Mathematics as an essential component of the sport training.

Authors:

Dr.C. Islay Pérez Martínez. Sports Science Department (BUSE)

MSc. Ariadna Quintana Díaz. University of Matanzas(Cuba)

In the sports, the mathematics have different applications; among them, we have the following examples:

• In the planning of the training.

• The control of the loads.

• The evaluation of the competitive results.

• The selection of talents.

• To predict results.

Equation to determine the heart frequency of training.

CRT = CRR + 0,6 (MCR – CRR)

• CRT = Cardiac rhythm of training B. /MIN.• MCR = Maximum cardiac rhythm. B. /MIN.• CRR = Cardiac rhythm of rest. B./ MIN.• MCR = 220 – AGE

EXAMPLE:• CRT = CRR + 0,6 (MCR –CRR)• CRT = 70 +0,6 ((220 –20) –70)• CRT = 70 + 0,6 (200 –70)• CRT = 70 + 0,6 * 130• CRT = 70 + 78

CRT = 148 B/MIN.

The work threshold (area of energy work) related with the power of the work, it determines the percentage

of the heart frequency of training regarding the maximum and the predominant energy system. According

to Menshikov and Volkov (1985) to the same one it corresponds the following values:

70% Aerobic work.

80% Mixed work.

90% Anaerobic workSCALE OF WORK:

Intensity Percent Energy Area

Maxim 96 – 100 Anaerobic

Sub - maxim 90 – 95 Anaerobic

Big 86 – 89 Mixed

Medium 80 – 85 Mixed

Sub - medium 71 – 79 Aerobic

Small 60 – 70 Aerobic

WINT = HEART FREQUENCY OF WORK - HEART FREQUENCY IN REST

MAXIMUM HEART FREQUENCY - HEART FREQUENCY IN REST

WINT = 160 - 70__ = 90__ = 90_

(220 – 20) – 70 200 – 70 130

WINT = 0,69

DETERMINATION OF THE COEFFICIENT OF INTENSITY

Aerobic Rhythm.

The aerobic rhythm is basic for a correct planning of the volume and the

intensity in the athletics races. With this calculation we can obtain an

information that allows us to dose and to adapt the training to each athlete's

individual characteristics.

Result of the race test: 13500 meters travelled in 45 minutes (2700 sec)

Speed (Aerobic Rhythm): 13500 / 2700 = 5 m/sec.

1000 m / 5 sec = 200 sec/km = 3:20 min/km.

Example of Rhythm of Training for an athlete with an aerobic rhythm of 3:20 min/km.

• Aerobic Rhythm: 3:20 min/km

Form of calculating the rhythm of training:

• 70% of the Aerobic Rhythm: 4:46 min/km (200 sec*100/70= 286 sec = 4:46 min/km)

• Rhythm of slow continuous race (70% of the aerobic rhythm) 4:46 min/km

• Rhythm of slow race of long distance (80-85% of the aerobic rhythm) 4:10-3:25 min/km.

• Rhythm of fast continuous race (90-95% of the aerobic rhythm) 3:42-3:25 min/km.

• Rhythm of training to extensive intervals (105-110% of the aerobic rhythm) 3:10-3:00 min/km.

Purpose Rhythm.

The purpose rhythm allows the athletes to train in a specific way for a certain race. Next the example of a runner of 1500 plane meters appears.

Example of as using it with an athlete that competes in 1500 meters:

Result of a test of 1500 meters: 1500 meters in 4:15 min.

Expectation of time: 4:15 min = 255 sec

Speed average in m/s: 1500 m / 255 sec = 5, 9 m/s.

Purpose rhythm for 100 m: 255 sec / 15 = 17 sec.

Purpose rhythm for 400 m: 17 sec * 4 = 68 sec.

Purpose rhythm for 1 km: 17 sec * 10 = 170 sec (2:50 min)

EXAMPLE OF A PROPOSAL FOR DISTRIBUTING THE LOADS.

For an athlete that executes 3000Km. in a season:

To Calculate the total of kilometers that should increase:

Proposal of increment 20% = 3000 + 600 = 3600 Kilometers.

It should be assigned to every month of training a volume in % regarding 100% that will be reached in the month of more work.

Oct. Nov. Dec. Jan. Feb. Mar. Apr. May. Jun. Jul. Aug.

70% 85% 100% 90% 85% 80% 90% 85% 80% 75% 70%

Add the % of every month and the total value is used to divide the total of kilometers.

3600 / 910 = 3.96.

To calculate the monthly quantity of kilometers, the previous figure should be multiplied by the %.

Example: OCTOBER 3.96 x 70 = 277 Km.

To calculate the total of kilometers of the weeks, some theoretical values are assigned to every week of the month in function of the structure of the micro cycle.

Medium Load High load Very high load Lower load

4 6 7 3

It should be added the values assigned for every week and the following operations are made: (4+6+7+3=20)

First Week: 277 x 4 / 20 = 55 km.

Second Week: 277x 6 / 20 = 83 km.

Third Week: 277 x 7 / 20 = 97 km.

Fourth Week: 277 x 3 / 20 = 42 km.

Total 277 km.

Control of the recovery using the rhythm of heart beaten.

• The control of the recovery after the training is extremely important because

it offers us the information about if the sportsman is assimilating the received

loads correctly. In this way, we can avoid to fall in a state of overtraining that

is harmful for the body.

• When the rhythm of beating returns to 100 bpm (beaten by minute) in:

• 3:00 minutes or less: Very good recovery.

• 5:00 minutes: Satisfactory recovery.

• More than 5:00 minutes: Poor recovery.

Rhythm of beating after 5 minutes (beaten per minute):

• 100-105: Very Good.

• 106-115: Good.

• 116-120: Satisfactory.

• 121-130: Poor.

• Higger than 130: Very poor.

Classification for the regulation of the exercises with weights. (A. Novikov y L. Matveyev. 1986)

Loads Repetitions %

Maximum 1 100

Sub-maxim 2 – 3 90 – 95

Big 4 – 8 80 – 89

Moderate 9 – 12 60 – 79

Medium 13 – 18 40 – 59

Small 19 – 25 25 – 39

Very small + 25 -25

• When the exercises with weights are planned you can use the following nomenclature:

% / Series = 80 / 3Repetitions 8

In this exercise, would be made 3 series of 8 repetitions with 80% of the maximum force, valid for 24 repetitions in total.

Tonnage: The tonnage is the result of multiplying the lifted weight, for the quantity of carried

out repetitions. Repetitions: The total of the repetitions is the sum of the repetitions carried out in a series, in a training,

week, month, etc.

Weight / series = 120/3 Repetitions 2

Example: 60 Kg 70 Kg 80 kg/2 90 kg/3 5 4 3 2

Tonnage: 60 x 5 + 70 x 4 + 80 x 2 x 3 + 90 x 3 x 2 = 300 + 280 + 480 + 540 = 1600 Kg

Repetitions: 5 + 4 + 2 x 3 + 3 x 2 = 21

Control of the load in the exercises with weights

Dosage example when the % of the maximum force is used.

To execute squatting from behind (maximum result 150 Kg).

60% 70% 80%/3 = 90kg 105kg 120kg/3 3 3 2 = 3 3 2

= 90 x 3 + 105 x 3 + 120 x 3 x 2

Tonnage = 305 Kg

Repetitions = 3 + 3 + 3 x 2

= 12 rept.

WeeksSessions of training   Dosage

Quantity of exercises Total

  Legs Thorax Arms Prom Legs Thorax Arms Prom Exercises Repetitions

1 1 3O/3 2O/3 IO/3 2O/3 I/45 I/45 I/45 I/45 4 180

    15 15 15 15          

  2 3O/3 2O/3 1O/3 2O/3 I/45 I/45 I/45 I/45 4 180

    15 15 15 15            

2 3 33/4 23/4 12/4 23/4 2/144 I/72 I/72 I/72 5 360

    18 18 18 18            

  4 36/4 26/4 14/4 26/4 2/144 I/72 I/72 I/72 5 360

    18 18 18 18            

3 5 38/5 28/5 17/5 28/5 2/180 2/180 I/90 I/90 6 540

    18 18 18 18            

  6 41/5 31/5 20/5 31/5 3/270 2/150 2/150 I/75 8 630

    18 18 18 18            

Example of planning of the load in exercises with weights.

Total of Hours 90 90 80 20 280

Total of minutes 5400 5400 4800 1200 16800

Component  Stage of General Preparation

Stage of Special Preparation

CompetitivePeriod

TransitPeriod

Total

General Physical Preparation 55% 21% 12% 70% 33%

2970 1134 576 840 5520

Special Physical Preparation

14% 23% 21% 10% 18%

756 1242 1008 120 3126

Technical Preparation 15% 27% 19% 15% 20%

810 1458 912 180 3360

Tactical Preparation 12% 25% 44% - 25%

648 1350 2112 - 4110

Theoretical Preparation 4% 4% 4% 5% 4%

216 216 192 60 684

Total 5400 5400 4800 1200 16800

DISTRIBUTION OF THE BLOCK OF DETAILED INFORMATION

HOW TO ACHIEVE A CORRECT DISTRIBUTION OF THE LOADS IN THE TABLE OF GENERAL INFORMATION?

WEEKS 9 MULTIPLY THE NUMBER OF WEEKS FOR THE PROPOSED

%HOURS 9.10= 90

MINUTES 90.60=5400

G. P. P. 55% 55.9= 495

2970

S. P. P. 14% 14.9= 126

756

TEC. P. 15% 15.9= 135

810

TACT. P. 12% 12.9= 108

648

P. TEO. P 4% 4.9= 36

216

TOTAL 5400 900/9= 100

STAGE OF GENERAL PREPARATION (EXAMPLE)

STAGES STAGE OF GENERAL PREPARATI ON

MESOCYCLOS INT. BAS. DES BAS. EST.

MI CROCYCLOS O O R O O R O CH R

No. WEEK: 1 2 3 4 5 6 7 8 9

MONTH

CALENDAR

% T % T % T % T % T % T % T % T % T

G.P. P 60 360 70 430 70 420 60 360 50 300 60 360 40 240 30 180 60 120

FORCE 30 108 30 129 30 126 30 108 30 90 30 108 30 72 30 54 30 36

SPEED 25 90 25 107 25 105 25 90 25 75 25 90 25 60 25 45 25 30

ENDURANCE 20 72 20 86 20 84 20 72 20 60 20 72 20 48 20 36 20 24

FLEXI BI LITY 15 54 15 64 15 63 15 54 15 45 15 54 15 36 15 27 15 18

COOR. CAP. 10 36 10 44 10 42 10 36 10 30 10 36 10 24 10 18 10 12

S. P. P. 15 90 15 90 10 60 15 90 15 90 10 60 15 90 20 120 10 60

SPECI AL FORCE 30 27 30 27 30 18 30 27 30 27 30 18 30 27 30 36 30 18

SPECI AL SPEED 30 27 30 27 30 18 30 27 30 27 30 18 30 27 30 36 30 18

SPECI AL ENDURANCE 20 18 20 18 20 12 20 18 20 18 20 12 20 18 20 24 20 12

SPECI AL FLEXI BI LITY 10 9 10 9 10 6 10 9 10 9 10 6 10 9 10 12 10 6

SPECI AL COOR. CAP. 10 9 10 9 10 6 10 9 10 9 10 6 10 9 10 12 10 6

TEC. PREP 10 60 10 60 10 60 10 60 20 120 20 120 25 150 20 120 20 120

F.T. Offensive 60 36 40 24 60 36 40 24 60 72 40 48 60 90 40 48 60 72

F.T. Deffensive 40 24 60 36 40 24 60 36 40 48 60 72 40 60 60 72 40 48

TACT. PREP. 10 60 5 30 5 30 10 60 15 90 5 30 20 120 30 180 5 30

OfFensive 60 36 40 12 60 18 40 24 60 54 60 18 40 48 40 72 60 18

Deff ensive 40 24 60 18 40 12 60 36 40 36 40 12 60 72 60 108 40 12

THEORETI CAL 5 30 - - 5 30 5 30 - - 5 30 - - - -- 5 30

- - - - - - - - - - - - - - - - - --

Selection of sport talents: • To identify sport talents several physical test are applied that evaluate the

capacities of force, speed, endurance and flexibility, the size and the weight

are also measured. To those results are applied statistical techniques and

they conform to databases that represent populations.

• For example, the following tables show the Percentiles of the speed test for

males and force test for the females. With these data when comparing the

individual results of the students you can determine if they possess the

suitable conditions for the practice of the sport.

Percentiles of the test of speed masculine sex.

Percentiles of the test of force - feminine sex.

AGE

AGE

The following table shows the values used in Cuba for the evaluation of the tests of physical efficiency in the Physical Education.

Age Size Flexibility Speed Press-up Abdominal Long Jump Endurance

M F M F M F M F M F M F M F6     46 47,5 6,0 6,2         140 125    

7 129,5 128,7 46 47,5 5,8 6,0         145 134    

8 135,4 134,8 46 47,5 5,6 5,9 21 22 33 30 154 141 1,23 1,25

9 140,6 140,9 46 47,5 5,3 5,6 21 24 39 32 154 149 1,19 1,22

10 145,5 147,2 46 47,5 5,2 5,4 23 27 43 35 172 156 1,16 1,19

11 150,3 153,5 46 47,5 5,2 5,3 24 30 49 38 180 164 1,14 1,17

12 156,0 160,1 46 47,5 4,25 4,4 24 30 50 39 190 174 1,12 1,15

13 164,7 163,6 46 47,5 6,6 6,8 29 30 50 42 202 180 2,34 2,40

14 171,0 165,6 46 47,5 6,8 7,2 29 30 53 41 211 186 2,29 2,35

15 177,1 166,9 46 47,5 6,9 7,5 35 30 58 41 222 191 2,24 2,30

16 179,9 167,4 46 47,5 5,8 7,5 39 30 58 41 233 192 2,19 2,25

17 181,1 167,8 46 47,5 6,7 7,6 39 30 58 42 238 197 2,14 2,20

18 181,3 167,9 46 47,5 7,0 8,0 40 27 41 30 240 279 2,09 2,15

19 181,4 168,0 46 47,5 7,0 8,0 40 27 41 30 240 175 2,04 2,10

Conclusions:

• Mathematics has a paramount importance in most of the fields of science; in

sports, specifically, it has many applications. In this paper, we have shown

some examples of the application of mathematics to the world of sports.

• It is demonstrated that a successful process of sport formation cannot be

lead without making use of mathematics. In sport the loads without control

are dangerous for health and this control is exercised with the use of

mathematics.

“The load without control is unthinkable and dangerous”

Peter Tschiene

Thank you very much for your attention