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
New contributor
$endgroup$
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
New contributor
$endgroup$
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
4 hours ago
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
New contributor
$endgroup$
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
New contributor
New contributor
edited 3 hours ago
Aaron Stevens
9,95931741
9,95931741
New contributor
asked 4 hours ago
MichaelFoxMichaelFox
62
62
New contributor
New contributor
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
4 hours ago
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
4 hours ago
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
4 hours ago
$begingroup$
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
4 hours ago
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 $7times10^{-6} rm m$
$endgroup$
$begingroup$
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
3 hours ago
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.
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "151"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
MichaelFox is a new contributor. Be nice, and check out our Code of Conduct.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f456934%2fwhat-is-the-point-of-generalizing-a-more-specific-result-to-an-order-of-magnitud%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
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 $7times10^{-6} rm m$
$endgroup$
$begingroup$
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
3 hours ago
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 $7times10^{-6} rm m$
$endgroup$
$begingroup$
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
3 hours ago
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 $7times10^{-6} rm m$
$endgroup$
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 3 hours ago
Aaron StevensAaron Stevens
9,95931741
9,95931741
$begingroup$
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
3 hours ago
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
3 hours ago
$begingroup$
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
3 hours ago
$begingroup$
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
3 hours ago
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.
$endgroup$
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.
$endgroup$
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.
$endgroup$
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 3 hours ago
ChemomechanicsChemomechanics
4,79631023
4,79631023
add a comment |
add a comment |
MichaelFox is a new contributor. Be nice, and check out our Code of Conduct.
MichaelFox is a new contributor. Be nice, and check out our Code of Conduct.
MichaelFox is a new contributor. Be nice, and check out our Code of Conduct.
MichaelFox is a new contributor. Be nice, and check out our Code of Conduct.
Thanks for contributing an answer to Physics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f456934%2fwhat-is-the-point-of-generalizing-a-more-specific-result-to-an-order-of-magnitud%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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
4 hours ago