Math Is Fun Forum

This post on Trig functions and CI scale.

Open a new window with the virtual slide rule, and just look at the mid-scales in the sliding bar. Above the line there's a "T" symbol and below it an "S" symbol. Check the right end you will find that the T scale ends at 45 and that S ends at 90. Yes, T represents Tangent function, by degree. And S represent Sine function, by degree.

To be precise,

T=Arctan(C/10)
S=Arcsin(C/10)

or C= 10Tan(T)
   C=10Sin(S)

So when C=10, T=45 and S=90
when C=1, T=arcsin0.1 and S=arcsin0.1, yes, the domain of arc funtions is incomplete. However there is a solution, which will be discussed later.

Sin(30)=0.5 10Sin(30)=5
Tan(30)=1.414/2=0.707 10Tan(30)=7.07, verify some values yourself.

Also you can note the scale value in red, backwards increasing- they are Cot and Cos.

The real problem comes when the given trig-function is not given by decimals, but fractions instead.

For example, Sinθ=3/5 and Tanθ=3/4. We don't have to transfer them to decimals.
Using the simple fraction calculation discussed earlier, we can get 4/5*10 by
C     5    10(right 1)
D     3     3/5*10
However this is an ineffective calculation because the S scale is a funtion of C instead of a function of D and the intermediate result 4/5*10 on D has little use.
Then we can change the roles C and D play.
C   3    3/5*10 
D   5    10(right 1)
4/5*10 is got by 4*(10/5), however you can just think

It's simple algebra, having discussed in the previous page.
But it's of great convenience. Upper is 3, Lower is 5, then let 3 match 5 and 10 match the enlarged trig function value.
When C=10*3/5, S=Arcsin(3/5).
Check the Arctan(3/4) both 37°.

You may encounter 9/16 problem
but 9/1.6=9/16*10
so
C 9/1.6  9
D   1     16

This way you can calculate many values, however some values are just too small.

It doesn't matter because there is "ST" as a compensation. ST means "small tangents"
It calculates the arctangents on [0.01,0.1] So ST=Arctan(C/100)

The angles for ST just compensate T, [0.6, 5.8]U[5.7,45]
How about you have Sines of small value? almost equal to tangents of the same angle. So you may pretend to calculate arcsin by calculating arctan. In addition some slide rules provide sqrt(1-x[sup]2[/sup]) etc, by which you may get an accurate result after shifting sine value to tangent value.

Next, the CI thing and DI thing.
CI=10/C so you can inversions immediately from CI scale and C scale. (DI=10/D)
Another privilage is
CI[sub]1[/sub]D[sub]1[/sub]=CI[sub]2[/sub]D[sub]2[/sub]
This is from
D[sub]1[/sub]/C[sub]1[/sub]= D[sub]2[/sub]/C[sub]2[/sub]
eg.
CI    1   3
D     9   3
or
C      1   3
DI     9   3

You may find it simplier to operate, but I find backward scale and distant scales hard. They also has too large too small problems as CD scales. Almost the same problem
Now calculate 2*7 by both ways and you will discover the off-scale is the same severe.

If you mean only Standard maths is worth your time, yes I am not "qualified".

Uh, don't forget this
simulated slide rule

Clicking while pressing "Shift" will open a new window with the new page, rather than changing to the new page.

Sorry, but this post is an exercise post. Because I feel it necessary for you, a fan of the slide rule, to master the existing C and D scales before you forget the knowledge. So an exercise is necessary. And in this worksheet carefully designed by me, problems appeared earlier may have clues for later ones. So just finish it sequentially and you will master the whole of C and D scales.

I Simple Multiplication
2×2
2×3
2×4
2×1
1.2×4.5
1.2×4.5×1.7
1.8×1.05×3.2
1.04³×20 (suppose you have deposited 20 bucks in the bank for 3 years and the interest rate is 4%)

II Simple fractions
2×4/3
1×4/3
4/3
1/3×4
1/5×6
4/3×9/7
4/3×7/9
2×6/10
10/3×5/7

III Large results (use division by 10 wisely)
12×45
12×45×17
2×6
2×9
7/2×9
7/2×9/2

IV Small results (use multiplication by 10 wisely)
0.12×0.0045
1/3×5/7
1/3×4/7
1/3×2/7
1/5×1/9

Yes, the perpendicular distance is the shortest.
But there are pratical problems.
Your feet is hard to adjust to the very correct angle at the beginning.
And you may hit the ball to early, the ball is too tough.
Also after each hitting you are supposed to go back to the original level, otherwise you will get closer and closer to the net compulsively. This means perpendicular moving bears long road back.

tell me

Haha!!!

By induction!!!

Okay, mathematica6 do presume 0.999...=1. So the result from some software or a calculator cannot substantiate 0.999...=1. That's it. But can a software substantiate 0.999...<>1?

Softwares are made by humans, thus they only reflect the coders' respective recognition. So let sofware be software, debate be debate.

Is the all dead of heat proof enough for global warming? Sorry, but it is tooooooooo late.

Eight big planets.

Note some slide rules mark 10 on C and D as "1", at the right end.
It is for convience, telling you that it can be treated as "1" regardless of decimal point.
Indeed, when you match 10 with any number, it means the number divided by 10.

Tips for "off scale" problem of fraction multiplication.

2*7 is off scale, to solve it, we calculate 2*7/10

How about 2*22/3?
Too large.
So 1/10 is needed for calculation.
we just calculate one tenth of it.

you can do 2*2.2/3
or 2*22/10*(1/3) etc.

How about 2*2/9?
This time too small, we calculate ten times of it instead
2*2*(10/9) etc

And there is also a way to begin with a fraction. But this fraction should be larger than 1.

For example, 3/4*9/5
inverse its sequence to let a fraction larger than 1 sit the first
9/5*3/4
OK, now think 9/5 as 1*9/5, how do you do?
match 5 on C to 1(left 1) on D, and 9 on C points out the result on D. Move the cursor to 9 on C, and the
cursor now record exact value of 9/5. Now just multiply this value by 3/4.

Here is a point. The cursor can play as a physical storage for one number, but this number is usuallly
on D scale.

Just multiply 1.4*2.4*2.9
You will notice you don't have to read 1.4*2.4 out or check it, the cursor helps you.
Now swap the roles of C and D.
match 1.4 on C to 1 on D(slide leftwards), the value on C above 2.4 on D is the value of 1.4*2.4
Now what?
If you want to multiply 2.9 then you have to match 1.4*2.4 on C to 1 on D and this time the cursor should
move leftwards. The cursor has to follow the C scale, the sliding part. So it's a huge disadvantage
to get the result on C.

Therefore, let's derive a principle:
Usually get the result on D, so that the cursor can stay on static part of the rule and store the result.
By this principle you can do successive multiplications and divisions such as 1.4*2.4*2.9 etc.

There is exception however, in occasions that the next calculation is on the sliding part of the rule.
Just check the middle bar and the functions on them.
For example, the Log function is on the sliding part.
Find log(5/3)
If we get 5/3 on C it's convenient to read the corresponding log on Log scale.
So this time 1 on C-3 on D, and 5 on D-the result 5/3 on C, move the cursor to the result 5/3 and log(5/3) is
on log scale.
Or,
match 5 on C - 3 on D,   1 on D - 5/3 on C. It means 5 divided by 3 in conventional ways.
But look, 5-3, doesn't it look like
5
3?

Yes, (5/3):1=5:3
or
5/3  =  5
1         3

This means C[sub]1[/sub]/D[sub]1[/sub]=C[sub]2[/sub]/D[sub]2[/sub]
which can be easily derived by C[sub]1[/sub]/C[sub]2[/sub]= D[sub]1[/sub]/D[sub]2[/sub]

Back to the topic, the slide rule.

From Post 14, we know every distance on a logarithm scale represents a ratio between the two values on each end.

Now look at the slide rule above.
Measure with your eyes the distances on D scale between 1 and 2, between 2 and 4, between 4 and 8-are they the same?
Correct, the same, the reason for this has been proved in Post 14.

Now another trick:
Measure with your eyes the distances on D scale between 1 and 1.5, between 2 and 3, between 4 and 6, between 6 and 9.
Are they the same?

The same, too. This result verify the second sentence in this post again. And, we know a new method to do multiplication.
We don't have to move the index on C above the a and read the number on D under C to get 1.5a.
Instead, we can move 2 on C above a, and read 3/2a out right under 3 on C.
That's of great importance because sometimes we need to calculate 4/3a, etc. 1.33 is not accurate, whereas 3 and 4 are.

Sim Pickette's N-600ES LL Speed Rule
Click above to do some multiplication with a fraction your self.

From the same property we get another clue: the ratio? the ratios?
Yes, between two different numbers we can get two ratios, reciprocal to each other, one larger than 1, one less than 1.
Back to the previous example. Knowing how to evaluate 2/3a?
Simple match 3 on C to a on D, and read the value on D right under 2 on C.
Ratios are less than 1 when distance is from right to left, or more accurately, displacement from right to left.

By this method, can you calculate 2 times 6/10?
Yes, this is the only way to calculate 2 times 6 when you have only C and D.
Just try 2 times 6 in conventional way. You will find 6 on C matches blank or no scale on D.
Actually conventional method inhibits 2 times any number larger than 5 less than 10.
Only by fraction multiplication can you calculte those forbidden numbers.

The last for this post
can you calculate 6/5?
Note: 6/5= 6*(1/5)
can you calculate 6/7?
Note: 6*(10/7)=10(6/7)

The acceleration is along the slope, at a=gCosθ
The total distance is s=h/Cosθ

So the time for travel is

t=√(2s/a)= √[2h/(Cosθ g Cosθ) = Secθ √(2h/g)

The longer the distance, the larger the angle θ, and the larger the secθ, finally the longer the time.