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Gauging heights by timing falling objects

Discussion in 'Archives - Yahoo Canyons Group' started by Benny R, Mar 3, 2006.

  1. Benny R

    Benny R Guest

    Due to a discussion on gauging cliff heights on the uutah forum and a massive boredom attack at work, I took the time to figure out what the heights would be if you're using the "drop a rock and time it" method of measuring a cliff. I know this has been gone over before, and the moral of the story is don't drop rocks where people are! It may not be the best way to gauge distances, but, in case you're ever in the middle of nowhere on top of a high cliff, positive there's no one below, and wondering... this may come in useful. Interesting at best.

    Sorry to those of you that prefer metric system, you'll have to do the conversions.

    Looks like its time for a high school physics refresher course!

    Gravity is an acceleration, not a velocity. 32 ft/(second squared).

    After doing the calculus the equation for distance(in feet) based on fall time (in seconds) works out to

    16 * (fall time seconds squared). Not too difficult to remember.

    After 1 second the rock will have fallen 16 feet. 2 seconds: 16 * 4 = 64 feet. 3 seconds: 16 * 9 = 144 feet 4 seconds: 16* 16 = 256 feet 5 seconds: 400 feet 6 seconds: 576 feet 7 seconds: 784 feet etc.

    Terminal velocity is around 293 ft/sec (200 mph) for a typical object with minimal atmospheric resistance (eg a rock).

    Terminal velocity would be reached after falling 1341 feet or 9.1 seconds at which point you could just add 293 feet for each second but would probably be too high to hear it crash if you're on THAT big of a cliff, and you're too high to check for people below so you shouldn't be throwing it ;).

    A rock falling from the top of el capitan (3000 ft or so) would take 13 seconds to hit. A body would take a second or two longer unless they were in dive bomber postition.

    Fun stuff. Good tool to have on hand, hope I didn't miscalculate something, that would be embarassing!

    Dang i'm a nerd

    Ben
  2. adkramoo

    adkramoo Guest

    And what does it say about me that I actually enjoyed reading this? Compared to the faulty pipe and my now flooded basement, this was a joy! ;-(

    --- In Yahoo Canyons Group, "Benny R" <benjaminroberts@...> wrote:
    Due to a discussion on gauging cliff heights on the uutah forum and a > massive boredom attack at work, I took the time to figure out what the > heights would be if you're using the "drop a rock and time it" method > of measuring a cliff. I know this has been gone over before, and the > moral of the story is don't drop rocks where people are! It may not be > the best way to gauge distances, but, in case you're ever in the > middle of nowhere on top of a high cliff, positive there's no one > below, and wondering... this may come in useful. Interesting at best.
    Sorry to those of you that prefer metric system, you'll have to do the > conversions.
    Looks like its time for a high school physics refresher course!
    Gravity is an acceleration, not a velocity. 32 ft/(second squared).
    After doing the calculus the equation for distance(in feet) based on > fall time (in seconds) works out to
    16 * (fall time seconds squared). Not too difficult to remember.
    After 1 second the rock will have fallen 16 feet. > 2 seconds: 16 * 4 = 64 feet. > 3 seconds: 16 * 9 = 144 feet > 4 seconds: 16* 16 = 256 feet > 5 seconds: 400 feet > 6 seconds: 576 feet > 7 seconds: 784 feet > etc.
    Terminal velocity is around 293 ft/sec (200 mph) for a typical object > with minimal atmospheric resistance (eg a rock).
    Terminal velocity would be reached after falling 1341 feet or 9.1 > seconds at which point you could just add 293 feet for each second but > would probably be too high to hear it crash if you're on THAT big of a > cliff, and you're too high to check for people below so you shouldn't > be throwing it ;).
    A rock falling from the top of el capitan (3000 ft or so) would take > 13 seconds to hit. A body would take a second or two longer unless > they were in dive bomber postition.
    Fun stuff. Good tool to have on hand, hope I didn't miscalculate > something, that would be embarassing!
    > Dang i'm a nerd
    Ben >
  3. As an old trundeler from way back I enjoyed this diversion immensely!

    Thanks for doing the math.

    Charly

    --- In Yahoo Canyons Group, "Benny R" <benjaminroberts@...> wrote:
    Due to a discussion on gauging cliff heights on the uutah forum and a > massive boredom attack at work, I took the time to figure out what the > heights would be if you're using the "drop a rock and time it" method > of measuring a cliff. I know this has been gone over before, and the > moral of the story is don't drop rocks where people are! It may not be > the best way to gauge distances, but, in case you're ever in the > middle of nowhere on top of a high cliff, positive there's no one > below, and wondering... this may come in useful. Interesting at best.
    Sorry to those of you that prefer metric system, you'll have to do the > conversions.
    Looks like its time for a high school physics refresher course!
    Gravity is an acceleration, not a velocity. 32 ft/(second squared).
    After doing the calculus the equation for distance(in feet) based on > fall time (in seconds) works out to
    16 * (fall time seconds squared). Not too difficult to remember.
    After 1 second the rock will have fallen 16 feet. > 2 seconds: 16 * 4 = 64 feet. > 3 seconds: 16 * 9 = 144 feet > 4 seconds: 16* 16 = 256 feet > 5 seconds: 400 feet > 6 seconds: 576 feet > 7 seconds: 784 feet > etc.
    Terminal velocity is around 293 ft/sec (200 mph) for a typical object > with minimal atmospheric resistance (eg a rock).
    Terminal velocity would be reached after falling 1341 feet or 9.1 > seconds at which point you could just add 293 feet for each second but > would probably be too high to hear it crash if you're on THAT big of a > cliff, and you're too high to check for people below so you shouldn't > be throwing it ;).
    A rock falling from the top of el capitan (3000 ft or so) would take > 13 seconds to hit. A body would take a second or two longer unless > they were in dive bomber postition.
    Fun stuff. Good tool to have on hand, hope I didn't miscalculate > something, that would be embarassing!
    > Dang i'm a nerd
    Ben >
  4. Koen

    Koen Guest

    > A rock falling from the top of el capitan (3000 ft or so) would take > 13 seconds to hit. A body would take a second or two longer unless > they were in dive bomber postition.
    Fun stuff. Good tool to have on hand, hope I didn't miscalculate > something, that would be embarassing!


    Thanks for calculating, it might even come in handy sometime :).

    But you don't take athmospheric resistance seriously enough, on a piece of rock it might not make much difference (but I woulnd't be surprised that it did at high velocities) but on a human body it certainly does !

    A few years ago we were ready to start rapping off a "big" waterfall off the side of a fjord - our altimeters said 950 meters, so that's about 3000 ft, right ? Along came a few BASE jumpers (who had the nerve to declare US crazy !) and off they jumped. We timed a guy (and have a video to prove it) who dropped for an astonishing 28 seconds (!!) before opening his parachute (so not even completely down) - the guy had no wingsuit but after falling about 10 seconds entered into a shallow "dive" away from the wall.

    So maybe a rock drops down 3000 ft in 13 seconds (in a void) but a more or less horizontal body takes half a minute.

    So when's you're next attack of boredom, we demand correct figures ;- ) !

    Koen
  5. --- In Yahoo Canyons Group, "adkramoo" <adkramoo@...> wrote:
    And what does it say about me that I actually enjoyed reading this? > Compared to the faulty pipe and my now flooded basement, this was a > joy! ;-( > Well maybe you could work on wet water disconnects!

    bruce from bryce
  6. Benny R

    Benny R Guest

    Yep the only figuring for atmospheric resistance is in the estimation of terminal velocity... in order to make that call (which I'm assuming would be an acceleration vector in the opposite direction) you'd likely need your altitude, barometric pressure, wind speed, I don't know what else! Best to pick an aerodynamic rock.

    Oh and if you're listening for the crash, I just realized that the speed of sound will cause it to get back you with some significant delay as well!

    Alas, the difficulty of sumarizing the real world with equations. Then again I'm an engineer not a physicist, so almost is good enough! Feel bad for those in the life sciences, not many nifty little equations to appease their brain on how things work.



    > Thanks for calculating, it might even come in handy sometime :).
    But you don't take athmospheric resistance seriously enough, on a > piece of rock it might not make much difference (but I woulnd't be > surprised that it did at high velocities) but on a human body it > certainly does !
    A few years ago we were ready to start rapping off a "big" waterfall > off the side of a fjord - our altimeters said 950 meters, so that's > about 3000 ft, right ? > Along came a few BASE jumpers (who had the nerve to declare US > crazy !) and off they jumped. > We timed a guy (and have a video to prove it) who dropped for an > astonishing 28 seconds (!!) before opening his parachute (so not > even completely down) - the guy had no wingsuit but after falling > about 10 seconds entered into a shallow "dive" away from the wall.
    So maybe a rock drops down 3000 ft in 13 seconds (in a void) but a > more or less horizontal body takes half a minute.
    So when's you're next attack of boredom, we demand correct figures ;- > ) !
    Koen >
  7. On Mar 3, 2006, at 6:06 PM, Benny R wrote: > Feel > bad for those in the life sciences, not many nifty little equations to > appease their brain on how things work.

    actually you'd be really impressed with the incredible equations that come out of studying the mathematics describing biology, from population ecology, to neurons, to biofluids, ...

    mathematical description of biology is one of the relatively new frontiers of science...it kinda goes by the names of biophysics and mathematical biology and will be all the rage in the years to come.

    really really cool stuff.

    stefan

    [sorry to the moderators as this is a stretch to this forum]
  8. Reed Seamons

    Reed Seamons Guest

    In order to make it useful to canyoneering though you would have to be more precise on your timing. Just one second difference adds 80 feet to your height. It would have to be in 1/10 of a second to get it accurate enough to judge whether or not you have enough rope. If you really wanted to get precise, you would have to take into account the time it takes for sound to travel back to your ears to determine that the rock hit. Geez, i think I am getting out of control with this topic.

    Reed



    >From: "Benny R" benjaminroberts@hotmail.com
    Reply-To: Yahoo Canyons Group
    To: Yahoo Canyons Group
    Subject: [from Canyons Group] Gauging heights by timing falling objects >Date: Fri, 03 Mar 2006 21:23:52 -0000
    Due to a discussion on gauging cliff heights on the uutah forum and a >massive boredom attack at work, I took the time to figure out what the >heights would be if you're using the "drop a rock and time it" method >of measuring a cliff. I know this has been gone over before, and the >moral of the story is don't drop rocks where people are! It may not be >the best way to gauge distances, but, in case you're ever in the >middle of nowhere on top of a high cliff, positive there's no one >below, and wondering... this may come in useful. Interesting at best.
    Sorry to those of you that prefer metric system, you'll have to do the >conversions.
    Looks like its time for a high school physics refresher course!
    Gravity is an acceleration, not a velocity. 32 ft/(second squared).
    After doing the calculus the equation for distance(in feet) based on >fall time (in seconds) works out to
    16 * (fall time seconds squared). Not too difficult to remember.
    After 1 second the rock will have fallen 16 feet. >2 seconds: 16 * 4 = 64 feet. >3 seconds: 16 * 9 = 144 feet >4 seconds: 16* 16 = 256 feet >5 seconds: 400 feet >6 seconds: 576 feet >7 seconds: 784 feet >etc.
    Terminal velocity is around 293 ft/sec (200 mph) for a typical object >with minimal atmospheric resistance (eg a rock).
    Terminal velocity would be reached after falling 1341 feet or 9.1 >seconds at which point you could just add 293 feet for each second but >would probably be too high to hear it crash if you're on THAT big of a >cliff, and you're too high to check for people below so you shouldn't >be throwing it ;).
    A rock falling from the top of el capitan (3000 ft or so) would take >13 seconds to hit. A body would take a second or two longer unless >they were in dive bomber postition.
    Fun stuff. Good tool to have on hand, hope I didn't miscalculate >something, that would be embarassing!
    >Dang i'm a nerd
    Ben


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  9. adkramoo

    adkramoo Guest

    --- In Yahoo Canyons Group, "bsilliman2001" <weabruce@...> wrote:
    --- In Yahoo Canyons Group, "adkramoo" <adkramoo@> wrote:

    And what does it say about me that I actually enjoyed reading this?
    Compared to the faulty pipe and my now flooded basement, this was a
    joy! ;-(
    > Well maybe you could work on wet water disconnects!

    Not that bad. Just a neos hike. ;-o Did have to move 20,000 slides of outdoor adventures, along with a literal ton of reading material.
  10. Nice try Benny (good answer in a beginner college physics question, but not one for real life.), but unfortunately, your equasions wouldn't work unless you are in a vaccuum and then in that case, you would have to eliminate your part on terminal velocity.

    Anyway gravity doesn't just start slowing an object once it hits terminal velocity, it slows it from the start. Otherwise your equasion would work, if there is no wind, if the object actually did have a terminal v of 200 mph, and if you also calculated the speed of sound into the equasion. Unfortunately the below won't even be that close because the object is going to fall at less than 32 f/s^2 from the start.

    Anyway here's a good question for you and the rest. A planet has no atmosphere (and thus is a vacuum). You have two objects of the same size. One is a styrofoam ball. One is a steel ball. You drop them at the exact same time. Which ball hits the ground first?

    A) The steel ball because it's heavier. B) They will hit at the same time in a vacuum. C) The styrofoam ball because it's lighter.

    An average ten year old might get the question right, but most college kids get it wrong.

    <<<<<After doing the calculus the equation for distance(in feet) based on fall time (in seconds) works out to

    16 * (fall time seconds squared). Not too difficult to remember.

    After 1 second the rock will have fallen 16 feet. 2 seconds: 16 * 4 = 64 feet. 3 seconds: 16 * 9 = 144 feet 4 seconds: 16* 16 = 256 feet 5 seconds: 400 feet 6 seconds: 576 feet 7 seconds: 784 feet etc.>>>>>





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  11. davewyo1

    davewyo1 Guest

    --- In Yahoo Canyons Group, scott patterson <kesscokim@...> wrote: Which ball hits the ground first?

    > B) They will hit at the same time in a vacuum.

    Dave
  12. Tim Hoover

    Tim Hoover Guest

    Nope. A)

    Tim

    --- davewyo1 davewyo1@yahoo.com> wrote:

    > --- In Yahoo Canyons Group, scott patterson > <kesscokim@...> wrote: > Which ball hits the ground first?
    > B) They will hit at the same time in a vacuum.
    Dave



  13. Rob Heineman

    Rob Heineman Guest

    Scott meant to say drag slows it down, gravity always accelerates objects toward each other.

    In determining drag, a guy I know at MIT once told me that drag increase with the cube of the velocity. And we all now that solving equations with third order polynomials just isn't fun. But he was an electrical engineer, and is now a software engineer and life flight paramedic, so I'm uncertain he's fully up on his Aero/Astro engineering. After all, helicopters aren't supposed to fly, and they crash regularly just to prove it.

    As drag will slow the rocks down, Bennie's basic equation will overestimate the height of the cliff.

    Overestimation is good.

    Too much rope usually works out way better than not enough.

    Unless you're trying to hang yourself.

    Which brings us right back to: TIE A KNOT IN THE END OF THE ROPE IF YOU'RE NOT SURE IT'S DOWN.

    My, how a little knowledge can be a dangerous thing!



    --- In Yahoo Canyons Group, scott patterson <kesscokim@...> wrote: >

    > Anyway gravity doesn't just start slowing an object once it hits terminal velocity, it slows it from the start.
  14. davewyo1

    davewyo1 Guest

    --- In Yahoo Canyons Group, Tim Hoover <frisbeedog02@...> wrote:
    Nope. > A)
    Tim

    No it's B http://tinyurl.com/loh3f Dave
  15. Hank Moon

    Hank Moon Guest

    *** How do you figure that? Even if the steel ball were heavier (and there is nothing in the info given to imply that it is), they still accelerate at the same rate. Assuming they are dropped from the same height at the same time onto an even surface, etc. they will hit at the same time.

    Nope. A)

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    Thank you, Petzl America
  16. Tim Hoover

    Tim Hoover Guest

    --- Hank Moon hmoon@petzl.com> wrote:


    *** How do you figure that? Even if the steel ball > were heavier (and > there is nothing in the info given to imply that it > is), they still > accelerate at the same rate. Assuming they are > dropped from the same > height at the same time onto an even surface, etc. > they will hit at the > same time. >

    Hank - Scott's original post stated that the balls were the same size. Unless there is some super-heavy form of styrofoam that I don't know about, the steel ball will be heavier.

    Despite Dave's link which really doesn't address the question that I think Scott's post was asking, the answer is A).

    Why? You and Dave and the guy who wrote the stuff that Dave pointed to, have probably only considered half of the problem. That is, the effect of the planet on the ball. Don't forget that the ball also exerts a gravitional force on the planet. That force is proportional to the mass of the ball. Steel ball = more massive = more force = more acceleration.

    Now, unless there is some other effect that Scott is thinking of in his thought experiment that I have overlooked, the answer is A).

    Tim

  17. davewyo1

    davewyo1 Guest

    --- In Yahoo Canyons Group, Tim Hoover <frisbeedog02@...> wrote: >> Now, unless there is some other effect that Scott is > thinking of in his thought experiment that I have > overlooked, the answer is A).
    Tim

    Tim; You may very well be correct.My high school physics is outdated.I guess there's still debate,but there's certainly reason to doubt. http://tinyurl.com/e5dyf Dave
  18. Rob Heineman

    Rob Heineman Guest

    --- In Yahoo Canyons Group, "Hank Moon" <hmoon@...> wrote:
    > *** How do you figure that? Even if the steel ball were heavier (and > there is nothing in the info given to imply that it is), they still > accelerate at the same rate. Assuming they are dropped from the same > height at the same time onto an even surface, etc. they will hit at the > same time.


    It's the static charge generated by a swarthy canyoneer's knuckles. This charge will attract the styrofoam ball, while the steel ball remains relatively unaffected.

    Of course, when the canyoneer's blood starts to boil in the vacuum, the whole experiment tends to get messy.
  19. Tim Hoover

    Tim Hoover Guest

    --- Rob Heineman heineman@alum.mit.edu> wrote:

    > It's the static charge generated by a swarthy > canyoneer's knuckles. > This charge will attract the styrofoam ball, while > the steel ball > remains relatively unaffected.
    Of course, when the canyoneer's blood starts to boil > in the vacuum, the > whole experiment tends to get messy.


    Actually Rob, that is one of the effects I considered. However, I think that to build up a sufficient static charge, the canyoneer would need to rigorously rub his balls with his swarthy knuckles and there was certainly nothing in Scott's original post to suggest such activity. Or is this simply implied when refering to canyoneers?

    Hmmm, maybe there is still room for doubt.

    Tim

  20. onkaluna

    onkaluna Guest

    Hank - Scott's original post stated that the balls were the same size. Unless there is some super-heavy form of styrofoam that I don't know about, the steel ball will be heavier.

    *** Just plain styrofoam...I'm thinking it's theoretically possible to make a hollow steel sphere of same radius and weight as the styroball. Not sure of that, but if I get bored enough this weekend, I'll calc it up to see.

    Why? You and Dave and the guy who wrote the stuff that Dave pointed to, have probably only considered half of the problem. That is, the effect of the planet on the ball. Don't forget that the ball also exerts a gravitional force on the planet. That force is proportional to the mass of the ball. Steel ball = more massive = more force = more acceleration.

    *** The attractive force between the (possibly) unequal masses is sometimes called "weight" and (so far as I remember from HS fizix) has no effect on the rate of acceleration in a vacuum. If you have any info (ideally, web links) to support your view, please provide - I love having my illusions busted up.

    hank
Similar Threads: Gauging heights
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Archives - Yahoo Canyons Group Gauging heights by timing falling objects-Correct anwser. Mar 3, 2006