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# Terminal Velocity

Wed 29 Oct, 2008 10:59 pm
For an average sized person, if you jumped off of a building how many feet would you fall before you reached terminal velocity?

I realized this likely can very, but any estimates of what one might expect would be appreciated. Thanks.
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Type: Question • Score: 6 • Views: 6,916 • Replies: 24
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roger

1
Wed 29 Oct, 2008 11:48 pm
@SCoates,
I'll make a guess of abou 70'. Terminal velocity for this person is only about 115 mph, depending on how he's dressed.
BillRM

1
Thu 30 Oct, 2008 02:31 pm
@SCoates,
Assuming terminal V is 172 ft/sec
Using d=1/2 *A*T^2 A=32 FT/SEC*SEC
V=A*T
I come up with roughly 400 feets
ebrown p

1
Thu 30 Oct, 2008 02:41 pm
@BillRM,
This math assumes that acceleration is constant (until it is zero). This is clearly a very wrong assumption-- the acceleration will be quite a bit lower than 32/ft/sec*sec for most of the trip.

The real number will be quite a bit less than 400 feet.
BillRM

2
Thu 30 Oct, 2008 04:20 pm
@SCoates,
Thank you for giving me a problem that cause me to think!

You never fully reach terminal velocity only approach it on a curve.

After writing a fast and dirty basic program I got the following information.

At 400 feets you are at 76 percent of terminal veleocity

At 766 feets you are at 90 percent of terminal velocity

At 1074 feets you are at 95 percent of terminal velocity

At 1807 feets you are at 99 percent of terminla velocity

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BillRM

1
Thu 30 Oct, 2008 04:26 pm
@ebrown p,
Ebrown you mean more not less then 400 feets as with a lesser average acceleration it would of course take a longer distance to reach a given velocity.

Please see the results of my fast and dirty basic program.
ebrown p

1
Thu 30 Oct, 2008 05:29 pm
@BillRM,
Yeah it's more. It looks like my quick logic was flawed.
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SCoates

1
Thu 30 Oct, 2008 08:52 pm
@roger,
He's dressed like a shifty 70's pimp. Does that matter?
gungasnake

1
Thu 30 Oct, 2008 08:59 pm
@SCoates,
It would depend on altitude (from which you jumped) and air density as a function of altitude and also temperature, humidity and a couple of other factors. Ballistics programs are complex and attempt to account for a number of variables.
BillRM

1
Thu 30 Oct, 2008 09:23 pm
@gungasnake,
Gungasnake all the factors you had name are fairly secondary in the case of a man jumping off a building ,in my opinion and would not change the answer by a percent or so at most.

Now if you are trying to figure the flight path of an artillery round then the factors you name would indeed all come into play.

The first electronic computer in this country was created during WW2 to solve such problems while the first electronic computer in England was created to deal with the Nazis machine codes.
roger

1
Thu 30 Oct, 2008 09:35 pm
@SCoates,
SCoates wrote:

He's dressed like a shifty 70's pimp. Does that matter?

If he was dressed fly, terminal velocity would be lower, and take longer to achieve.
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gungasnake

1
Thu 30 Oct, 2008 09:36 pm
@BillRM,
If you mean an ordinary building of several stories fairly close to sea level, you're right.
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roger

1
Thu 30 Oct, 2008 09:48 pm
@BillRM,
I seem to recall that early naval computer was analog, and mechanical to boot. Have not seen the reference in quite a while though.

I don't think the 32ft/sec/sec applies to a human body. Terminal velocity really is in the area of 115 mph, due to wind resistance and low density. My guess of 70' could be way off. It was only a guess, but remember that wind resistance increases as the square of the change in velocity.
Thomas

1
Thu 30 Oct, 2008 10:00 pm
@SCoates,
Scoates wrote:
For an average sized person, if you jumped off of a building how many feet would you fall before you reached terminal velocity?

Trick question. You never reach terminal velocity. You only approach it asymptotically.
0 Replies

rosborne979

1
Thu 30 Oct, 2008 10:43 pm
@roger,
Quote:
I don't think the 32ft/sec/sec applies to a human body. Terminal velocity really is in the area of 115 mph, due to wind resistance and low density. My guess of 70' could be way off. It was only a guess, but remember that wind resistance increases as the square of the change in velocity.

32ft/sec/sec applies to everything (near Earth), but ONLY in a vacuum. In an atmosphere (which is a friction condition) the acceleration of an object decreases as it approaches "terminal velocity", eventually reaching zero (effectively, although Thomas is correct, it only approaches zero asymptotically).

This is a very difficult calculation to set up. Even if you assume an "average" human body with average clothing and ignore variations in altitude, air pressure, temperature and wind and other real-world variables it's still complex.

Skydivers typically reach terminal velocity in about 12 seconds of free fall. So if terminal velocity is about 115mph (about 169 ft/sec), then if he fell at full terminal velocity for 12 seconds he would be 2028 feet away. Since he was falling at some speed less than terminal velocity most of the time, the answer must be less than 2028 ft.

If I had to guestimate the final answer, I would put it around 1200 feet somewhere.

rosborne979

1
Thu 30 Oct, 2008 10:52 pm
@rosborne979,
Bear in mind that there is huge variation in terminal velocity depending on what the person is wearing. If his parachute is open for example, then his terminal velocity is going to be more like 10mph instead of 115mph.

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ebrown p

1
Thu 30 Oct, 2008 11:00 pm
I have always thought the term "terminal velocity" was unfortunate... considering the other meaning of "terminal".

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BillRM

1
Thu 30 Oct, 2008 11:30 pm
@rosborne979,
First as a young man one of my hobbies was jumping out of aircrafts for no good reason and I was more concern about height above ground then my terminal velocity in free fall.

Be that as it may you can get a fairly good idea with fairly simple computer programs and I do remember writting such programs that taken into account air pressure and surface area to weight ratios etc

The one that generated the result I just posted was written with the assume that the terminal velocity was indeed 115 mph to find the disatance needed to reach that speed.

If anyone wish to look at the math or run it here it is.

START:
CLS
TIMEC = .0001
GOSUB ENGINE

END

ENGINE:
vt = 172
vx = 0
d = 0
G = 32
TIMESEC = 0
n = .05

DO UNTIL vx / vt > .99
IF vx / vt > n THEN PRINT STR\$(INT(vx / vt * 100)) + "%", STR\$(INT(vx / 1.466)) + " MPH", STR\$(INT(d)) + " feet", STR\$(TIMESEC) + " SEC": n = n + .05
A = G * (1 - (vx / vt) ^ 2)
TIMESEC = TIMESEC + TIMEC
d = d + vx * TIMEC
vx = vx + A * (TIMEC)
LOOP
RETURN

0 Replies

BillRM

1
Thu 30 Oct, 2008 11:51 pm
@roger,
Roger:
I seem to recall that early naval computer was analog, and mechanical to boot. Have not seen the reference in quite a while though.
------------------------------------------------------------------------------
Yes and no the naval mechanical firing computers were devices using build in firing tables that were created first by hand by having a number of young ladies with mechanical adding machines and log/log tables working away and then the first electronic/tube computer was employ to sadly replaced the young women.

Happily, the young ladies was then given the task of setting up the computer for runs. As this was a task of changing wiring around in plug boards, it was not really programming but as the hardware got better it approach programming.

As a side note there was a number of marriages between the engineers and scientists and the young ladies working on the project. War is hell.

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roger

1
Fri 31 Oct, 2008 12:15 am
@SCoates,
Dang it, Sam. Why didn't you ask how fast he would be going when he hit?

And yeah, Bill. The 115 mph seems to have been an Army figure representing the parachutist who absentmindedly neglected to deploy the chute.

Interesting that they reprogramed the computer by rewiring it. I'm using a language program that changes the keyboard as soon as you load it. Life is good.

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