GBlab4PibalBalloon
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School
University of Colorado, Boulder *
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Course
1070
Subject
Geography
Date
Dec 6, 2023
Type
docx
Pages
5
Uploaded by BrigadierGorilla17107
Lab 4 Report
Name: __Grant Bowditch__________________
Partners: _Chase, Nico, Carter
Please paste your Excel data and graph below this line. Highlight everything you need in Excel,
copy, and then use ‘Paste Special’ to paste as an image (e.g., PNG, PDF, etc.). You can take a
screenshot as well and paste it here, but please make sure it is easily viewable by your TA.
Question 1 (5pts):
Consider this situation. Your
elevation angle is 45º with
your balloon directly east rising
at
5 m/s. The horizontal wind is
straight west to east. How fast
will these west winds have to
blow for you to have to reduce
the elevation angle on your
next measurement? Explain
your reasoning.
The winds will have to be
anything greater than 5 m/s.
The reasoning for this is that 5
m/s is the speed at which the
balloon is rising, and to displace it off its path it needs a force greater than
this in another direction. This will lower the elevation angle because the
winds will have pushed it further east onto a new path, requiring the
theodolite to reach a further distance on the horizon than before.
Question 2 (15pts):
The scale height of the atmosphere (
H
) is a variable that meteorologists
often use to do conversions from altitude to pressure. Scale height of the
atmosphere refers to the height one must go to in the atmosphere to achieve
a density or pressure that is roughly 37% of the surface density or pressure.
Strictly,
H
is a function of temperature since that influences the changes in
density or pressure with height, but for our purposes we can
consider the
scale height (
H
) to be a constant 8000 meters
.
The equation to calculate the pressure at a given altitude for the balloon is
P
balloon
=
P
sealevel
∙e
(
−
Z
H
)
where
P
sealeve
l
= 1013mb,
H
=8000m,
e
is the base of the natural logarithm
and is 2.718, and
Z
=the balloon’s height above sea level
.
a.
Given that you launched your balloon at 1640m above sea level, calculate
the expected pressure at the launch site. Note that you need to calculate (-
Z/H) first, then calculate e
(-Z/H)
, then multiply by 1013mb. Show your
calculation.
(-Z/H) = (-.205)
2.718^(-.205) = .8147 x 1013 =
825.26 mb
b.
How well does this match with the pressure observed by your TA?
844.3 – 825.26 = ~
10.04 mb difference
c.
At what height (in meters) above sea level did you stop tracking your
balloon at?
2,140 meters above sea level
d. Using the equation given above, at what pressure level did you stop
tracking your balloon at? Show your work.
Pressure = 1013 x 2.718^(-.2675) = 1013 x 0.7653 =
775.26 mb
e.
Long’s Peak has an average pressure of 588mb. Was the pressure at your
balloon’s highest height higher or lower than the pressure at the top of
Long’s Peak? Given this, would your balloon have flown over Long's or would
it have hit it?
The pressure at the balloon's highest viewed point was higher than the
pressure at the top of Long’s Peak, by 187.26 mb mb. Given this, it would
have hit it because pressure decreases as altitude increases, so the altitude
of the top of Long’s Peak is higher than the balloon's highest altitude.
Question 3 (15pts):
Normally, wind speeds are weakest near the ground (due to friction with
trees, hills, building, etc.) and increase with height, but this isn’t always the
case.
Find an area of your wind speed vs. height graph where the wind speed
changes the most over a 300 meter change in altitude.
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a.
Which layer did you choose (example: 3000-3300 meters above the
surface)?
1840-2140 meters above the surface
b.
Calculate the change in wind speed per meter for this layer. Do this by
taking the wind speed at the top of the chosen layer and subtracting the
wind speed at the bottom of layer. Then divide this number by 300 meters.
Note that you just calculated a measure of vertical ‘spin’ or more simply wind
speed generated shear in the atmosphere. The units are knots/m.
(13.8-3.2)/300 = .0353 knots/m
Find an area of your wind direction vs. height graph where the wind direction
changes the most over a 300 meter change in altitude.
c.
Which layer did you choose (example: 2700-3000 meters above the
surface)?
1740-2040 meters above the surface
d.
Calculate the change in wind direction per meter for this layer. Do this by
taking the wind direction (in degrees) at the top of the chosen layer and
subtracting the wind direction at the bottom of layer. Then divide this
number by 300 meters. Note that you just calculated a value of
horizontal
‘spin’ or more simply directional wind shear in the atmosphere. The units will
be º/m.
(12-351)/300 = -1.13 º/m
e.
Small craft pilots tend to avoid the zones you just identified. Why would
this be the case?
These zones of the atmosphere have quick changes in wind speed and direction and can cause
these small planes to easily lose control.
Conclusion Question 1 (10 points):
Give 1 experimental or physical source of error for this lab and explain how it
could be reduced in a subsequent experiment
In the lab background, it says that the balloon is 30g and filled with 28g of
helium; the diameter of the balloon is 2/3 meter and from that, we assume it
rises 200m every 60 seconds. This is not a very specific measurement and it
is definitely possible that it was filled up either too much or too little, which
could have caused the balloon to rise at a faster or slower rate – both
causing the data to change. A way to reduce this would be to use a more
exact measurement for the diameter and a more precise way of actually
measuring it.
Conclusion Question 2 (10 points):
Give another experimental or physical source of error for this lab and explain
how it could be reduced in a subsequent experiment
Another possible error could be caused while moving the theodolite. One
person was assigned to track down the balloon while moving the scope on
the theodolite. It is entirely possible that while doing this, the theodolite
either moved away from true north or was knocked off balance causing the
balancing bubbles to be off, and thus the measurements to be skewed. A
way to reduce this could be to somehow stake the theodolite into the ground
or use some mechanism to keep it sturdier. Another way to at least reduce
the chance of this a little bit would be to place the theodolite on flat concrete
rather than bumpy grass.
Conclusion Question 3 (5 points):
The pibal balloon system is relatively inexpensive and easy to set up. The
limitation is that most people will lose track of the balloon by 15,000 feet
above the surface (in very optimal conditions).
Identify an area of research, recreation, or emergency response where cheap
low-level wind data would be valuable. How would this data be useful for
your chosen application?
The pibal balloon system could be applied to surfing as recreation. Wind
speeds and direction have a profound effect on many different aspects of
waves in the ocean, and this system is great at tracking those two things. A
higher wind speed can create bigger waves and offshore winds can produce
a better wave shape and longer intervals. The data from this system would
be useful in predicting the surf nearby if you didn’t have access to a surf
report.