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Estimating the size of a gap
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How do people measure gaps like Chad's or the cliff that Pierre did? is it the same formula they use for calculating the distance to the sun? Same thing with pictures where they say hees about 17 ft out
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i just picture a bunch of basketball hoops. 3 basketball hoops = 30 ft. haha
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the same way the measure mountains, they use a triangle/algebreric formula some how.
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well before i drop cliffs/ hit booters i just test speed by thrwoing a snowball.
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idk but when they say that people are a certain height out of the pipe its usually off, i noticed at the xgames that all the spectators were between 7-9ft tall according to the height thing
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im tlkin bout w/out measuring poles. like when they hit the hip at june
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Physics D= -gt^2/2 + ht g = acceleration due to gravity t = time and h=height then solve for height if you know time
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u actaully have to hit it to do that
just use trig
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You cant use trig unless you have measurements. And you dont have to hit it a snow ball or any object will fall at the same speed.
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i imagine things like Chads, Pyramid, and World Record Cliff drops (Pierre's) they use an actual measuring tape. In Teddybear Crisis i think there is a shot of them measure Chads. But yeah..... i'm sure some geometrical/algebraic equations are used.
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chads they defiently used measuring tape its in some movie where it show the measurement
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well cliffs are easy all you do is put in d=1/2 at^2 where d=distance(your variable) a= acceloration of gravity (9.8) and t= how long it takes a guy to fall of the cliff.
and i don't know about chads.
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i read somewhere that for pierres 245ft drop they used some electronic measuring device.. probably like a transmitter at the top and a reciever at the bottom and they calculate the distance between eachother
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well i think everyone is over thinking this ..tape measure?
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i just estimate a normal person being 5-6 feet, so i guess how many people could stand head to toe
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ill stand below a cliff or on the side of a jump and im 5'6 so ill just guess and hope im rite
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Usually with gaps like Chad's, they just get as much speed as they can and it normally works out pretty well.
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HAHA wise words from earl
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like that thing in the futureshop comercial, thats a laser pointer and hes all like, the pole is 8 feet, the ceiling is 20 feet, your wife is 16 feet...14 feet...
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well its simple 8th grade math for the cliffs, you stand on the bottom of the cliff 100 feet away with a protractor and a string attached with a bolt and the string and you see what degree the top of the cliff is and use pathagoere and theroem
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I think you are leaving out terminal velocity, so you would therefore have to encompass the persons mass and the wind resistance
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lol. I think you mean the Pythagorean Theorem, and no, that wouldn't work. You'd have to use trig ratios, but that is the easiest way to do it. That, or the law of cosines. I'll guarntee you they don't throw snowballs off and do the calculations. That shit is for high school students.
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as one said before, it isnt (on cliffs) "1/2* a*t^2" because you miss out air friction and the time problems because of the forward movement (but it doesnt influence the time at all, we all know since this thread about a mountainbike drop a good time ago)
they def. measured chads with a tape and bigger things can be easily calculated with (i dont know the english term) those two triangles that are overlapping and all three angles are the same, so you can calculate the sides and so on, its pretty easy, dunno if its used...
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laser things sounds most likly
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Simple trig for mountain drops, comeon sine cosine tangent.
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Yeah but in situations like Chad's and Pierre it's not reasonable to use that.
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wind resistance would be neglagable and terminal velocity is when a body has reached its maxium velocity in camparision to wind resistance and for a human is around 75mph anyone dropping a clif at 175mph is going to die anyway do it doesent really matter what happens.
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maybe they just estimate.
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