Borbin the 🐱

It was in one of the first physics courses when the Prof used 22/7 in an equation as an approximation. Within a few steps, he calculated the result on the board, leaving us students astounding about this.

So simple, and yet so close:

22 / 7 = 3,1̅4̅2̅8̅5̅7

𝜋 = 3,14159265358

22 / 7 - 𝜋 ≈ 0,00126

The math department would not had approved this, but not us engineers!

Dragonfly made it savely to our garden in the middle of the hot weather.

Captains view

The amount of oil in final drive of the CJ750 / M72 motorcycle needs to be very precise, otherwise the oil is pushed out through the main seal. The reason for this is the small volume of the final drive gear. Even the slightest overfilling push out oil when the final drive gets warm during driving.
To solve the problem, I had installed a small cylindric chamber (1) to collect the oil. But there was still some oil splashed out of the breathing hole so I added another chamber on top of it (2).

With the recommended 0.15l oil, there was still some splashing. Especially driving uphill.
To get around this problem once and for all, I have made an even larger chamber (3) holding a fair amount of oil that allows a slight overfilling to ensure I have always enough oil.
The thread size is M14x1.5.

The CJ750 / M72 motorcycle does not have a neutral indicator. With a simple switch extension to the gearbox lever you can add a neutral indicator.

It starts with the switch plate that gets attached to the gearbox lever cut out from stainless steel.

Now with the parts fabricated, the switch holder gets attached to the gearbox.

You can use any ball switch for automotive purpose.

The neutral lamp is added to a simple mini dashboard, that also contains the high beam and oil lamp.

A regular expression defines a search pattern for strings. Replace is possible for text matches, but repeated substring replace in substrings is difficult. Especially if the text you want to replace is also outside of the initial substrings. Using cascaded match evaluators, this replacements gets very easy. For example to replace quoted text withing square brackets.
Please note the middle text "text" in this example which must not be replaced.

["text1"] "text" ["text2"]
Begin[Pre"1txet"Post]End "text" Begin[Pre"2txet"Post]End

First the text is splitted by the primary substrings defined by square brackets.
Then these substrings are recomposed according to the business logic. In this example the string gets enclosed in a "Pre" / "Post" and then reverted. The primary substrings gets enclosed in "Begin" / "End".

string inputText = "[\"text1\"] "text" [\"text2\"]";

string replacedText = Regex.Replace(
    inputText,
    $"(\\[)(.*?)(\\])",
    arrayMatch =>
    {
        return 
        $"Begin" +
        arrayMatch.Groups[1].Value +
        Regex.Replace(
            arrayMatch.Groups[2].Value,
            $"(\".*?\")+",
            contentMatch =>
            {
                return $"Pre" + new string(contentMatch.Groups[1].Value.Reverse().ToArray()) + "Post";
            }) +

        arrayMatch.Groups[3].Value +
        $"End";
    });

For Math126 we were talking about the Taylor series for

It looks difficult at first glance, but it is actually quite simple. Let me explain:

Using this as a base:

The derivative of this term is:

Which gets to:

Now move the nominator in the sum to get the solution:

Time for a small PowerShell script for the Taylor series:

function f([double]$x)
{
    [Math]::Pow($x/(1-$x), 2)
}

function T([double]$x, [int]$k)
{
    0..$k | % { $sum = 0 } { $sum += $_ * [Math]::Pow($x, $_ + 1) } { $sum }
}


[int]$K = 16

for([int]$xi = -5; $xi -le 5; $xi += 1)
{
    [double]$x = $xi / 10

    [PSCustomObject]@{
        'x' = $x
        'f(x)' = f $x
        'T(x)' = T $x $K
    }
}

   x                f(x)                T(x)
   -                ----                ----
-0,5   0,111111111111111   0,111068725585938
-0,4  0,0816326530612245  0,0816318326348185
-0,3  0,0532544378698225  0,0532544328723274
-0,2  0,0277777777777778  0,0277777777741005
-0,1 0,00826446280991736 0,00826446280991734
   0                   0                   0
 0,1  0,0123456790123457  0,0123456790123457
 0,2              0,0625  0,0624999999943475
 0,3   0,183673469387755   0,183673459741776
 0,4   0,444444444444445   0,444442421037629
 0,5                   1   0,999862670898438

The convergence is | x | < 1
T₁₅ is used for this test.

 k             Tk(x)
 -             -----
 0                 0
 1              0,25
 2               0,5
 3            0,6875
 4            0,8125
 5          0,890625
 6            0,9375
 7        0,96484375
 8        0,98046875
 9      0,9892578125
10       0,994140625
11    0,996826171875
12    0,998291015625
13  0,99908447265625
14     0,99951171875
15 0,999740600585938