Manual transmission: device, types, features. Riddles related to physical phenomena in nature

Which I found on the DataGenetics website. Please send all errors in this article in private messages.

In this problem, there are 100 prisoners in the prison, each numbered from 1 to 100. The jailer decides to give the prisoners a chance to be released, he tells them the conditions of the test, and if all the prisoners pass the test, then they will be released. If at least one of them fails the test, then all the prisoners will die.

A task

The jailer goes to secret room and prepares 100 boxes with lids. On each box, he marks numbers from 1 to 100. Then he brings 100 paper tablets, according to the number of prisoners, and numbers these tablets from 1 to 100. After that, he shuffles 100 tablets and places one tablet in each box, closing the lid . The prisoners do not see how the jailer performs all these actions.

The competition begins, the jailer takes each prisoner one by one to the room with boxes and tells the prisoners that they must find a box that will contain a plate with the number of the prisoner. The prisoners are trying to find the plate with their number by opening the boxes. Each is allowed to open up to 50 boxes; if each of the prisoners finds his number, then the prisoners will be released, if at least one of them does not find his number in 50 attempts, then all the prisoners will die.

In order for prisoners to be released, ALL prisoners must pass the test successfully.

So what is the chance that the prisoners will be pardoned?

  • After the prisoner opens the box and checks the plate, it is placed back in the box and the lid is closed again;
  • Places of the plates cannot be changed;
  • Prisoners cannot leave clues to each other or interact with each other in any way once the trial has begun;
  • Prisoners are allowed to discuss strategy before the trial begins.

What is the best strategy for prisoners?

Additional question:

If a friend of the prisoners (not a participant in the test) will be able to enter the secret room before the start of the test, examine all the tablets in all boxes and (optionally, but not required) swap two tablets from two boxes (in this case, the comrade will not have the opportunity to both inform the prisoners about the result of his actions), then what strategy should he take to increase the chances of the prisoners to escape?

Solution improbable?

At first glance, this task seems almost hopeless. It seems that the chance for each of the prisoners to find their tablet is microscopically small. In addition, prisoners cannot exchange information with each other during the trial.

The odds of one prisoner are 50:50. There are 100 boxes in total and he can open up to 50 boxes looking for his sign. If he opens the boxes at random and opens half of all the boxes, he will find his tablet in the open half of the boxes, or his tablet will remain in the closed 50 boxes. His chances of success are ½.

Let's take two prisoners. If both choose boxes at random, the chances for each of them will be ½, and for two ½x½=¼.
(for two prisoners, success will be in one case out of four).

For three prisoners, the odds are ½ × ½ × ½ = ⅛.

For 100 prisoners, the odds are: ½ × ½ × … ½ × ½ (multiply 100 times).


This equals

Pr ≈ 0.000000000000000000000000000008

So it's a very small chance. In this scenario, most likely, all the prisoners will be dead.

Incredible answer

If each prisoner opens the boxes at random, they are unlikely to pass the test. There is a strategy where prisoners can expect to be successful more than 30% of the time. This is a stunningly incredible result (if you haven't heard of this math problem previously).

More than 30% for all 100 prisoners! Yes, this is even more than the chances for two prisoners, provided that they open the boxes at random. But how is this possible?

It is clear that one for each prisoner, the chances cannot be higher than 50% (after all, there is no way for communication between prisoners). But do not forget that the information is stored in the location of the plates inside the boxes. No one shuffles the tablets between visits to the room by individual prisoners, so we can use that information.

Solution

First, I'll tell you the solution, then I'll explain why it works.

The strategy is extremely easy. The first of the prisoners opens the box with the number that is written on his clothes. For example, prisoner number 78 opens the box with the number 78. If he finds his number on the plate inside the box, that's great! If not, he looks at the number on the plate in "his" box and then opens the next box with that number. Having opened the second box, he looks at the number of the tablet inside this box and opens the third box with this number. Then we simply transfer this strategy to the remaining boxes. For clarity, look at the picture:


Eventually, the prisoner will either find his number or reach the 50 box limit. At first glance, this looks pointless compared to simply choosing a box at random (and for one individual prisoner it does), but since all 100 prisoners will use the same set of boxes, it makes sense.

The beauty of this math problem- not only to know the result, but also to understand why this strategy works.

So why does the strategy work?

Each box contains one plate - and this plate is unique. This means that the plate is in a box with the same number, or it points to a different box. Since all plates are unique, there is only one plate for each box pointing to it (and only one way to get to that box).


If you think about it, the boxes form a closed circular chain. One box can be part of only one chain, since inside the box there is only one pointer to the next one and, accordingly, in the previous box there is only one pointer to this box (programmers can see the analogy with linked lists).

If the box does not point to itself (the box number is equal to the plate number in it), then it will be in the chain. Some chains may consist of two boxes, some are longer.


Since all prisoners start with a box with the same number on their clothes, they are by definition placed on the chain that contains their nameplate (there is only one nameplate that points to this box).

Exploring the boxes along this chain in a circle, they are guaranteed to eventually find their sign.

The only question remains whether they will find their tablet in 50 moves.


Chain length

In order for all prisoners to pass the test, the maximum chain length must be less than 50 boxes. If the chain is longer than 50 boxes, prisoners with numbers from those chains will fail the test - and all prisoners will be dead.

If the maximum length of the longest chain is less than 50 boxes, then all prisoners will pass the test!

Think about this for a second. It turns out that there can only be one chain that is longer than 50 boxes in any layout of the plates (we have only 100 boxes, so if one chain is longer than 50, then the rest will be shorter than 50 in total).


Long chain hand odds

Once you've convinced yourself that the maximum chain length must be less than or equal to 50 to succeed, and there can only be one long chain in any set, we can calculate the probability of passing the challenge:

Some more math

So what do we need to figure out the probability of a long chain?

For a chain of length l, the probability that the boxes will be outside this chain is:

There is (l-1) in this collection of numbers! ways to arrange the signs.

The remaining signs can be located (100-l)! ways (do not forget that the length of the chain does not exceed 50).

Given this, the number of permutations that contain the chain exact length l: (>50)


It turns out that there are 100(!) ways to arrange the plates, so that the probability of existence of a chain of length l is equal to 1/l. By the way, this result does not depend on the number of boxes.

As we already know, there can only be one case in which there is a chain with a length > 50, so the probability of success is calculated by this formula:

Result

31.18% - the probability that the size of the longest chain will be less than 50 and each of the prisoners will be able to find their tablet, given the limit of 50 attempts.

The probability that all prisoners will find their plates and pass the test is 31.18%

Below is a graph showing the probabilities (on the y-axis) for all chains of length l (on the x-axis). Red means all "failures" (given curve here is just a 1/l plot). Green color means "success" (the calculation is a little more complicated for this part of the graph, since there are several ways to determine maximum length <50). Общая вероятность складывается из зеленых столбцов в 31.18% шанс на спасение.


Harmonic number (this part of the article is for geeks)

In mathematics, the nth harmonic number is the sum of the reciprocals of the first n consecutive numbers of the natural series.


Let's calculate the limit if instead of 100a boxes we have an arbitrary a large number of boxes (let's assume we have 2n boxes in total).


The Euler-Mascheroni constant is a constant defined as the limit of the difference between the partial sum of a harmonic series and the natural logarithm of a number.

As the number of prisoners increases, if the overseer allows prisoners to open half of all the boxes, then the chance of salvation tends to 30.685%

(If you made a decision in which the prisoners randomly guess the boxes, then as the number of prisoners increases, the probability of being saved tends to zero!)

Additional question

Anyone else remember the extra question? What can our helpful comrade do to increase our chances of survival?

Now we already know the solution, so the strategy here is simple: he must examine all the signs and find the longest chain of boxes. If the longest chain is less than 50, then he does not need to change the tablets at all, or change them so that the longest chain does not become longer than 50. However, if he finds a chain longer than 50 boxes, all he has to do is swap the contents of two boxes from that chain to break that chain into two shorter chains.

As a result of this strategy, there will be no long chains and all prisoners are guaranteed to find their sign and salvation. So, by swapping two signs, we reduce the probability of salvation to 100%!


Remember that the volume of a cuboid (or an ordinary box) is equal to the product of its length, width and height. If your box is rectangular or square, all you need to know is its length, width, and height. To obtain the volume, it is necessary to multiply the results of measurements. The calculation formula in abbreviated form is often presented as follows: V \u003d L x W x H.
Problem example: "If the length of the box is 10 cm, the width is 4 cm, and the height is 5 cm, then what is its volume?"
V = L x W x H
V=10cmx4cmx5cm
V \u003d 200 cm 3
The "height" of the box may be referred to as the "depth". For example, a task could contain the following information: "The length of the box is 10 cm, the width is 4 cm, and the depth is 5 cm."

2
Measure the length of the box. If you look at the box from above, it will appear before your eyes in the form of a rectangle. The length of the box will be the longest side of this rectangle. Record the result of the measurement of this side as the value of the "length" parameter.
Be sure to use the same unit of measure when taking measurements. If you measured one side in centimeters, then the other sides must also be measured in centimeters.

3
Measure the width of the box. The width of the box will be represented by the other, shorter, side of the rectangle visible from above. If you visually connect the sides of the box measured in length and width, then they will appear in the form of the letter "G". Record the value of the last measurement as "width".
Width is always the shorter side of the box.

4
Measure the height of the box. This is the last parameter you haven't measured yet. It represents the distance from the top edge of the box to the bottom. Record the value of this measurement as "height".
Depending on which side you put the box on, the specific sides you designate as "length", "width" or "height" may be different. However, it doesn't matter, you just need the measurements from three different sides.

5
Multiply the results of the three measurements together. As already mentioned, the formula for calculating volume is as follows: V = Length x Width x Height; therefore, to obtain the volume, you simply need to multiply all three sides. Be sure to specify the units you used in the calculation so that you do not forget what exactly the values ​​\u200b\u200bmean.

6
When indicating volume units, do not forget to indicate the third power " 3 ". The calculated volume has a numerical expression, but without the correct indication of the units of measurement, your calculations will be meaningless. To correctly reflect volume units, they must be specified in the cube. For example, if all sides were measured in centimeters, then the volume units would be specified as "cm3".
An example of a problem: "If a box has a length of 2 m, a width of 1 m, and a height of 3 m, then what is its volume?"
V = L x W x H
V = 2m x 1m x 4m
V \u003d 8 m 3
Note: Specifying cubic units of volume allows you to understand how many such cubes can be placed inside the box. If we refer to the previous example, this means that eight cubic meters are placed in a box.

Calculation of the volume of boxes of other shapes

Determine the volume of the cylinder. The cylinder is a round tube with circles at both ends. To determine the volume of a cylinder, the formula is used: V = π x r 2 x h, where π = 3.14, r is the radius of the round side of the cylinder, and h is its height.
To determine the volume of a cone, or a pyramid with a round base, the same formula is used, but multiplied by 1/3. That is, the volume of the cone is calculated by the formula: V = 1/3 (π x r 2 x h)

2
Determine the volume of the pyramid. A pyramid is a figure that has a flat base and sides converging at the top to one point. To determine the volume of a pyramid, you need to take 1/3 of the product of the area of ​​\u200b\u200bits base and height. That is, the calculation formula is as follows: Volume of the pyramid = 1/3 (Base area x Height).
In most cases, pyramids have a square or rectangular base. In such a situation, the area of ​​the base is calculated by multiplying the length of the base by the width.

To determine the volume of a box of complex shapes, add the volumes of its individual parts. For example, you may need to measure the volume of a box shaped like the letter "L". In this case, the box will have more sides to measure. If you break this box into two parts, you can measure the volume of these two parts in the standard way, and then add the resulting values. In the case of the L-shaped box, the longer part can be considered as a separate long rectangular box, and the shorter part as a square (or almost square) box attached to it.
If your box has very complex shapes, then know that there are many ways to determine the volume of objects of any shape.

    Write a program that compares two integers entered from the keyboard. The program must indicate which number is greater, or, if the numbers are equal, display an appropriate message.

    Type two integers and press Enter.
    -> 34 67
    34 less than 67

    with Ada.Text_IO; use Ada.Text_IO; with Ada.Integer_Text_IO; use Ada.Integer_Text_IO; procedure main is A, B: Integer; begin Put_Line("Enter two integers on one line and press Enter."); put("-> "); Get(A); Get(B); --Enter 2 numbers if A > B then Put(Item =>; A, Width =>; 1); put("greater than"); Put(Item => B, Width => 1); elsif A< B then Put(Item =>A, Width => 1); put("less than"); Put(Item => B, Width => 1); else Put("Entered numbers are equal!"); end if; end main;

    Three integers are given. Find the largest of them (the program should output exactly one integer). The largest in this problem is understood as a number that is not less than any other.


  • -> 1 2 3
    Maximum of three numbers: 3
  • with Ada.Text_IO ; use Ada.Text_IO ; with Ada.Integer_Text_IO ; use Ada.Integer_Text_IO ; procedure main is A, B, C: Integer; max: Integer; begin Put_Line( "Enter three integers on one line and press Enter.") ; Put("-> "); Get(A); Get(B) ; Get(C); --Enter three integers max:=A; --default we consider that the number A is the maximum if B > max then --If B is greater than the maximum, then max:=B; --maximum number is B end if ; if C > max then --If C is greater than the maximum, then max:=C; --maximum number is C end if ; Put( "Maximum of three numbers:"& Integer"image(max) ); end main;

    with Ada.Text_IO; use Ada.Text_IO; with Ada.Integer_Text_IO; use Ada.Integer_Text_IO; procedure main is A, B, C: Integer; max: Integer; begin Put_Line("Enter three integers on one line and press Enter."); put("-> "); Get(A); Get(B); Get(C); --Enter three integers max:= A; --by default, we assume that the number A is the maximum if B > max then --If B is greater than the maximum, then max:= B; --maximum number is B end if; if C > max then --If C is greater than max, then max:= C; --maximum number is equal to C end if; Put("Maximum of three numbers:" & Integer"image(max)); end main;

    Given three natural numbers A, B, C. Determine if a triangle exists with these sides. If the triangle exists, print a message that a triangle with such sides exists, otherwise print that the triangle does not exist.

  • Type in the three sides of the triangle and press Enter.
    -> 3 4 5
    A triangle with sides 3, 4, 5 exists.

    • A triangle is three points that do not lie on the same line. A triangle only exists if the sum of any two of its sides is greater than the third.

  • with Ada.Integer_Text_IO ; use Ada.Integer_Text_IO ; --For reading integers with Ada.Text_IO ; use Ada.Text_IO ; --To display strings procedure main is a, b, c: Integer; begin Put_Line( "Enter three sides of a triangle and press Enter.Enter.") ; Put("-> "); Get(a); Get(b) ; Get(c); --Read the sides of the triangle if a + b > c and then b + c > a and then c + a > b then --Check all conditions in one line Put( "Triangle with Sides"& Integer"image(a) & "," & Integer"image(b) & "," & Integer"image(c) & " exists" ) ; else Put( "Triangle with Sides"& Integer"image(a) & "," & Integer"image(b) & "," & Integer"image(c) & " does not exist" ) ; end if ; end main;

    with Ada.Integer_Text_IO; use Ada.Integer_Text_IO; --To read integers with Ada.Text_IO; use Ada.Text_IO; --To print strings procedure main is a, b, c: Integer; begin Put_Line("Enter the three sides of the triangle and press Enter.Enter."); put("-> "); get(a); get(b); get(c); --Read triangle sides if a + b > c and then b + c > a and then c + a > b then --Check all conditions in one line Put("Triangle with sides" & Integer"image(a) & " ," & Integer"image(b) & "," & Integer"image(c) & " exists"); else Put("Triangle with sides" & Integer"image(a) & "," & Integer"image( b) & "," & Integer"image(c) & " does not exist"); end if; end main;

    Three integers are given. Determine how many of them match. The program should output one of the numbers: 3 (if all are the same), 2 (if two are the same), or 0 (if all are different).

  • Type three integers and press Enter.
    -> 1 2 3
    0
  • with Ada.Text_IO ; use Ada.Text_IO ; with Ada.Integer_Text_IO ; use Ada.Integer_Text_IO ; procedure Main is A, B, C: Integer; begin Put_Line( "Enter three integers and press Enter.") ; Put("-> "); Get(A); Get(B) ; Get(C); if A = B and then A = C then --If all three numbers match Put(Item => 3 , Width => 1 ) ; elsif A = B or A = C or B = C then --If two numbers match Put(Item => 2 , Width => 1 ) ; else --If a same numbers No Put(Item => 0 , Width => 1 ) ; end if ; endMain;

    with Ada.Text_IO; use Ada.Text_IO; with Ada.Integer_Text_IO; use Ada.Integer_Text_IO; procedure Main is A, B, C: Integer; begin Put_Line("Enter three integers and press Enter."); put("-> "); Get(A); Get(B); Get(C); if A = B and then A = C then --If all three numbers match Put(Item => 3, Width => 1); elsif A = B or A = C or B = C then --If two numbers match Put(Item => 2, Width => 1); else --If there are no identical numbers Put(Item => 0, Width => 1); end if; endMain;

    The chess rook moves horizontally or vertically. Given two various cells chessboard, determine if the rook can move from the first cell to the second in one move. The program receives as input four numbers from 1 to 8 each, specifying the column number and row number, first for the first cell, then for the second cell. The program should print "YES" if it is possible to get to the second from the first cell by the move of the rook, or "NO" otherwise.


  • 4 4
    5 5
    NO
  • ) ; Put() ; Get(A); Get(B) ; Put() ; Get(C); Get(D); if A = C or B = D then Put("YES" ) ; else Put("NO" ) ; end if ; endMain;

    with Ada.Integer_Text_IO; use Ada.Integer_Text_IO; with Ada.Text_IO; use Ada.Text_IO; procedure Main is subtype checkBoard is Integer range 1..8; A, B, C, D: checkBoard; begin Put_Line("Enter the column and row numbers for the two cells:"); Put("Enter the column and row numbers for the first cell and press: "); Get(A); Get(B); Put("Enter the column and row numbers for the second cell and press: "); Get(C); Get(D); if A = C or B = D then Put("YES"); else Put("NO"); end if; endMain;

    The chess king moves horizontally, vertically and diagonally, but only 1 square. Given two different cells on a chessboard, determine if the king can get from the first cell to the second in one move. The program receives as input four numbers from 1 to 8 each, specifying the column number and row number, first for the first cell, then for the second cell. The program should print "YES" if it is possible to get to the second from the first cell by the move of the king, or "NO" otherwise.

  • Enter the column and row numbers for the two cells:
    Enter the column and row numbers for the first cell and press: 4 4
    Enter the column and row numbers for the second cell and press: 5 5
    YES
  • "Enter the column and row numbers for the two cells:") ; Put( "Enter the column and row numbers for the first cell and press: ") ; Get(A); Get(B) ; Put( "Enter the column and row numbers for the second cell and press: ") ; Get(C); Get(D); if abs (A - C)<= 1 and then abs (B - D) <= 1 then -- the abs() command returns the absolute -- value (modulus) of number Put("YES"); else Put("NO" ) ; end if ; end main;

    with Ada.Integer_Text_IO; use Ada.Integer_Text_IO; with Ada.Text_IO; use Ada.Text_IO; procedure main is subtype checkBoard is Integer range 1..8; A, B, C, D: checkBoard; begin Put_Line("Enter the column and row numbers for the two cells:"); Put("Enter the column and row numbers for the first cell and press: "); Get(A); Get(B); Put("Enter the column and row numbers for the second cell and press: "); Get(C); Get(D); if abs(A - C)<= 1 and then abs(B - D) <= 1 then -- команда abs() возвращает абсолютное --значение (модуль) числа Put("ДА"); else Put("НЕТ"); end if; end main;

    The chess bishop moves diagonally. Given two different cells of a chessboard, determine if the bishop can get from the first cell to the second in one move. The program receives as input four numbers from 1 to 8 each, specifying the column number and row number, first for the first cell, then for the second cell. The program should output "YES" if it is possible to get to the second from the first cell by the bishop's move, or "No" otherwise.

  • Enter the column and row numbers for the two cells:
    Enter the column and row numbers for the first cell and press: 4 4
    Enter the column and row numbers for the second cell and press: 5 5
    YES
  • with Ada.Integer_Text_IO ; use Ada.Integer_Text_IO ; with Ada.Text_IO ; use Ada.Text_IO ; procedure main is subtype checkBoard is Integer range 1 ..8 ; A, B, C, D: checkBoard; begin Put_Line( "Enter the column and row numbers for the two cells:") ; Put( "Enter the column and row numbers for the first cell and press: ") ; Get(A); Get(B) ; Put( "Enter the column and row numbers for the second cell and press: ") ; Get(C); Get(D); if abs (a - c) = abs (b - d) then Put("YES" ) ; else Put("NO" ) ; end if ; end main;

    with Ada.Integer_Text_IO; use Ada.Integer_Text_IO; with Ada.Text_IO; use Ada.Text_IO; procedure main is subtype checkBoard is Integer range 1..8; A, B, C, D: checkBoard; begin Put_Line("Enter the column and row numbers for the two cells:"); Put("Enter the column and row numbers for the first cell and press: "); Get(A); Get(B); Put("Enter the column and row numbers for the second cell and press: "); Get(C); Get(D); if abs(a - c) = abs(b - d) then Put("YES"); else Put("NO"); end if; end main;

    The chess queen moves diagonally, horizontally or vertically. Given two different cells of a chessboard, determine if the queen can get from the first cell to the second in one move.

    Input data format:
    The program receives as input four numbers from 1 to 8 each, specifying the column number and row number, first for the first cell, then for the second cell.
    Output format:
    The program should output YES if it is possible to get from the first cell to the second by the queen's move, or NO otherwise.

    Example 1:
    Enter the column and row numbers for the two cells:
    Enter the column and row numbers for the first cell and press: 1 1
    Enter the column and row numbers for the second cell and press: 2 2
    YES

    Example 2:
    Enter the column and row numbers for the two cells:
    Enter the column and row numbers for the first cell and press: 1 1
    Enter the column and row numbers for the second cell and press: 2 3
    NO

  • with Ada.Integer_Text_IO ; use Ada.Integer_Text_IO ; with Ada.Text_IO ; use Ada.Text_IO ; procedure Main is subtype checkBoard is Integer range 1 ..8 ; A, B, C, D: checkBoard; begin Put_Line( "Enter the column and row numbers for the two cells:") ; Put( "Enter the column and row numbers for the first cell and press: ") ; Get(A); Get(B) ; Put( "Enter the column and row numbers for the second cell and press: ") ; Get(C); Get(D); if abs (A - C) = abs (B - D) or A = D or B = C then Put("YES" ) ; else Put("NO" ) ; end if ; endMain;

    with Ada.Integer_Text_IO; use Ada.Integer_Text_IO; with Ada.Text_IO; use Ada.Text_IO; procedure Main is subtype checkBoard is Integer range 1..8; A, B, C, D: checkBoard; begin Put_Line("Enter the column and row numbers for the two cells:"); Put("Enter the column and row numbers for the first cell and press: "); Get(A); Get(B); Put("Enter the column and row numbers for the second cell and press: "); Get(C); Get(D); if abs(A - C) = abs(B - D) or A = D or B = C then Put("YES"); else Put("NO"); end if; endMain;

    The chess horse moves in the letter “G” - two squares vertically in any direction and one square horizontally, or vice versa. Given two different cells of a chessboard, determine if the knight can get from the first cell to the second in one move. The program receives as input four numbers from 1 to 8 each, specifying the column number and row number, first for the first cell, then for the second cell. The program should output YES if the knight can move from the first cell to the second one, or NO otherwise.

    Example 1:
    Enter the column and row numbers for the two cells:
    Enter the column and row numbers for the first cell and press: 1 1
    Enter the column and row numbers for the second cell and press: 1 4
    NO

    Example 2:
    Enter the column and row numbers for the two cells:
    Enter the column and row numbers for the first cell and press: 1 1
    Enter the column and row numbers for the second cell and press: 8 8
    NO

  • with Ada.Text_IO ; use Ada.Text_IO ; with Ada.Integer_Text_IO ; use Ada.Integer_Text_IO ; procedure main is subtype checkBoard is Integer range 1 ..8 ; A, B, C, D: Integer; begin Put_Line( "Enter the column and row numbers for the two cells:") ; Put( "Enter the column and row numbers for the first cell and press: ") ; Get(A); Get(B) ; Put( "Enter the column and row numbers for the second cell and press: ") ; Get(C); Get(D); if abs (A - C) = 2 and then abs (B - D) = 1 then Put("YES" ) ; elsif abs (A - C) = 1 and then abs (B - D) = 2 then Put("YES" ) ; else Put("NO" ) ; end if ; end main;

    with Ada.Text_IO; use Ada.Text_IO; with Ada.Integer_Text_IO; use Ada.Integer_Text_IO; procedure main is subtype checkBoard is Integer range 1..8; A, B, C, D: Integer; begin Put_Line("Enter the column and row numbers for the two cells:"); Put("Enter the column and row numbers for the first cell and press: "); Get(A); Get(B); Put("Enter the column and row numbers for the second cell and press: "); Get(C); Get(D); if abs(A - C) = 2 and then abs(B - D) = 1 then Put("YES"); elsif abs(A - C) = 1 and then abs(B - D) = 2 then Put("YES"); else Put("NO"); end if; end main;

    The chocolate has the form of a rectangle divided into N×M slices. Chocolate can be broken once in a straight line into two parts. Determine if exactly K slices can be broken off from a chocolate bar in this way. The program receives three numbers as input: N, M, K. The program must output one of two words: "YES" or "No".

    Example 1:
    4
    2
    6
    YES

    Example 2:
    Number of slices horizontally: 2
    Number of slices vertically: 10
    How many slices to separate: 7
    NO

  • with Ada.Text_IO ; use Ada.Text_IO ; with Ada.Integer_Text_IO ; use Ada.Integer_Text_IO ; procedure Main is N, M, K: Integer; begin Put( "Number of slices horizontally:") ; Get(N) ; Put( "Number of slices vertically:") ; Get(M) ; Put( "How many slices to separate:") ; Get(K) ; if K > M * N then --If a chocolate bar is asked to break off more than the chocolate bar itself put("NO"); elsif K rem N = 0 and then K >= N then - Break off horizontally Put("YES"); elsif K rem M = 0 and then K >= M then - Break off vertically Put("YES"); else Put("NO" ) ; end if ; endMain;

    with Ada.Text_IO; use Ada.Text_IO; with Ada.Integer_Text_IO; use Ada.Integer_Text_IO; procedure Main is N, M, K: Integer; begin Put("Number of slices horizontally: "); Get(N); Put("Number of slices vertically: "); Get(M); Put("How many slices to separate: "); Get(K); if K > M * N then --If the chocolate bar is asked to break off more than the chocolate bar itself Put("NO"); elsif K rem N = 0 and then K >= N then -- Break off horizontally Put("YES"); elsif K rem M = 0 and then K >= M then -- Break off vertically Put("YES"); else Put("NO"); end if; endMain;

    Yasha swam in a pool measuring N × M meters and got tired. At this point, he found himself X meters from one of the long ledges (not necessarily the nearest one) and Y meters from one of the short ledges. What is the minimum distance Yasha must swim to get out of the pool onto the side? The program receives the numbers N, M, X, Y as input. The program should output the number of meters that Yasha needs to swim to the side.

  • Pool Width: 23
    Pool length: 52
    Distance from Yasha to the long side: 8
    Distance from Yasha to a short side: 43
    You need to swim at least to get out of the pool: 8

    • It is possible that to solve the problem you will need to swap 2 variables. This algorithm looks something like this:

    a, b, tmp: Integer; -- Declaring variables. Two main and one auxiliary a:= 3; --Initialization of variable a b:= 5; --Initialization of variable b --Algorithm itself: tmp:= a; --Now tmp = 3 and a = 3 a:= b; --Now a = 5 and b = 5; b:=tmp; --Now b = 3

  • with Ada.Text_IO ; use Ada.Text_IO ; with Ada.Integer_Text_IO ; use Ada.Integer_Text_IO ; procedure Main is N, M, X, Y: Integer; -- N - short edge, M - long edge: -- X - Distance to one of the long sides -- Y - Distance to one of the short sides Tmp: Integer; begin Put( "Pool Width: ") ; Get(N) ; Put( "Pool Length: ") ; Get(M) ; Put( "Distance from Yasha to a long side:") ; Get(X) ; Put( "Distance from Yasha to the short side:") ; Get(Y) ; if N > M then --If the sides are mixed up during input, then we change their places: Tmp:=M; --Save the length M to a temporary variable M:=N; --Assign the variable M a new value N:=Tmp; --Restore the length M in variable N end if ; Tmp:=X; --Assume the minimum distance is X if abs(N-X)< X then --If the distance to the second long edge is less than X, then Tmp:= N - X; --minimum distance equal to the distance to the second long side end if ; if Y< Tmp then --If the distance to the short edge is less than that found above -- minimum, then Tmp:=Y; --Minimum distance is Y end if ; if abs (M - Y)< Tmp then --If you swim closer to the second short side, then Tmp:= abs (M - Y) ; --minimum distance is equal to the distance to the second short ledge end if ; Put( "You need to swim at least to get out of the pool: ") ; Put(Item => Tmp, Width => 1 ) ; endMain;

    with Ada.Text_IO; use Ada.Text_IO; with Ada.Integer_Text_IO; use Ada.Integer_Text_IO; procedure Main is N, M, X, Y: Integer; -- N - short ledge, M - long ledge: -- X - Distance to one of the long ledges -- Y - Distance to one of the short ledges Tmp: Integer; begin Put("Pool Width: "); Get(N); Put("Pool length: "); Get(M); Put("Distance from Yasha to the long side: "); Get(X); Put("Distance from Yasha to the short edge: "); Get(Y); if N > M then --If the edges are mixed up during input, then we change their places: Tmp:= M; --Save the length M into a temporary variable M:= N; --Assign a new value to the variable M N:= Tmp; --Restore the length M in the variable N end if; Tmp:=X; --Assume the minimum distance is X if abs(N - X)< X then --Если до второго длинного бортика расстояние меньше X, то Tmp:= N - X; --минимальное расстояние равно расстоянию до второго длинного бортика end if; if Y < Tmp then --Если до короткого бортика расстояние меньше найденного выше --минимального, то Tmp:= Y; --Минимальное расстояние равно Y end if; if abs(M - Y) < Tmp then --Если до второго короткого бортика плыть ближе, то Tmp:= abs(M - Y); --минимальное расстояние равно расстоянию до второго короткого бортика end if; Put("Нужно проплыть минимум, чтобы выбраться из бассейна: "); Put(Item =>Tmp, Width => 1); endMain;

    The electronic clock shows the time in the h:mm:ss format (from 0:00:00 to 23:59:59), that is, first the number of hours is recorded, then the two-digit number of minutes is mandatory, then the two-digit number of seconds is mandatory. The number of minutes and seconds, if necessary, are padded to two-digit number zeros. N seconds have passed since the beginning of the day. Output what the clock shows. The input is a natural number N, not exceeding 10 7 (10000000). Output the answer to the problem.

    Input example 1:
    3602
    Output example 1:
    1:00:02

    Input example 2:
    129700
    Output example 2:
    12:01:40

  • with Ada.Long_Integer_Text_IO ; use Ada.Long_Integer_Text_IO ; with Ada.Text_IO ; use Ada.Text_IO ; procedure Main is subtype Sub_LI is Long_Integer range 1 ..10000000 ; N: Sub_LI; h, m, s: Long_Integer; begin Get(N) ; h:= N / 3600 ; - We get the clock. The remainder of the division is discarded N:= N - h * 3600 ; --Get remaining seconds (minus hours) if h > 24 then --Since the clock cannot show > 24, we put everything in a readable form h:= h rem 24 ; --Remainder of dividing by 24 will give the exact number of hours elsif h = 24 then h:= 0 ; end if ; m:= N / 60 ; --Get minutes s:= N rem 60 ; --Get seconds Put(Item => h, Width => 1 ) ; Put(":"); --Output hours and ":" if m< 10 then --If the number of minutes is less than 10, output a leading 0 Put(Item => 0 , Width => 1 ) ; end if ; Put(Item => m, Width => 1 ) ; Put(":"); --Output minutes and ":" if s< 10 then --If the number of seconds is less than 10, output a leading 0 Put(Item => 0 , Width => 1 ) ; end if ; Put(Item => s, Width => 1 ) ; --Output seconds end Main;

    with Ada.Long_Integer_Text_IO; use Ada.Long_Integer_Text_IO; with Ada.Text_IO; use Ada.Text_IO; procedure Main is subtype Sub_LI is Long_Integer range 1..10000000; N: Sub_LI; h, m, s: Long_Integer; begin Get(N); h:= N / 3600; - We get the clock. The remainder of the division is discarded N:= N - h * 3600; --Get remaining seconds (minus hours) if h > 24 then --Since clocks can't read > 24, put everything in readable form h:= h rem 24; --The remainder of dividing by 24 will give the exact number of hours elsif h = 24 then h:= 0; end if; m:= N / 60; --Get minutes s:= N rem 60; --Get seconds Put(Item => h, Width => 1); Put(":"); --Output hours and ":" if m< 10 then --Если количество минут меньше 10, выводим ведущий 0 Put(Item =>0, Width => 1); end if; Put(Item => m, Width => 1); Put(":"); --Output minutes and ":" if s< 10 then --Если количество секунд меньше 10, выводим ведущий 0 Put(Item =>0, Width => 1); end if; Put(Item => s, Width => 1); --Output seconds end Main;

  • Three numbers are given. Arrange them in ascending order.
  • Input example:
    1 2 1
    Sample output:
    1 1 2
  • with Ada.Integer_Text_IO ; use Ada.Integer_Text_IO ; with Ada.Text_IO ; use Ada.Text_IO ; procedure Main is A, B, C: Integer; min, mid, max: Integer; begin Get(A) ; Get(B) ; Get(C); --Are looking for minimum value min:=A; if B< min then min:= B; end if ; if C < min then min:= C; end if ; --Are looking for maximum value max:=A; if B > max then max:= B; end if ; if C > max then max:= C; end if ; --Looking for the average value mid:=A; if B > min and B< max then mid:= B; end if ; if C >min and C< max then mid:= C; end if ; Put(Item =>min, Width => 1 ) ; put(" "); Put(Item => mid, width => 1 ) ; put(" "); Put(Item => max, Width => 1 ) ; endMain;

    with Ada.Integer_Text_IO; use Ada.Integer_Text_IO; with Ada.Text_IO; use Ada.Text_IO; procedure Main is A, B, C: Integer; min, mid, max: Integer; begin Get(A); Get(B); Get(C); --Looking for the minimum value min:= A; if B< min then min:= B; end if; if C < min then min:= C; end if; --Ищем максимальное значение max:= A; if B >max then max:= B; end if; if C > max then max:= C; end if; --Looking for the average value mid:= A; if B > min and B< max then mid:= B; end if; if C >min and C< max then mid:= C; end if; Put(Item =>min, Width => 1); put(" "); Put(Item => mid, width => 1); put(" "); Put(Item => max, Width => 1); endMain;

    There are two boxes, the first is A1×B1×C1, the second is A2×B2×C2. Determine if one of these boxes can be placed inside the other, provided that the boxes can only be rotated 90 degrees around the edges. The program receives numbers A1, B1, C1, A2, B2, C2 as input. The program should output one of the following lines:
    - "Boxes are equal" if the boxes are the same,
    - "First box less than a second"if the first box can be placed in the second,
    - "First box more than a second" if the second box can be placed inside the first one.

    Example 1:
    First box dimensions: 1 2 3
    Second box dimensions: 3 2 1
    Boxes are equal

    Example 2:
    First box dimensions: 2 2 3
    Second box dimensions: 3 2 1
    The first box is larger than the second

  • with Ada.Integer_Text_IO ; use Ada.Integer_Text_IO ; with Ada.Text_IO ; use Ada.Text_IO ; procedure main is A1, B1, C1, A2, B2, C2: Integer; min, max, mid, tmp: Integer; begin Get(A1) ; Get(B1) ; Get(C1) ; Get(A2) ; Get(B2) ; Get(C2) ; --Bring the faces in line with the lengths A1 => A2, B1 => B2, C1 => C2: --A1 and A2 are the longest, C1 and C2 are the shortest -- "Spin" the first box: min:=A1; mid:=B1; max:=C1; if B1< min then mid:= min; min:= B1; end if ; if C1 < min then max:= min; min:= C1; end if ; if mid >max then tmp:= mid; mid:=max; max:=tmp; end if ; A1:=min; B1:=mid; C1:=max; -- "Spin" the second box: min:=A2; mid:=B2; max:=C2; if B2< min then mid:= min; min:= B2; end if ; if C2 < min then max:= min; min:= C2; end if ; if mid >max then tmp:= mid; mid:=max; max:=tmp; end if ; A2:=min; B2:=mid; C2:=max; --Checking the correspondence of boxes and displaying the result: if A1 = A2 and then B1 = B2 and then C1 = C2 then Put_Line("The boxes are equal" ) ; elsif A1 >= A2 and then B1 >= B2 and then C1 >= C2 then Put_Line( "The first box is larger than the second") ; Elsif A1<= A2 and then B1 <= B2 and then C1 <= C2 then Put_Line("The first box is smaller than the second") ; end if ; end main;

    with Ada.Integer_Text_IO; use Ada.Integer_Text_IO; with Ada.Text_IO; use Ada.Text_IO; procedure main is A1, B1, C1, A2, B2, C2: Integer; min, max, mid, tmp: Integer; begin Get(A1); Get(B1); Get(C1); Get(A2); Get(B2); Get(C2); --Bring the sides in line with the lengths A1 => A2, B1 => B2, C1 => C2: --A1 and A2 - the longest, C1 and C2 - the shortest -- "Twist" the first box: min:= A1 ; mid:=B1; max:=C1; if B1< min then mid:= min; min:= B1; end if; if C1 < min then max:= min; min:= C1; end if; if mid >max then tmp:= mid; mid:=max; max:=tmp; end if; A1:=min; B1:=mid; C1:=max; -- "Spin" the second box: min:= A2; mid:=B2; max:=C2; if B2< min then mid:= min; min:= B2; end if; if C2 < min then max:= min; min:= C2; end if; if mid >max then tmp:= mid; mid:=max; max:=tmp; end if; A2:=min; B2:=mid; C2:=max; --Check if the boxes match and output the result: if A1 = A2 and then B1 = B2 and then C1 = C2 then Put_Line("Boxes are equal"); elsif A1 >= A2 and then B1 >= B2 and then C1 >= C2 then Put_Line("First box is bigger than second"); Elsif A1<= A2 and then B1 <= B2 and then C1 <= C2 then Put_Line("Первая коробка меньше второй"); end if; end main;

    Write a program that calculates the cost of a long-distance telephone conversation (the price of one minute is determined by the distance to the city where the subscriber is located). The initial data for the program are the area code and the duration of the call. Below are the codes of some cities and the recommended screen view while the program is running:

  • Calculate the cost of a telephone conversation.
    Enter initial data:
    City code -> 423
    Duration (integer minutes) -> 3
    Vladivostok city
    Price per minute: 4 rubles.
    Call cost: 12 rubles.
  • with Ada.Integer_Text_IO ; use Ada.Integer_Text_IO ; with Ada.Text_IO ; use Ada.Text_IO ; procedure Main is Code, Len: Integer; begin Put_Line ("Calculation of the cost of a telephone conversation.") ; Put_Line ("Enter initial data:") ; Put ("City code -> ") ; Get ( Code ) ; Put ("Duration (integer minutes) -> ") ; Get ( len ) ; case Code is when 423 => Put_Line ("Vladivostok city") ; Put_Line ("Price of a minute: 4 rubles.") ; Put ("Call cost: ") ; Put ( Item => len * 4 , Width=> 1 ) ; Put_Line (" rub.") ; when 095 => Put_Line ("Moscow city") ; Put_Line ("Price of a minute: 2 rubles.") ; Put ("Call cost: ") ; Put ( Item => len * 2 , Width=> 1 ) ; Put_Line (" rub.") ; when 815 => Put_Line ("City: Murmansk") ; Put_Line ("Price of a minute: 3 rubles.") ; Put ("Call cost: ") ; Put ( Item => len * 3 , Width=> 1 ) ; Put_Line (" rub.") ; when 846 => Put_Line ("Samara city") ; Put_Line ("Price of a minute: 1 rub.") ; Put ("Call cost: ") ; Put ( Item => len, Width => 1 ) ; Put_Line (" rub.") ; when others=> Put ("There is no city with this code in the database! Try again.") ; end case; end main;

    with Ada.Integer_Text_IO; use Ada.Integer_Text_IO; with Ada.Text_IO; use Ada.Text_IO; procedure Main is Code, Len: Integer; begin Put_Line("Calculating the cost of a phone call."); Put_Line("Enter initial data:"); Put("City code -> "); get(Code); Put("Duration (integer minutes) -> "); get(len); case Code is when 423 => Put_Line("City: Vladivostok"); Put_Line("Price per minute: 4 rubles."); Put("Call cost: "); Put(Item => len * 4, Width => 1); Put_Line(" rub."); when 095 => Put_Line("City: Moscow"); Put_Line("Price per minute: 2 rubles."); Put("Call cost: "); Put(Item => len * 2, Width => 1); Put_Line(" rub."); when 815 => Put_Line("City: Murmansk"); Put_Line("Price per minute: 3 rubles."); Put("Call cost: "); Put(Item => len * 3, Width => 1); Put_Line(" rub."); when 846 => Put_Line("City: Samara"); Put_Line("Price per minute: 1 rub."); Put("Call cost: "); Put(Item => len, Width => 1); Put_Line(" rub."); when others => Put("There is no city with this code in the database! Try again."); end case; endMain;

The section briefly describes the operators if and case, function abs() and an algorithm for swapping variables.

Definition.

This is a hexagon whose bases are two equal square, and the side faces are equal rectangles

Side rib- this is common side two adjacent side faces

Prism Height- this is a cut perpendicular to bases prisms

Prism Diagonal- a segment connecting two vertices of the bases that do not belong to the same face

Diagonal plane is the plane that passes through the diagonal of the prism and its side ribs

Diagonal section - the boundaries of the intersection of the prism and the diagonal plane. The diagonal section of a regular quadrangular prism is a rectangle

Perpendicular section (orthogonal section)- this is the intersection of a prism and a plane drawn perpendicular to its side edges

Elements of a regular quadrangular prism

The figure shows two regular quadrangular prisms, which are marked with the corresponding letters:

  • Bases ABCD and A 1 B 1 C 1 D 1 are equal and parallel to each other
  • Side faces AA 1 D 1 D, AA 1 B 1 B, BB 1 C 1 C and CC 1 D 1 D, each of which is a rectangle
  • Side surface- the sum of the areas of all side faces of the prism
  • Total surface - the sum of the areas of all bases and side faces (the sum of the area of ​​the side surface and bases)
  • Side ribs AA 1 , BB 1 , CC 1 and DD 1 .
  • Diagonal B 1 D
  • Base diagonal BD
  • Diagonal section BB 1 D 1 D
  • Perpendicular section A 2 B 2 C 2 D 2 .

Properties of a regular quadrangular prism

  • The bases are two equal squares
  • The bases are parallel to each other
  • The sides are rectangles.
  • Side faces are equal to each other
  • Side faces are perpendicular to the bases
  • Lateral ribs are parallel to each other and equal
  • Perpendicular section perpendicular to all side ribs and parallel to the bases
  • corners perpendicular section- straight
  • The diagonal section of a regular quadrangular prism is a rectangle
  • Perpendicular (orthogonal section) parallel to the bases

Formulas for a regular quadrangular prism

Instructions for solving problems

When solving problems on the topic " correct quadrangular prism " implies that:

Correct prism- a prism at the base of which lies regular polygon, and the side edges are perpendicular to the base planes. That is, a regular quadrangular prism contains at its base square. (see above the properties of a regular quadrangular prism) Note. This is part of the lesson with tasks in geometry (section solid geometry - prism). Here are the tasks that cause difficulties in solving. If you need to solve a problem in geometry, which is not here - write about it in the forum. To indicate the action of extracting square root symbol is used in problem solving√ .

A task.

In a regular quadrangular prism, the base area is 144 cm 2 and the height is 14 cm. Find the diagonal of the prism and the area full surface.

Solution.
A regular quadrilateral is a square.
Accordingly, the side of the base will be equal to

√144 = 12 cm.
Where does the diagonal of the base come from? rectangular prism will be equal to
√(12 2 + 12 2 ) = √288 = 12√2

Diagonal right prism forms with the diagonal of the base and the height of the prism right triangle. Accordingly, according to the Pythagorean theorem, the diagonal of a given regular quadrangular prism will be equal to:
√((12√2) 2 + 14 2 ) = 22 cm

Answer: 22 cm

A task

Find the total surface area of ​​a regular quadrangular prism if its diagonal is 5 cm and the diagonal of the side face is 4 cm.

Solution.
Since the base of a regular quadrangular prism is a square, then the side of the base (denoted as a) is found by the Pythagorean theorem:

A 2 + a 2 = 5 2
2a 2 = 25
a = √12.5

The height of the side face (denoted as h) will then be equal to:

H 2 + 12.5 \u003d 4 2
h 2 + 12.5 = 16
h 2 \u003d 3.5
h = √3.5

The total surface area will be equal to the sum of the lateral surface area and twice the base area

S = 2a 2 + 4ah
S = 25 + 4√12.5 * √3.5
S = 25 + 4√43.75
S = 25 + 4√(175/4)
S = 25 + 4√(7*25/4)
S \u003d 25 + 10√7 ≈ 51.46 cm 2.

Answer: 25 + 10√7 ≈ 51.46 cm 2.

Cars with a manual transmission, which is abbreviated as manual transmission, until recently amounted to absolute majority among other vehicles with different .

Moreover, a mechanical (manual) box remains a fairly common device for changing and transmitting engine torque today. Next, we will talk about how the "mechanics" is arranged and works, what the gearbox scheme of this type looks like, and also what advantages and disadvantages this solution has.

Read in this article

Manual transmission diagram and features

To begin with, this type of gearbox is called mechanical due to the fact that such a unit involves manual gear shifting. In other words, on cars with manual transmission, the driver himself switches gears.

We go further. The "mechanics" box is stepped, that is, the torque changes in steps. Many motorists know that the gearbox actually has gears and shafts, but not everyone understands how the unit works.

So, a stage (it is also a transmission) is a pair of gears (driving and driven gear) interacting with each other. Each such stage provides rotation with one or another angular velocity, that is, it has its own gear ratio.

Under the gear ratio should be understood as the ratio of the number of teeth of the driven gear to the number of teeth on the drive gear. In this case, different stages of the box receive different gear ratios. The lowest gear (low gear) has the largest gear ratio, and the most high step(upshift) has the smallest gear ratio.

It becomes clear that the number of steps is equal to the number of gears on a particular box (four-speed gearbox, five-speed, etc.) previously 4-speed mechanical boxes gear gradually faded into the background.

Manual transmission device

So, although there can be many designs of such a box with certain features, however, initial stage two main types can be distinguished:

  • three-shaft gearboxes;
  • two-shaft boxes;

A three-shaft manual gearbox is usually installed on rear-wheel drive cars, while a two-shaft gearbox is placed on front-wheel drive passenger cars. At the same time, the device of mechanical gearboxes of both the first and second types can differ markedly.

Let's start with a three-shaft mechanical box. This box contains:

  • the drive shaft, which is also called the primary;
  • intermediate shaft gearbox;
  • driven shaft (secondary);

Gears with synchronizers are installed on the shafts. The gearshift mechanism is also included in the gearbox. Specified constituent elements located in the gearbox housing, which is also called the gearbox housing.

The task of the drive shaft is to create a connection with the clutch. The drive shaft has slots for the clutch disc. As for the torque, the specified torque from the input shaft is transmitted through the gear, which is in rigid engagement with it.

Affecting the work of the intermediate shaft, this shaft is located parallel to the input shaft of the gearbox, a group of gears is installed on it, which is in rigid engagement. In turn, the driven shaft is mounted on the same axis as the drive shaft.

Such an installation is implemented using an end bearing on the drive shaft. This bearing includes the driven shaft. The group of gears (gear block) on the driven shaft does not have a rigid engagement with the shaft itself and therefore rotates freely on it. In this case, the group of gears of the intermediate shaft, the driven shaft and the gear of the drive shaft are in constant engagement.

Synchronizers (synchronizer couplings) are installed between the gears of the driven shaft. Their task is to align the angular velocities of the gears of the driven shaft with the angular velocity of the shaft itself through the force of friction.

Synchronizers are in rigid engagement with the driven shaft, and also have the ability to move along the shaft in the longitudinal direction due to the spline connection. Modern gearboxes have synchronizer clutches in all gears.

If we consider the gearshift mechanism on three-shaft gearboxes, often this mechanism is installed on the unit body. The design includes a control lever, sliders and forks.

The box body (crankcase) is made of aluminum or magnesium alloys, it is necessary for installing shafts with gears and mechanisms, as well as a number of other parts. There is also gear oil (gearbox oil) in the gearbox housing.

  • To understand how a three-shaft type mechanical (manual) gearbox works, let's in general terms Let's take a look at how it works. When the gear lever is in the neutral position, there is no transmission of torque from the engine to the vehicle's drive wheels.

After the driver moves the lever, the fork will move the synchronizer clutch of one or another gear. The synchronizer will then align angular velocities desired gear and driven shaft. Then the gear ring of the clutch will engage with a similar gear ring, which will ensure that the gear is locked on the driven shaft.

We also add that the reverse gear of the car is provided by the reverse gear of the gearbox. In this case, a reverse idle gear mounted on a separate axle allows the direction of rotation to be reversed.

Two-shaft manual gearbox: device and principle of operation

Having dealt with what a three-shaft gearbox consists of, let's move on to two-shaft gearboxes. This type The gearbox has two shafts in its device: primary and secondary. The input shaft is the driving one, the secondary is the driven one. Gears and synchronizers are fixed on the shafts. Also in the crankcase of the box is the main gear and differential.

The drive shaft is responsible for connecting with the clutch, and there is also a gear block on the shaft in rigid engagement with the shaft. The driven shaft is located parallel to the drive shaft, while the gears of the driven shaft are in constant engagement with the gears of the drive shaft, and also rotate freely on the shaft itself.

Also, the drive gear of the main gear is rigidly fixed on the driven shaft, and synchronizer couplings are located between the gears of the driven shaft. We add, in order to reduce the size of the gearbox, as well as increase the number of gears, in modern gearboxes, 2 or even 3 shafts can often be installed instead of one driven shaft.

On each such shaft, the gear of the main gear is rigidly fixed, while such a gear has a rigid engagement with the driven gear. It turns out that the design actually implements 3 main gears.

The main gear itself, as well as the differential in the gearbox device, transmit torque from the secondary shaft to the drive wheels. In this case, the differential can also provide such rotation of the wheels when the drive wheels rotate at different angular speeds.

As for the gearshift mechanism, on two-shaft gearboxes it is taken out separately, that is, outside the body. The box is connected to the switching mechanism by cables or special rods. The most common connection is with cables.

The shift mechanism of the 2-shaft box itself has a lever, which is connected by cables to the selector lever and the gear shift lever. These levers are connected to the central shift rod, which also has forks.

  • If we talk about the principle of operation of a two-shaft manual gearbox, it is similar to the principle of a three-shaft gearbox. The differences are in how the gearshift mechanism works. In a nutshell, the lever can carry out both longitudinal and transverse movements relative to the axis of the car. During transverse movement gear selection occurs, as the force is applied to the gear selector cable, which acts on the gear selector lever.

Further, the lever moves longitudinally, and the force goes to the gearshift cable. The corresponding lever horizontally moves the stem with the forks, the fork on the stem displaces the synchronizer, which leads to blocking of the driven shaft gear.

Finally, we note that also mechanical boxes different types have additional blocking devices that prevent the inclusion of two gears at the same time or an unexpected disengagement of the gear.

Read also

Depressing the clutch before starting the engine: when to depress the clutch and in what cases it is not recommended to do so. Helpful Hints and recommendations.

  • Causes of difficult gear shifting on a running engine. Transmission oil and level in the gearbox, wear of synchronizers and gears of the box, clutch.



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