [(length of your finger)*(fraction of your finger covering the person)]


    If you know the distance and not their height, just rearrange.

    EDIT: Just realised what I said is exactly the same as what JC_2K3 said.


  • Closed Accounts Posts: 7,794 ✭✭✭JC 2K3


    Heh, yup.

    Forget that crap about needing to know about the human eye to get a ratio though(I was tired). You can easily work it out once you get the ratio of apparant height:actual height once and apply it to every other situation.


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