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Posted

Hey!!! I'm a second year student, just starting an Atmospheric Science course/lab duo. My Instructor for our lectures is amazing, but the on for our lab is God awful! She reads our manual to us and doesn't explain why things "are", even when prodded by student questioning. Anyways, she recited a formula, told us it was in our lab manual, which- it isn't!- and she didn't explain where the numbers came from. I'm going to post a picture of the question and data below to perhaps communicate better, the problem. THANK YOU FOR HELPING!!!

:confused:

Here is the data,

DSCF8157.jpg

Here it is specifically for the example problem:

DSCF8159.jpg

This is the equation given to us,

DSCF8166.jpg

I don’t understand where the three comes from. Is it 3km less 0km equals 3, is it the number of units to get from 0 to 3km

I just need to know how/why she arrived at 3 being the height so that I can do the following equations.

DSCF8164.jpg

Because the way I see it, it has to be 5 for the first answer (the height, not the answer to the entire problem), but the second could be 5 or ten, and the next, 10 or twenty, etc.

 

Please help

Shann

Posted

I take that your actual question is where the 3 km came from? That is indeed the difference 3 km - 0 km = 3 km, the upper boundary minus the lower one (which in that very case happens to be 0 km).

 

The reason for that is the keyword "average". I don't know what level of math to expect from you but I'll save myself time to give an intiutive explanation - if you're not sufficiently familiar with calculus, just ask for further clarifications.

 

Here we go:

Taking an average <V> of N values [math]V_{1\dots N}[/math] is done by [math] \left< V \right> = \frac{1}{N} \sum_{i=1}^N V_i [/math]. If you have a continuous distribution f(x), its average over some interval [math] \left[x_0, x_1 \right] [/math] is [math]\frac{1}{x_1-x_0} \int _{x_0} ^{x_1} dx f(x)[/math].

 

In your case, you have a pressure field with the P(h) depending on height h. The pressure gradient, g(h), is the derivative of the pressure with respect to height, i.e. dP(h)/dh. Hence, the average pressure gradient between height [math]h_0[/math] and [math]h_1[/math] according to above is

[math] \left< g \right> = \frac{1}{h_1 - h_0} \int_{h_0}^{h_1} dh \ g(h) = \frac{1}{h_1 - h_0} \int_{h_0}^{h_1} dh \ \frac{dP(h)}{dh} = \frac{1}{h_1 - h_0} \left( P(h_1) - P(h_0) \right) [/math].

Posted

oh my! i'm a mature student, we didn't have this math when i went to school, alot of that looked like a foreign language to me. Still, thank you for your help!:eek:

Posted

A note ahead: Saying what level of education you are in can be very non-saying to people not living in <whatever country you live in> unless it is an internationally-comparable level (e.g. a PhD course is roughly comparable all over the world), age might be a better indicator for lower levels.

 

Anyways, a very simple example that hopefully helps you:

I'm going to drive to Göttingen tomorrow. The distance is 250 km. Suppose I leave home at 10:00 and arrive at 13:00.

- What will be my average velocity (including units) and why? What would be my average velocity if I left at 11:00 and arrived at 14:00?

- Can you see how this question, specifically the times and velocity, relate to your problem, specifically to the heights and the pressure gradient?

- Bonus question: Does an average velocity of whatever mean I travelled with that average velocity all the time?

Posted

Simply speaking, this looks like delta P over the distance in question, very reasonable even if you never seen the formula. Part of your problem may be that 701.. mb is under 3 km, not 2 km

 

by the way, how do you write summation/integral symbols ?

Posted
by the way, how do you write summation/integral symbols ?

It's a TeX (a typesetting language) plugin. TeX mode is enabled by the tag [ math] (no space) and closed by [ /math]. Most of the commands are quite straighforward, e.g. the sum symbol is \sum, a greek capital gamma is \Gamma. You can click on the images of the mathematical term to get a pop-up window that shows the code. A simple tutorial should be lying around somewhere in these forums, too.

Posted

If you want to know the pressure at any altitude that is NOT in your table... the lazy and not-100%-correct-option is:

 

Put all numbers in a spreadsheet (such as Excel), plot, add trendline, and click on the trendline to show the formula. Then use the formula to calculate the new numbers.

 

It's not very scientific, but it's quick and dirty and will probably give you a better result than when you mess up some tricky math... :D

Posted
It's not very scientific, but it's quick and dirty and will probably give you a better result than when you mess up some tricky math... :D

Better results?

 

Listen, I don't know much on the subject but if there's something that can give you better result than math, then I'd honestly want to see it!

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