• Support PF! Buy your school textbooks, materials and every day products Here!

Mental calculation and its importance in physics

  • Physics
  • Thread starter nilfound
  • Start date
  • #1
6
0

Main Question or Discussion Point

I have been struggling with being able to mentally calculate math problems for quite a long time. I am worried I will fall behind if I don't master this skill. How important is it to have a firm grasp on this skill if I am choosing to pursue a career in physics?
 

Answers and Replies

  • #2
11,823
5,445
Can you be more specific?

Are you trying to estimate orders of magnitude?

Or more accurately sums and products of number in scientific notation?



I used to worry about this when I was in grade school during 4th grade when we learned our multiplication tables. I kept looking for some kind of magical solution and found the Trachtenberg system. It didn't help because I feared learning it would put me at odds with what the teacher taught.

What truly amazed me was the utter simplicity of the system (little memorization) but at the same time you had to keep a running sum in your head and mine was a tad small at the time. Actually most people can hold about 7 digits at a time in their head so it takes some hard training to accomplish this.

https://en.wikipedia.org/wiki/Trachtenberg_system

https://human-memory.net/short-term-working-memory/

There is also the Chisanbop speed math used in Korea where your fingers act like an abacus. Kids trained in it can do math quite fast.

https://en.wikipedia.org/wiki/Chisanbop

Other kinds of speed math are presented as various parlor tricks/recipes in books by Bill Handley, Danica and others:

https://www.amazon.com/dp/0471467316/?tag=pfamazon01-20&tag=pfamazon01-20

https://www.amazon.com/gp/product/0452289491/?tag=pfamazon01-20&tag=pfamazon01-20

There's also the humourous but wrong Abbott and Costello math routine:


and lastly this modern classic of adding 2+2 = 22 where the teacher wins out in the end:

 
  • Like
Likes JC2000, Nick-stg and sysprog
  • #3
6
0
I am referring to calculating sums and products in problems in physics, ranging from kinematics to electromagnetism and onwards, where you have to calculate quantities with the given derived equations. For instance, if on an exam you need to calculate the paths of motion of an object, how to quickly mentally sum and calculate the kinematic equations for the object with the proper units/conversions.
 
  • #4
11,823
5,445
Are you allowed a calculator?

Is this primarily about dimensional analysis?

A concrete example would be good where you show what is tripping you up.

https://www.khanacademy.org/math/al...on/v/dimensional-analysis-units-algebraically

Another thing that may be driving you crazy is how physics is taught using specific equations for each case in kinematics. This means you must memorize what equation applies under what conditions. As you get more into physics this approach is scrapped in favor of Lagrangian and Hamiltonian theory and you use your understanding of Calculus and Differential Equations and physics basic principles to derive those equations.
 
Last edited:
  • Like
Likes JC2000 and berkeman
  • #5
1,459
874
I have been struggling with being able to mentally calculate math problems for quite a long time. I am worried I will fall behind if I don't master this skill. How important is it to have a firm grasp on this skill if I am choosing to pursue a career in physics?
I have always found it useful to carry around enough stuff in my head to produce a number (say +/-20%) without running to book to look up values for constants. Sometimes I don't know enough theory to carry it off but within the labyrinthine passages I can usually find enough arcane values to get out the number I need without interrupting my walk. This has nothing to do with facility with arithmetic, but one does need to know how it all fits together.
One of the great joys of my life was to know (just a little bit) the great scientist, steadfastly moral man and one of my heroes: Hans A. Bethe. The stories of his ability to arrive at a largely correct answer from the concentrated power of thought are legend. May I suggest some biograpical research on your part. Nobody did it better
 
  • Like
Likes sysprog
  • #7
1,459
874
Yes ...T here was one less well-received joke where he was so fed up with the "numerology" that seemed to be spreading among scientists (maybe Eddington....don't quote me) that Bethe "proved" that the value of absolute zero (in Celcius!!) had to be (2(1/α)-1) where α is the fine structure constant. Some folks were not amused I guess. Sounds pretty funny to me.
 
  • Like
Likes sysprog
  • #8
196
204
I don't think you have to be a powerful calculator to be successful in science. But there is fun in challenging yourself to compute things in your head and searching for tricks.

In Surely You're Joking, Mr. Feynman!, he mentions an incident with Hans Bethe where Hans shows Feynman how to compute squares near 50 by simply thinking of it as ##(50-n)=2500-100n+n^2##. Thinking of squares or cubes as binomial expansions can simplify calculations.

I was trying to find a way to calculate square roots mentally, and found that if you want to know the square root of some number n and know the two nearest squares that bound it, a and b, you can get a decent approximation with ##\sqrt n = a + \left( \frac {n-a^2} {b^2-a^2} \right)##. If you know that last fraction, you can calculate it in your head. I don't know or really think there's anything mathematically right to it, it just seemed to work for the bunch of square roots I was calculating while working one day.

For example, for 215, you know ##14^2=196## and ##15^2=225##. Then ##\sqrt 215 = 14 + \left( \frac {215-196} {225-196} \right) = 14 + \frac {19} {29}##. You know ##\frac {19} {30}## must be somewhere close to 0.66, a little less, since ##\frac {2} {3} = 0.6666...##. We know ##\frac {1} {30} = 0.03333....##, so subtract 0.03333... from 0.6666... to get 0.63333... . So then ##\frac {19} {29}## must be a little bigger than 0.63333..., say closer to 0.65 or 0.66. We'll use the halfway point and say 0.655. So then our approximation is ##\sqrt 215 = 14.655.## A calculator shows ##\sqrt 215 = 14.6628783##. So not bad for just knowing a little about fractions.
 
  • Like
Likes sysprog
  • #9
symbolipoint
Homework Helper
Education Advisor
Gold Member
5,963
1,075
I have been struggling with being able to mentally calculate math problems for quite a long time. I am worried I will fall behind if I don't master this skill. How important is it to have a firm grasp on this skill if I am choosing to pursue a career in physics?
NOT IMPORTANT, but nice if you can do them.
 
  • Like
Likes sysprog
  • #10
Dr_Nate
Science Advisor
251
138
I learned that I was losing silly points on exams for doing steps in my head. There was a noticeable increase in my grades once I started writing out more steps. By doing this on the exam, I downloaded the mental effort required to store this stuff into my short term memory onto the paper. My suggestion is not to waste your time on mental calculations beyond using it to estimate quantities.

If you still want to learn how to do more mental math, you could learn the abacus and visualize the results like some Japanese students are able to.
 
  • Like
Likes FactChecker and sysprog
  • #11
Dr_Nate
Science Advisor
251
138
I am referring to calculating sums and products in problems in physics, ranging from kinematics to electromagnetism and onwards, where you have to calculate quantities with the given derived equations. For instance, if on an exam you need to calculate the paths of motion of an object, how to quickly mentally sum and calculate the kinematic equations for the object with the proper units/conversions.
I can almost guarantee you that your shortcoming is not mental math. If you were to upload a few examples of your attempts at solutions, I think we could find your true weaknesses.
 
  • Like
Likes sysprog
  • #12
1,471
785
I like doing mathematics in my head: {me + 'nice pretty of-sufficient-age girl person' = fun}. :cool:
 
  • #13
140
14
Sorry to jump into the convo!

Yes ...T here was one less well-received joke where he was so fed up with the "numerology" that seemed to be spreading among scientists (maybe Eddington....don't quote me) that Bethe "proved" that the value of absolute zero (in Celcius!!) had to be (2(1/α)-1) where α is the fine structure constant. Some folks were not amused I guess. Sounds pretty funny to me.
Can you provide me with a reference to this formula? Does it actually work? Thanks!
 
  • #14
FactChecker
Science Advisor
Gold Member
5,591
2,073
In my experience, the most common problem that people have is that they try to do too much in their heads, especially when multiple steps are involved. The most common advice I give is to use the Keep It Simple Stupid (KISS) method.

Some people can do amazing things in their heads. Good for them. If you are like that, good. If you are not like that, do not pretend that you are.
 
  • Like
Likes symbolipoint
  • #15
1,459
874
The original is in German (and I don't know any other ready source):
Beck, G, Bethe, H. and Riezler, W. (1931) Remarks on Quantum Theory of the Absolute Zero Temperature. Die Naturwissenschaften, 2, 38-39. (In German)

It is written in the best scientific prose, was accepted for publication and it is completely a spoof.

I remember seeing an English translation, on paper, more than 40 yrs ago. That's my best reference, sorry!
 
  • Haha
Likes Replusz
  • #16
140
14
Thanks! :D
 
  • Like
Likes hutchphd
  • #17
33,641
5,307
I have always found it useful to carry around enough stuff in my head to produce a number (say +/-20%) without running to book to look up values for constants.
Or running to a calculator.
I believe that being able to come up with a "ball park" or "seat of the pants" estimate is a very useful skill. If your +/-20% estimate disagrees wildly with a calculated result, one possibility is that the calculated result was based on a mistyped number or incorrect formula.
For example, suppose I want to reseal my asphalt driveway (which I actually do want to do), and I estimate its area as 12' wide by 28' long, and that each 5 gal. container covers 96 sq. ft. A rough estimate of the area is 360 sq. ft., so 4 containers of sealer ought to do the job.

If I then use a calculator to get the area and come up with 216 sq. ft. (due to mistyping one of the dimensions of the driveway), I would think that I could get by with two containers of the sealer. Without having done an initial estimate, I wouldn't have any way of knowing that my calculation was off.

This is something of a contrived example, but my point is that doing a rough estimate first is useful in showing that a calculated result is reasonable or not.
NOT IMPORTANT, but nice if you can do them.
I disagree that being able to do a mental calculation is not important.
 
  • Like
Likes hutchphd
  • #18
1,459
874
As a former occasional teacher of intro physics for engineers, there was no quicker route to my professorial wrath than to report a result to seven significant figures that was off by 3 orders of magnitude. Real world stuff.
 
  • Like
Likes symbolipoint and Vanadium 50
  • #19
FactChecker
Science Advisor
Gold Member
5,591
2,073
It certainly is very beneficial to be able to do mental "sanity checks" on a calculation. It is also good to be able to do preliminary planning with a mental understanding of orders of magnitudes.
 
  • #20
symbolipoint
Homework Helper
Education Advisor
Gold Member
5,963
1,075
I disagree that being able to do a mental calculation is not important.
How so? The simpler, the more likely more people can do them; the less simpler, the less likely more people can do them. Why is mental calculation important? What kind?
 
  • #21
FactChecker
Science Advisor
Gold Member
5,591
2,073
As a former occasional teacher of intro physics for engineers, there was no quicker route to my professorial wrath than to report a result to seven significant figures that was off by 3 orders of magnitude. Real world stuff.
The mistake is probably the same no matter how much accuracy is given, and there is nothing wrong with trying to be accurate.
 
  • #22
33,641
5,307
How so? The simpler, the more likely more people can do them; the less simpler, the less likely more people can do them. Why is mental calculation important? What kind?
My point about mental calculations is not to do the same problem mentally, but rather to round the numbers to mentally do a simpler problem. In my example about the asphalt sealer, instead of doing 12 X 28 in my head, I simply multiplied 12 and 30 in my head to get 360 sq. ft. Then, instead of dividing by 98 sq. ft coverage of each sealer bucket, I mentally divided 360 by 100, which when rounded, comes out to 4 buckets of sealer.
 
  • #23
symbolipoint
Homework Helper
Education Advisor
Gold Member
5,963
1,075
My point about mental calculations is not to do the same problem mentally, but rather to round the numbers to mentally do a simpler problem. In my example about the asphalt sealer, instead of doing 12 X 28 in my head, I simply multiplied 12 and 30 in my head to get 360 sq. ft. Then, instead of dividing by 98 sq. ft coverage of each sealer bucket, I mentally divided 360 by 100, which when rounded, comes out to 4 buckets of sealer.
Fine. That much is understood and commonly taken as valuable. I thought you were trying to support something else.
 
  • #24
196
204
You don't need to be fast in computing large numbers like 12 X 28. But being able to simplify calculations, make approximations, and handle common squares, cubes, square roots can be really handy. I call it having good number sense, and feel that any physicist or physics major should have some of it.

Being able to recognize that (12)(28)=(2)(2)(2)(2)(3)(7)=(21)(2)(2)(2)(2)=(42)(2)(2)(2)=(84)(2)(2)=(168)(2)=336 is an example. Or (12)(28)= 12(8) + 12(20)= 96 + 240= 336. This can really simplify long calculations.

I saw a youtube channel that wanted to simplify ##\sqrt {3 - 2\sqrt 2}##. After struggling to simplify it I decided to tackle it with an approximation without the use of a calculator. Write it ##\sqrt{1+2-2\sqrt 2}##. Using the approximation ##\sqrt 2 = 1.4142##, we find ##1+2(1-\sqrt2)=1-0.8284=1-\frac {82} {100} - \frac {84} {10,000} = \frac {429} {2500}##. We approximate ##\sqrt 429 = 20 + \frac {429-400} {441-400} = 20 + \frac {29} {41}##. Again, approximate ##\frac {29} {41} = \frac 7 {10}##, and we find that ##20 + \frac {7} {10} = \frac {207} {10}##. Thus, we can write the whole square root as ##\sqrt \frac {429} {2500} = \frac {207} {500} = \frac {1} {100} ( \frac {200} {5} + \frac {7} {5} ) = \frac {41.4} {100} = 0.414##. From this I could deduce that the answer is ##\sqrt 2 -1##. It's not rigorous of course, but it shows how it is good to know you're way around numbers.

Even handier in physics is knowing your derivatives and integrals, cold. Being able to run through integrals such as ##\int\sqrt x dx## or ##\int \frac {1} {a^2 - u^2} du## without any difficulty can simplify your life when studying or working through difficult problems. Of course, check your work. It's always better to plod along with pencil and paper. But if you're practiced to the point you know you can do it in your head, you sure as well can do it on paper too.
 
  • #25
988
125
phdSimpleMathVhardMath.gif
This is a thing
 
  • Like
Likes FactChecker and DrClaude

Related Threads on Mental calculation and its importance in physics

Replies
7
Views
2K
Replies
2
Views
968
  • Last Post
Replies
4
Views
3K
Replies
3
Views
758
Replies
1
Views
961
  • Last Post
Replies
1
Views
3K
Replies
16
Views
4K
Replies
14
Views
1K
Top