Mixing
What is the point of generalizing a more specific result to an order of magnitude?
$begingroup$
In my textbook, an example wants me to find an estimate of the number of cells in a human brain.
It gives the volume of the brain as $8 times 10^{-3} rm m^3$ which it then estimates further as $1 times 10^{-2} rm m^3$. It follows the same process for the volume of a cell, which it ultimately estimates as $1 times 10^{-15} rm m^3$. It then divides these two quantities to get $1 times 10^{13} rm {cells}$.
I'm not sure if this example is simply illustrative of orders if magnitude or what, but what is the point of using the order of magnitude estimate for the volume of the brain and the volume of the cell as opposed to just using the original results to find the number of cells? Isn't it being needlessly less accurate?
The book also seems to imply that it is commonplace to do this. I understand the how but not the why. Thanks.
estimation volume order-of-magnitude
edited Jan 26 at 21:17
Aaron Stevens
13.5k42250
asked Jan 26 at 20:56
MichaelFoxMichaelFox
112
$endgroup$
add a comment |
$begingroup$
In my textbook, an example wants me to find an estimate of the number of cells in a human brain.
It gives the volume of the brain as $8 times 10^{-3} rm m^3$ which it then estimates further as $1 times 10^{-2} rm m^3$. It follows the same process for the volume of a cell, which it ultimately estimates as $1 times 10^{-15} rm m^3$. It then divides these two quantities to get $1 times 10^{13} rm {cells}$.
I'm not sure if this example is simply illustrative of orders if magnitude or what, but what is the point of using the order of magnitude estimate for the volume of the brain and the volume of the cell as opposed to just using the original results to find the number of cells? Isn't it being needlessly less accurate?
The book also seems to imply that it is commonplace to do this. I understand the how but not the why. Thanks.
estimation volume order-of-magnitude
edited Jan 26 at 21:17
Aaron Stevens
13.5k42250
asked Jan 26 at 20:56
MichaelFoxMichaelFox
112
$endgroup$
1
$begingroup$
Well, you try keeping track of more significant figures when you're trying to do this on the back of a napkin. It's not better, it's just what people might do to save a little time if they're only after an estimate.
$endgroup$
– knzhou
Jan 26 at 21:01
add a comment |
$begingroup$
In my textbook, an example wants me to find an estimate of the number of cells in a human brain.
It gives the volume of the brain as $8 times 10^{-3} rm m^3$ which it then estimates further as $1 times 10^{-2} rm m^3$. It follows the same process for the volume of a cell, which it ultimately estimates as $1 times 10^{-15} rm m^3$. It then divides these two quantities to get $1 times 10^{13} rm {cells}$.
I'm not sure if this example is simply illustrative of orders if magnitude or what, but what is the point of using the order of magnitude estimate for the volume of the brain and the volume of the cell as opposed to just using the original results to find the number of cells? Isn't it being needlessly less accurate?
The book also seems to imply that it is commonplace to do this. I understand the how but not the why. Thanks.
estimation volume order-of-magnitude
edited Jan 26 at 21:17
Aaron Stevens
13.5k42250
asked Jan 26 at 20:56
MichaelFoxMichaelFox
112
$endgroup$
In my textbook, an example wants me to find an estimate of the number of cells in a human brain.
It gives the volume of the brain as $8 times 10^{-3} rm m^3$ which it then estimates further as $1 times 10^{-2} rm m^3$. It follows the same process for the volume of a cell, which it ultimately estimates as $1 times 10^{-15} rm m^3$. It then divides these two quantities to get $1 times 10^{13} rm {cells}$.
I'm not sure if this example is simply illustrative of orders if magnitude or what, but what is the point of using the order of magnitude estimate for the volume of the brain and the volume of the cell as opposed to just using the original results to find the number of cells? Isn't it being needlessly less accurate?
The book also seems to imply that it is commonplace to do this. I understand the how but not the why. Thanks.
estimation volume order-of-magnitude
estimation volume order-of-magnitude
edited Jan 26 at 21:17
Aaron Stevens
13.5k42250
asked Jan 26 at 20:56
MichaelFoxMichaelFox
112
edited Jan 26 at 21:17
Aaron Stevens
13.5k42250
asked Jan 26 at 20:56
MichaelFoxMichaelFox
112
edited Jan 26 at 21:17
Aaron Stevens
13.5k42250
edited Jan 26 at 21:17
Aaron Stevens
13.5k42250
edited Jan 26 at 21:17
Aaron Stevens
13.5k42250
13.5k42250
asked Jan 26 at 20:56
MichaelFoxMichaelFox
112
asked Jan 26 at 20:56
MichaelFoxMichaelFox
112
asked Jan 26 at 20:56
MichaelFoxMichaelFox
112
112
1
$begingroup$
Well, you try keeping track of more significant figures when you're trying to do this on the back of a napkin. It's not better, it's just what people might do to save a little time if they're only after an estimate.
$endgroup$
– knzhou
Jan 26 at 21:01
add a comment |
1
$begingroup$
Well, you try keeping track of more significant figures when you're trying to do this on the back of a napkin. It's not better, it's just what people might do to save a little time if they're only after an estimate.
$endgroup$
– knzhou
Jan 26 at 21:01
1
1
$begingroup$
Well, you try keeping track of more significant figures when you're trying to do this on the back of a napkin. It's not better, it's just what people might do to save a little time if they're only after an estimate.
$endgroup$
– knzhou
Jan 26 at 21:01
$begingroup$
Well, you try keeping track of more significant figures when you're trying to do this on the back of a napkin. It's not better, it's just what people might do to save a little time if they're only after an estimate.
$endgroup$
– knzhou
Jan 26 at 21:01
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
First, the question in the book is specifically focused on estimation. So it seems like the main reason is to serve as an example of how to do order of magnitude estimates.
Second, for this specific problem order of magnitude estimates are really all you can do. Not all brains have the same volume, and not all cells have the same volume. In addition, cells are constantly dividing/dying. There is no answer to "how many cells make up a human brain" because of this. The best you can do is make an estimate based on orders of magnitude.
And third, to tie this back to physics, calculations like these are very important. If you want to know how plausible an experiment or hypothesis is, you can do a quick order of magnitude estimate to see what the size of certain numbers are in your system. For example, in the biophysics lab I am in we were looking at using lasers to activate certain genes in fruit flies. We used order of magnitude estimates to determine how much energy was being delivered to the flies over a given time period based on an estimate of the laser spot size and frequency.
It is always important to know the size of the numbers involved in the problems you are trying to tackle. I would even argue that they are more important than specific numbers. Specific numbers are only good for, well, specific things, but orders of magnitude estimates help you get a grasp on the scales of the important parameters of your system where it's not important, for example, to distinguish between $5times10^{-6} rm m$ and $7times10^{-6} rm m$
answered Jan 26 at 21:11
Aaron StevensAaron Stevens
13.5k42250
$endgroup$
$begingroup$
Thank you for your comment. I understand now that it is definitely only illustrative. What confused me was that they got the pre-order of magnitude result from estimates, so it was an estimate of an estimate, with the understanding that the pre-order of magnitude estimate was based on approximates dimensions according to the text. So it seemed pointless but it was obviously just an example in retrospect. Thanks. Physics is not my major and while I enjoy layman physics, I am jusy trying to survive the course :p
$endgroup$
– MichaelFox
Jan 26 at 21:21
$begingroup$
@MichaelFox No problem. Please considering upvoting any answer that is helpful and selecting one answer as the accepted answer to help future readers know what was the most helpful.
$endgroup$
– Aaron Stevens
Jan 27 at 2:28
add a comment |
$begingroup$
An immediate advantage is that the calculation can be done in your head. Quickly, what's
$$frac{(8times 10^{13})(3times 10^{-3})(2times 10^{5})}{(2times 10^{7})(5times 10^{8})}+3times 10^{-2}?$$
How about
$$frac{(10^{14})(10^{-3})(10^{5})}{(10^{7})(10^{9})}+ 10^{-2}?$$
In the second example, I can add the exponents and get 0 (meaning $10^0=1$) and see that $10^{-2}$ is negligible in comparison. The actual answer is 4.83, which is within an order of magnitude of the estimate achieved in a few seconds by mental calculation.
answered Jan 26 at 21:23
ChemomechanicsChemomechanics
5,23631124
$endgroup$
add a comment |
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2 Answers
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$begingroup$
First, the question in the book is specifically focused on estimation. So it seems like the main reason is to serve as an example of how to do order of magnitude estimates.
Second, for this specific problem order of magnitude estimates are really all you can do. Not all brains have the same volume, and not all cells have the same volume. In addition, cells are constantly dividing/dying. There is no answer to "how many cells make up a human brain" because of this. The best you can do is make an estimate based on orders of magnitude.
And third, to tie this back to physics, calculations like these are very important. If you want to know how plausible an experiment or hypothesis is, you can do a quick order of magnitude estimate to see what the size of certain numbers are in your system. For example, in the biophysics lab I am in we were looking at using lasers to activate certain genes in fruit flies. We used order of magnitude estimates to determine how much energy was being delivered to the flies over a given time period based on an estimate of the laser spot size and frequency.
It is always important to know the size of the numbers involved in the problems you are trying to tackle. I would even argue that they are more important than specific numbers. Specific numbers are only good for, well, specific things, but orders of magnitude estimates help you get a grasp on the scales of the important parameters of your system where it's not important, for example, to distinguish between $5times10^{-6} rm m$ and $7times10^{-6} rm m$
answered Jan 26 at 21:11
Aaron StevensAaron Stevens
13.5k42250
$endgroup$
$begingroup$
Thank you for your comment. I understand now that it is definitely only illustrative. What confused me was that they got the pre-order of magnitude result from estimates, so it was an estimate of an estimate, with the understanding that the pre-order of magnitude estimate was based on approximates dimensions according to the text. So it seemed pointless but it was obviously just an example in retrospect. Thanks. Physics is not my major and while I enjoy layman physics, I am jusy trying to survive the course :p
$endgroup$
– MichaelFox
Jan 26 at 21:21
$begingroup$
@MichaelFox No problem. Please considering upvoting any answer that is helpful and selecting one answer as the accepted answer to help future readers know what was the most helpful.
$endgroup$
– Aaron Stevens
Jan 27 at 2:28
add a comment |
$begingroup$
First, the question in the book is specifically focused on estimation. So it seems like the main reason is to serve as an example of how to do order of magnitude estimates.
Second, for this specific problem order of magnitude estimates are really all you can do. Not all brains have the same volume, and not all cells have the same volume. In addition, cells are constantly dividing/dying. There is no answer to "how many cells make up a human brain" because of this. The best you can do is make an estimate based on orders of magnitude.
And third, to tie this back to physics, calculations like these are very important. If you want to know how plausible an experiment or hypothesis is, you can do a quick order of magnitude estimate to see what the size of certain numbers are in your system. For example, in the biophysics lab I am in we were looking at using lasers to activate certain genes in fruit flies. We used order of magnitude estimates to determine how much energy was being delivered to the flies over a given time period based on an estimate of the laser spot size and frequency.
It is always important to know the size of the numbers involved in the problems you are trying to tackle. I would even argue that they are more important than specific numbers. Specific numbers are only good for, well, specific things, but orders of magnitude estimates help you get a grasp on the scales of the important parameters of your system where it's not important, for example, to distinguish between $5times10^{-6} rm m$ and $7times10^{-6} rm m$
answered Jan 26 at 21:11
Aaron StevensAaron Stevens
13.5k42250
$endgroup$
$begingroup$
Thank you for your comment. I understand now that it is definitely only illustrative. What confused me was that they got the pre-order of magnitude result from estimates, so it was an estimate of an estimate, with the understanding that the pre-order of magnitude estimate was based on approximates dimensions according to the text. So it seemed pointless but it was obviously just an example in retrospect. Thanks. Physics is not my major and while I enjoy layman physics, I am jusy trying to survive the course :p
$endgroup$
– MichaelFox
Jan 26 at 21:21
$begingroup$
@MichaelFox No problem. Please considering upvoting any answer that is helpful and selecting one answer as the accepted answer to help future readers know what was the most helpful.
$endgroup$
– Aaron Stevens
Jan 27 at 2:28
add a comment |
$begingroup$
First, the question in the book is specifically focused on estimation. So it seems like the main reason is to serve as an example of how to do order of magnitude estimates.
Second, for this specific problem order of magnitude estimates are really all you can do. Not all brains have the same volume, and not all cells have the same volume. In addition, cells are constantly dividing/dying. There is no answer to "how many cells make up a human brain" because of this. The best you can do is make an estimate based on orders of magnitude.
And third, to tie this back to physics, calculations like these are very important. If you want to know how plausible an experiment or hypothesis is, you can do a quick order of magnitude estimate to see what the size of certain numbers are in your system. For example, in the biophysics lab I am in we were looking at using lasers to activate certain genes in fruit flies. We used order of magnitude estimates to determine how much energy was being delivered to the flies over a given time period based on an estimate of the laser spot size and frequency.
It is always important to know the size of the numbers involved in the problems you are trying to tackle. I would even argue that they are more important than specific numbers. Specific numbers are only good for, well, specific things, but orders of magnitude estimates help you get a grasp on the scales of the important parameters of your system where it's not important, for example, to distinguish between $5times10^{-6} rm m$ and $7times10^{-6} rm m$
answered Jan 26 at 21:11
Aaron StevensAaron Stevens
13.5k42250
$endgroup$
First, the question in the book is specifically focused on estimation. So it seems like the main reason is to serve as an example of how to do order of magnitude estimates.
Second, for this specific problem order of magnitude estimates are really all you can do. Not all brains have the same volume, and not all cells have the same volume. In addition, cells are constantly dividing/dying. There is no answer to "how many cells make up a human brain" because of this. The best you can do is make an estimate based on orders of magnitude.
And third, to tie this back to physics, calculations like these are very important. If you want to know how plausible an experiment or hypothesis is, you can do a quick order of magnitude estimate to see what the size of certain numbers are in your system. For example, in the biophysics lab I am in we were looking at using lasers to activate certain genes in fruit flies. We used order of magnitude estimates to determine how much energy was being delivered to the flies over a given time period based on an estimate of the laser spot size and frequency.
It is always important to know the size of the numbers involved in the problems you are trying to tackle. I would even argue that they are more important than specific numbers. Specific numbers are only good for, well, specific things, but orders of magnitude estimates help you get a grasp on the scales of the important parameters of your system where it's not important, for example, to distinguish between $5times10^{-6} rm m$ and $7times10^{-6} rm m$
answered Jan 26 at 21:11
Aaron StevensAaron Stevens
13.5k42250
answered Jan 26 at 21:11
Aaron StevensAaron Stevens
13.5k42250
answered Jan 26 at 21:11
Aaron StevensAaron Stevens
13.5k42250
answered Jan 26 at 21:11
Aaron StevensAaron Stevens
13.5k42250
13.5k42250
$begingroup$
Thank you for your comment. I understand now that it is definitely only illustrative. What confused me was that they got the pre-order of magnitude result from estimates, so it was an estimate of an estimate, with the understanding that the pre-order of magnitude estimate was based on approximates dimensions according to the text. So it seemed pointless but it was obviously just an example in retrospect. Thanks. Physics is not my major and while I enjoy layman physics, I am jusy trying to survive the course :p
$endgroup$
– MichaelFox
Jan 26 at 21:21
$begingroup$
@MichaelFox No problem. Please considering upvoting any answer that is helpful and selecting one answer as the accepted answer to help future readers know what was the most helpful.
$endgroup$
– Aaron Stevens
Jan 27 at 2:28
add a comment |
$begingroup$
Thank you for your comment. I understand now that it is definitely only illustrative. What confused me was that they got the pre-order of magnitude result from estimates, so it was an estimate of an estimate, with the understanding that the pre-order of magnitude estimate was based on approximates dimensions according to the text. So it seemed pointless but it was obviously just an example in retrospect. Thanks. Physics is not my major and while I enjoy layman physics, I am jusy trying to survive the course :p
$endgroup$
– MichaelFox
Jan 26 at 21:21
$begingroup$
@MichaelFox No problem. Please considering upvoting any answer that is helpful and selecting one answer as the accepted answer to help future readers know what was the most helpful.
$endgroup$
– Aaron Stevens
Jan 27 at 2:28
$begingroup$
Thank you for your comment. I understand now that it is definitely only illustrative. What confused me was that they got the pre-order of magnitude result from estimates, so it was an estimate of an estimate, with the understanding that the pre-order of magnitude estimate was based on approximates dimensions according to the text. So it seemed pointless but it was obviously just an example in retrospect. Thanks. Physics is not my major and while I enjoy layman physics, I am jusy trying to survive the course :p
$endgroup$
– MichaelFox
Jan 26 at 21:21
$begingroup$
Thank you for your comment. I understand now that it is definitely only illustrative. What confused me was that they got the pre-order of magnitude result from estimates, so it was an estimate of an estimate, with the understanding that the pre-order of magnitude estimate was based on approximates dimensions according to the text. So it seemed pointless but it was obviously just an example in retrospect. Thanks. Physics is not my major and while I enjoy layman physics, I am jusy trying to survive the course :p
$endgroup$
– MichaelFox
Jan 26 at 21:21
$begingroup$
@MichaelFox No problem. Please considering upvoting any answer that is helpful and selecting one answer as the accepted answer to help future readers know what was the most helpful.
$endgroup$
– Aaron Stevens
Jan 27 at 2:28
$begingroup$
@MichaelFox No problem. Please considering upvoting any answer that is helpful and selecting one answer as the accepted answer to help future readers know what was the most helpful.
$endgroup$
– Aaron Stevens
Jan 27 at 2:28
add a comment |
$begingroup$
An immediate advantage is that the calculation can be done in your head. Quickly, what's
$$frac{(8times 10^{13})(3times 10^{-3})(2times 10^{5})}{(2times 10^{7})(5times 10^{8})}+3times 10^{-2}?$$
How about
$$frac{(10^{14})(10^{-3})(10^{5})}{(10^{7})(10^{9})}+ 10^{-2}?$$
In the second example, I can add the exponents and get 0 (meaning $10^0=1$) and see that $10^{-2}$ is negligible in comparison. The actual answer is 4.83, which is within an order of magnitude of the estimate achieved in a few seconds by mental calculation.
answered Jan 26 at 21:23
ChemomechanicsChemomechanics
5,23631124
$endgroup$
add a comment |
$begingroup$
An immediate advantage is that the calculation can be done in your head. Quickly, what's
$$frac{(8times 10^{13})(3times 10^{-3})(2times 10^{5})}{(2times 10^{7})(5times 10^{8})}+3times 10^{-2}?$$
How about
$$frac{(10^{14})(10^{-3})(10^{5})}{(10^{7})(10^{9})}+ 10^{-2}?$$
In the second example, I can add the exponents and get 0 (meaning $10^0=1$) and see that $10^{-2}$ is negligible in comparison. The actual answer is 4.83, which is within an order of magnitude of the estimate achieved in a few seconds by mental calculation.
answered Jan 26 at 21:23
ChemomechanicsChemomechanics
5,23631124
$endgroup$
add a comment |
$begingroup$
An immediate advantage is that the calculation can be done in your head. Quickly, what's
$$frac{(8times 10^{13})(3times 10^{-3})(2times 10^{5})}{(2times 10^{7})(5times 10^{8})}+3times 10^{-2}?$$
How about
$$frac{(10^{14})(10^{-3})(10^{5})}{(10^{7})(10^{9})}+ 10^{-2}?$$
In the second example, I can add the exponents and get 0 (meaning $10^0=1$) and see that $10^{-2}$ is negligible in comparison. The actual answer is 4.83, which is within an order of magnitude of the estimate achieved in a few seconds by mental calculation.
answered Jan 26 at 21:23
ChemomechanicsChemomechanics
5,23631124
$endgroup$
An immediate advantage is that the calculation can be done in your head. Quickly, what's
$$frac{(8times 10^{13})(3times 10^{-3})(2times 10^{5})}{(2times 10^{7})(5times 10^{8})}+3times 10^{-2}?$$
How about
$$frac{(10^{14})(10^{-3})(10^{5})}{(10^{7})(10^{9})}+ 10^{-2}?$$
In the second example, I can add the exponents and get 0 (meaning $10^0=1$) and see that $10^{-2}$ is negligible in comparison. The actual answer is 4.83, which is within an order of magnitude of the estimate achieved in a few seconds by mental calculation.
answered Jan 26 at 21:23
ChemomechanicsChemomechanics
5,23631124
answered Jan 26 at 21:23
ChemomechanicsChemomechanics
5,23631124
answered Jan 26 at 21:23
ChemomechanicsChemomechanics
5,23631124
answered Jan 26 at 21:23
ChemomechanicsChemomechanics
5,23631124
5,23631124
add a comment |
add a comment |
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$begingroup$
Well, you try keeping track of more significant figures when you're trying to do this on the back of a napkin. It's not better, it's just what people might do to save a little time if they're only after an estimate.
$endgroup$
– knzhou
Jan 26 at 21:01