Commission Delegated Regulation (EU) No 1322/2014 of 19 September 2014 supplementing and amending Regulation (EU) No 167/2013 of the European Parliament and of the Council with regard to vehicle construction and general requirements for the approval of agricultural and forestry vehicles Text with EEA relevance

This could be done manually or with the use of external energy (hydraulic, pneumatic or electric actuators). In both cases the force shall be applied within a plane parallel to the trajectory of the roll-bar and passing through the grasping area, the angular speed of the roll-bar shall be roughly constant and less than 20 deg/s.

After the 500 cycles, the force applied when the roll-bar is in the upright position shall not exceed by more than 50 % the allowed force (Table 6.2).

The unlocking of the roll-bar shall be done following the operator manual.

After the completion of the 500 cycles there shall be no maintenance or adjustment on the locking system,

• Note 1:The preliminary test could be applied to automatic foldable ROPS systems as well. The test should be carried out before the ROPS strength test.

• Note 2:The preliminary test could be carried out by the manufacturer. In such a case the manufacturer shall provide the test station with a certificate stating that the test has been done according the test procedure and that there were no maintenance or adjustment on the locking system after the completion of the 500 cycles. The test station will check the performance of the device with one cycle from the lower position to the upright locked position and back.

Figure 6.20

Grasping area

Image omitted; see official source

Figure 6.21

Accessible zones

(Dimensions in mm)

Image omitted; see official source

Figure 6.22

Envelope of the accessible zones

(Dimensions in mm)

Image omitted; see official source

Figure 6.23

Accessible part of the grasping area

Image omitted; see official source

Figure 6.24

Points where the force requirement shall be measured

Image omitted; see official source

B4.   REQUIREMENTS FOR VIRTUAL TESTING

Computer programme⁽³⁾ (BASIC) for determining the continuous or interrupted roll over behaviour in case of a laterally overturning narrow track tractor with a protective frame mounted in front of the driver’s seat

Preliminary note:

The following programme is valid for its calculation methods. Presentation of the printed text as proposed (English language and layout) is indicative; the user will adapt the programme to available printing and other requirements specific to the testing station.

10 CLS

20 REM REFERENCE OF THE PROGRAM COD6ABAS.BAS 08/02/96

30 FOR I = 1 TO 10: LOCATE I, 1, 0: NEXT I

40 COLOR 14, 8, 4

50 PRINT ""

60 PRINT " CALCULATION FOR DETERMINING THE NON-CONTINUOUS ROLLING BEHAVIOUR "

70 PRINT "OF A LATERAL OVERTURNING NARROW TRACTOR WITH A ROLL-OVER PROTECTIVE "

80 PRINT " STRUCTURE MOUNTED IN FRONT OF THE DRIVER’S SEAT "

90 PRINT ""

100 A$ = INKEY$: IF A$ = "" THEN 100

110 COLOR 10, 1, 4

120 DIM F(25), C(25), CAMPO$(25), LON(25), B$(25), C$(25), X(6, 7), Y(6, 7), Z(6, 7)

130 DATA 6,10,10,14,14,17,19,21,11,11,12,12,13,13,14,14,15,15,16,16,17,17,18,18,19

140 DATA 54,8,47,8,47,12,8,12,29,71,29,71,29,71,29,71,29,71,29,71,29,71,29,71,29

150 DATA 12,30,31,30,31,25,25,25,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9

160 FOR I = 1 TO 25: READ F(I): NEXT

170 FOR I = 1 TO 25: READ C(I): NEXT

180 FOR I = 1 TO 25: READ LON(I): NEXT

190 CLS

200 FOR I = 1 TO 5: LOCATE I, 1, 0: NEXT I

210 PRINT "In case of misprint, push on the enter key up to the last field"

220 PRINT :LOCATE 6, 44: PRINT " TEST NR: ": PRINT

230 LOCATE 8, 29: PRINT " FRONT MOUNTED- PROTECTIVE STRUCTURE:": PRINT

240 PRINT " MAKE: ": LOCATE 10, 40: PRINT " TYPE: ": PRINT

250 LOCATE 12, 29: PRINT " TRACTOR :": PRINT : PRINT " MAKE:"

260 LOCATE 14, 40: PRINT " TYPE: ": PRINT : PRINT

270 PRINT " LOCATION: ": PRINT

280 PRINT " DATE: ": PRINT : PRINT " ENGINEER:"

290 NC = 1: GOSUB 4400

300 PRINT : PRINT : PRINT " In case of misprint, it is possible to acquire the data again"

310 PRINT : INPUT " Do you wish to acquire again the data ? (Y/N)"; Z$

320 IF Z$ = "Y" OR Z$ = "y" THEN 190

330 IF Z$ = "N" OR Z$ = "n" THEN 340

340 FOR I=1 TO 3:LPRINT : NEXT: LPRINT ; " TEST NR: "; TAB(10); CAMPO$(1)

350 LPRINT : LPRINT TAB(24); " FRONT MOUNTED PROTECTIVE STRUCTURE:"

360 LL = LEN(CAMPO$(2) + CAMPO$(3))

370 LPRINT TAB(36 - LL / 2); CAMPO$(2) + " - " + CAMPO$(3) : LPRINT

380 LPRINT TAB(32); " OF THE NARROW TRACTOR": LL = LEN(CAMPO$(4) + CAMPO$(5))

390 LPRINT TAB(36 - LL / 2); CAMPO$(4) + " - " + CAMPO$(5) : LPRINT

400 CLS

410 PRINT "In case of mistype, push on the enter key up to the last field"

420 PRINT

430 FOR I = 1 TO 7: LOCATE I, 1, 0: NEXT

440 LOCATE 8, 1: PRINT " CHARACTERISTIC UNITS:"

450 LOCATE 8, 29: PRINT "LINEAR (m): MASS (kg):MOMENT OF INERTIA (kg×m²):"

460 LOCATE 9, 1: PRINT " ANGLE (radian)"

470 LPRINT : PRINT

480 PRINT "HEIGHT OF COG H1=": LOCATE 11, 29: PRINT""

490 LOCATE 11, 40: PRINT "H. DIST. COG-REAR AXLE L3="

500 LOCATE 11, 71: PRINT""

510 PRINT "H. DIST. COG-FRT AXLE L2=": LOCATE 12, 29: PRINT""

520 LOCATE 12, 40: PRINT "HEIGHT OF THE REAR TYRES D3="

530 LOCATE 12, 71: PRINT""

540 PRINT "HEIGHT OF THE FRT TYRES D2=": LOCATE 13, 29: PRINT""

550 LOCATE 13, 40: PRINT "OVERALL HEIGHT(PT IMPACT) H6="

560 LOCATE 13, 71: PRINT""

570 PRINT "H.DIST.COG-LEAD.PT INTER.L6=": LOCATE 14, 29: PRINT""

580 LOCATE 14, 40: PRINT "PROTECTIVE STRUCT. WIDTH B6="

590 LOCATE 14, 71: PRINT""

600 PRINT "HEIGHT OF THE ENG.B. H7=": LOCATE 15, 29: PRINT""

605 LOCATE 15, 40: PRINT "WIDTH OF THE ENG. B. B7="

610 LOCATE 15, 71: PRINT""

615 PRINT "H.DIST.COG-FRT COR.ENG.B.L7=": LOCATE 16, 29: PRINT""

620 LOCATE 16, 40: PRINT "HEIGHT FRT AXLE PIVOT PT H0="

630 LOCATE 16, 71: PRINT""

640 PRINT "REAR TRACK WIDTH S =": LOCATE 17, 29: PRINT""

650 LOCATE 17, 40: PRINT "REAR TYRE WIDTH B0="

660 LOCATE 17, 71: PRINT""

670 PRINT "FRT AXLE SWING ANGLE D0=": LOCATE 18, 29: PRINT""

680 LOCATE 18, 40: PRINT "TRACTOR MASS Mc ="

690 LOCATE 18, 71: PRINT""

700 PRINT "MOMENT OF INERTIA Q =": LOCATE 19, 29: PRINT""

710 LOCATE 19, 40: PRINT""

720 LOCATE 19, 71: PRINT " ": PRINT : PRINT

730 H1 = 0: L3 = 0: L2 = 0: D3 = 0: D2 = 0: H6 = 0: L6 = 0: B6 = 0

740 H7 = 0: B7 = 0: L7 = 0: H0 = 0: S = 0: B0 = 0: D = 0: Mc = 0: Q = 0

750 NC = 9: GOSUB 4400

760 FOR I = 1 TO 3: PRINT "": NEXT

770 H1 = VAL(CAMPO$(9)): L3 = VAL(CAMPO$(10)): L2 = VAL(CAMPO$(11))

780 D3 = VAL(CAMPO$(12)): D2 = VAL(CAMPO$(13)): H6 = VAL(CAMPO$(14))

790 L6 = VAL(CAMPO$(15)): B6 = VAL(CAMPO$(16)): H7 = VAL(CAMPO$(17))

800 B7 = VAL(CAMPO$(18)): L7 = VAL(CAMPO$(19)): H0 = VAL(CAMPO$(20))

810 S = VAL(CAMPO$(21)): B0 = VAL(CAMPO$(22)): D0 = VAL(CAMPO$(23))

820 Mc = VAL(CAMPO$(24)): Q = VAL(CAMPO$(25)): PRINT : PRINT

830 PRINT "In case of mistype, it is possible to acquire again the data": PRINT

840 INPUT " Do you wish to acquire again the data ? (Y/N)"; X$

850 IF X$ = "Y" OR X$ = "y" THEN 400

860 IF X$ = "n" OR X$ = "N" THEN 870

870 FOR I = 1 TO 3: LPRINT : NEXT

880 LPRINT TAB(20); "CHARACTERISTIC UNITS :": LOCATE 8, 29

890 LPRINT "LINEAR (m) : MASS (kg) : MOMENT OF INERTIA (kg×m²) : ANGLE (radian)"

900 LPRINT

910 LPRINT "HEIGHT OF THE COG H1=";

920 LPRINT USING"####.####"; H1;

930 LPRINT TAB(40); "H. DIST. COG-REAR AXLE L3=";

940 LPRINT USING"####.####"; L3

950 LPRINT "H.DIST. COG-FRT AXLE L2=";

960 LPRINT USING"####.####"; L2;

970 LPRINT TAB(40); "HEIGHT OF THE REAR TYRES D3=";

975 LPRINT USING"####.####"; D3

980 LPRINT "HEIGHT OF THE FRT TYRES D2=";

990 LPRINT USING"####.####"; D2;

1000 LPRINT TAB(40); "OVERALL HEIGHT(PT IMPACT)H6=";

1010 LPRINT USING"####.####"; H6

1020 LPRINT "H.DIST.COG-LEAD PT INTER.L6=";

1030 LPRINT USING"####.####"; L6;

1040 LPRINT TAB(40); "PROTECTIVE STRUCT. WIDTH B6=";

1050 LPRINT USING"####.####"; B6

1060 LPRINT "HEIGHT OF THE ENG.B. H7=";

1070 LPRINT USING"####.####"; H7;

1080 LPRINT TAB(40); "WIDTH OF THE ENG. B. B7=";

1090 LPRINT USING"####.####"; B7

1100 LPRINT "H.DIST.COG-FRT COR.ENG.B.L7=";

1110 LPRINT USING"####.####"; L7;

1120 LPRINT TAB(40); "HEIGHT FRT AXLE PIVOT PT H0=";

1130 LPRINT USING"####.####"; H0

1140 LPRINT "REAR TRACK WIDTH S =";

1150 LPRINT USING"####.####"; S;

1160 LPRINT TAB(40); "REAR TYRE WIDTH B0=";

1170 LPRINT USING"####.####"; B0

1180 LPRINT "FRT AXLE SWING ANGLE D0=";

1185 LPRINT USING"####.####"; D0;

1190 LPRINT TAB(40); "TRACTOR MASS Mc = ";

1200 LPRINT USING "####.###"; Mc

1210 LPRINT "MOMENT OF INERTIA Q =";

1215 LPRINT USING"####.####"; Q

1220 FOR I = 1 TO 10: LPRINT : NEXT

1230 A0 = .588: U = .2: T = .2: GOSUB 4860

1240 REM * THE SIGN OF L6 IS MINUS IF THE POINT LIES IN FRONT

1250 REM * OF THE PLANE OF THE CENTRE OF GRAVITY.

1260 IF B6 > S + B0 THEN 3715

1265 IF B7 > S + B0 THEN 3715

1270 G = 9.8

1280 REM ***

1290 REM B2 VERSION (POINT OF IMPACT OF THE ROPS NEAR OF EQUILIBRIUM POINT)

1300 REM ***

1310 B = B6: H = H6

1320 REM POSITION OF CENTER OF GRAVITY IN TILTED POSITION

1330 R2 = SQR(H1 * H1 + L3 * L3)

1340 C1 = ATN(H1 / L3)

1350 L0 = L3 + L2

1360 L9 = ATN(H0 / L0)

1370 H9 = R2 * SIN(C1 - L9)

1380 W1 = H9 / TAN(C1 - L9)

1390 W2 = SQR(H0 * H0 + L0 * L0): S1 = S / 2

1400 F1 = ATN(S1 / W2)

1410 W3 = (W2 - W1) * SIN(F1)

1420 W4 = ATN(H9 / W3)

1430 W5 = SQR(H9 * H9 + W3 * W3) * SIN(W4 + D0)

1440 W6 = W3 - SQR(W3 * W3 + H9 * H9) * COS(W4 + D0)

1450 W7 = W1 + W6 * SIN(F1)

1460 W8 = ATN(W5 / W7)

1470 W9 = SIN(W8 + L9) * SQR(W5 * W5 + W7 * W7)

1480 W0 = SQR(W9 * W9 + (S1 - W6 * COS(F1)) ^ 2)

1490 G1 = SQR(((S + B0) / 2) ^ 2 + H1 * H1)

1500 G2 = ATN(2 * H1 / (S + B0))

1510 G3 = W0 - G1 * COS(A0 + G2)

1520 O0 = SQR(2 * Mc * G * G3 / (Q + Mc * (W0 + G1) * (W0 + G1) / 4))

1530 F2 = ATN(((D3 - D2) / L0) / (1 - ((D3 - D2) / (2 * L3 + 2 * L2)) ^ 2))

1540 L8 = -TAN(F2) * (H - H1)

1550 REM COORDINATES IN POSITION 1

1560 X(1, 1) = H1

1570 X(1, 2) = 0: X(1, 3) = 0

1580 X(1, 4) = (1 + COS(F2)) * D2 / 2

1590 X(1, 5) = (1 + COS(F2)) * D3 / 2

1600 X(1, 6) = H

1610 X(1, 7) = H7

1620 Y(1, 1) = 0

1630 Y(1, 2) = L2

1640 Y(1, 3) = -L3

1650 Y(1, 4) = L2 + SIN(F2) * D2 / 2

1660 Y(1, 5) = -L3 + SIN(F2) * D3 / 2

1670 Y(1, 6) = -L6

1680 Y(1, 7) = L7

1690 Z(1, 1) = (S + B0) / 2

1700 Z(1, 2) = 0: Z(1, 3) = 0: Z(1, 4) = 0: Z(1, 5) = 0

1710 Z(1, 6) = (S + B0) / 2 - B / 2

1720 Z(1, 7) = (S + B0) / 2 - B7 / 2

1730 O1 = 0: O2 = 0: O3 = 0: O4 = 0: O5 = 0: O6 = 0: O7 = 0: O8 = 0: O9 = 0

1740 K1 = Y(1, 4) * TAN(F2) + X(1, 4)

1750 K2 = X(1, 1)

1760 K3 = Z(1, 1)

1770 K4 = K1 - X(1, 1): DD1 = Q + Mc * K3 * K3 + Mc * K4 * K4

1780 O1 = (Q + Mc * K3 * K3 - U * Mc * K4 * K4 - (1 + U) * Mc * K2 * K4) * O0 / DD1

1790 REM TRANSFORMATION OF THE COORDINATES FROM THE POSITION 1 TO 2

1800 FOR K = 1 TO 7 STEP 1

1810 X(2, K) = COS(F2) * (X(1, K) - H1) + SIN(F2) * Y(1, K) - K4 * COS(F2)

1820 Y(2, K) = Y(1, K) * COS(F2) - (X(1, K) - H1) * SIN(F2)

1830 Z(2, K) = Z(1, K)

1840 NEXT K

1850 O2 = O1 * COS(F2)

1860 A2 = ATN(TAN(A0) / SQR(1 + (TAN(F2)) ^ 2 / (COS(A0)) ^ 2))

1870 C2 = ATN(Z(2, 6) / X(2, 6))

1880 T2 = T

1890 V0 = SQR(X(2, 6) ^ 2 + Z(2, 6) ^ 2)

1900 E1 = T2 / V0

1910 E2 = (V0 * Y(2, 4)) / (Y(2, 4) - Y(2, 6))

1920 T3 = E1 * E2

1930 E4 = SQR(X(2, 1) * X(2, 1) + Z(2, 1) * Z(2, 1))

1940 V6 = ATN(X(2, 1) / Z(2, 1))

1950 REM ROTATION OF THE TRACTOR FROM THE POSITION 2 TO 3

1960 FOR K = 1 TO 7 STEP 1

1970 IF Z(2, K) = 0 THEN 2000

1980 E3 = ATN(X(2, K) / Z(2, K))

1990 GOTO 2010

2000 E3 = -3.14159 / 2

2010 X(3, K) = SQR(X(2, K) * X(2, K) + Z(2, K) * Z(2, K)) * SIN(E3 + C2 + E1)

2020 Y(3, K) = Y(2, K)

2030 Z(3, K) = SQR(X(2, K) ^ 2 + Z(2, K) ^ 2) * COS(E3 + C2 + E1)

2040 NEXT K

2050 IF Z(3, 7) < 0 THEN 3680

2060 Z(3, 6) = 0

2070 Q3 = Q * (COS(F2)) ^ 2 + 3 * Q * (SIN(F2)) ^ 2

2080 V5 = (Q3 + Mc * E4 * E4) * O2 * O2 / 2

2090 IF -V6 > A2 THEN 2110

2100 GOTO 2130

2110 V7 = E4 * (1 - COS(-A2 - V6))

2120 IF V7 * Mc * G > V5 THEN 2320

2130 V8 = E4 * COS(-A2 - V6) - E4 * COS(-A2 - ATN(X(3, 1) / Z(3, 1)))

2140 O3 = SQR(2 * Mc * G * V8 / (Q3 + Mc * E4 * E4) + O2 * O2)

2150 K9 = X(3, 1)

2160 K5 = Z(3, 1)

2170 K6 = Z(3, 1) + E1 * V0

2180 K7 = V0 - X(3, 1)

2190 K8 = U: DD2 = Q3 + Mc * K6 * K6 + Mc * K7 * K7

2200 O4 = (Q3 + Mc * K5 * K6 - K8 * Mc * K7 * K7 - (1 + K8) * Mc * K9 * K7) * O3 / DD2

2210 N3 = SQR((X(3, 6) - X(3, 1)) ^ 2 + (Z(3, 6) - Z(3, 1)) ^ 2)

2220 N2 = ATN(-(X(3, 6) - X(3, 1)) / Z(3, 1))

2230 Q6 = Q3 + Mc * N3 ^ 2

2240 IF -N2 <= A2 THEN 2290

2250 N4 = N3 * (1 - COS(-A2 - N2))

2260 N5 = (Q6) * O4 * O4 / 2

2270 IF N4 * Mc * G > N5 THEN 2320

2280 O9 = SQR(-2 * Mc * G * N4 / (Q6) + O4 * O4)

2290 GOSUB 3740

2300 GOSUB 4170

2310 GOTO 4330

2320 GOSUB 3740

2330 IF L6 > L8 THEN 2790

2340 REM *

2350 REM ***

2355 REM B3 VERSION (POINT OF IMPACT OF THE ROPS IN FRONT OF EQUILIBRIUM POINT)

2360 REM ***

2370 O3 = 0: O4 = 0: O5 = 0: O6 = 0: O7 = 0: O8 = 0: O9 = 0

2380 E2 = (V0 * Y(2, 5)) / (Y(2, 5) - Y(2, 6))

2390 T3 = E2 * E1

2400 Z(3, 6) = 0

2410 Q3 = Q * (COS(F2)) ^ 2 + 3 * Q * (SIN(F2)) ^ 2

2420 V5 = (Q3 + Mc * E4 * E4) * O2 * O2 / 2

2430 IF -V6 > A2 THEN 2450

2440 GOTO 2470

2450 V7 = E4 * (1 - COS(-A2 - V6))

2460 IF V7 * Mc * G > V5 THEN 2760

2470 V8 = E4 * COS(-A2 - V6) - E4 * COS(-A2 - ATN(X(3, 1) / Z(3, 1)))

2480 O3 = SQR((2 * Mc * G * V8) / (Q3 + Mc * E4 * E4) + O2 * O2)

2490 K9 = X(3, 1)

2500 K5 = Z(3, 1)

2510 K6 = Z(3, 1) + T3

2520 K7 = E2 - X(3, 1)

2530 K8 = U: DD2 = Q3 + Mc * K6 * K6 + Mc * K7 * K7

2540 O4 = (Q3 + Mc * K5 * K6 - K8 * Mc * K7 * K7 - (1 + K8) * Mc * K9 * K7) * O3 / DD2

2550 F3 = ATN(V0 / (Y(3, 5) - Y(3, 6)))

2560 O5 = O4 * COS(F3)

2570 REM TRANSFORMATION OF THE COORDINATES FROM THE POSITION 3 TO 4

2580 REM POSITION 4

2590 FOR K = 1 TO 7 STEP 1

2600 X(4, K) = X(3, K) * COS(F3) + (Y(3, K) - Y(3, 5)) * SIN(F3)

2610 Y(4, K) = (Y(3, K) - Y(3, 5)) * COS(F3) - X(3, K) * SIN(F3)

2620 Z(4, K) = Z(3, K)

2630 NEXT K

2640 A4 = ATN(TAN(A0) / SQR(1 + (TAN(F2 + F3)) ^ 2 / (COS(A0)) ^ 2))

2650 M1 = SQR(X(4, 1) ^ 2 + Z(4, 1) ^ 2)

2660 M2 = ATN(X(4, 1) / Z(4, 1))

2670 Q5 = Q * (COS(F2 + F3)) ^ 2 + 3 * Q * (SIN(F2 + F3)) ^ 2

2680 IF -M2 < A4 THEN 2730

2690 M3 = M1 * (1 - COS(-A4 - M2))

2700 M4 = (Q5 + Mc * M1 * M1) * O5 * O5 / 2

2710 IF M3 * Mc * G > M4 THEN 2760

2720 O9 = SQR(O5 * O5 - 2 * Mc * G * M3 / (Q5 + Mc * M1 * M1))

2730 GOSUB 3740

2740 GOSUB 4170

2750 GOTO 4330

2760 GOSUB 3740

2770 GOSUB 4240

2780 GOTO 4330

2790 REM

2795 REM B1 VERSION (POINT OF IMPACT OF THE ROPS BEHIND OF EQUILIBRIUM POINT)

2800 REM

2810 REM *

2820 O3 = 0: O4 = 0: O5 = 0: O6 = 0: O7 = 0: O8 = 0: O9 = 0

2830 Z(3, 6) = 0

2840 Q3 = Q * (COS(F2)) ^ 2 + 3 * Q * (SIN(F2)) ^ 2

2850 V5 = (Q3 + Mc * E4 * E4) * O2 * O2 / 2

2860 IF -V6 > A2 THEN 2880

2870 GOTO 2900

2880 V7 = E4 * (1 - COS(-A2 - V6))

2890 IF V7 * Mc * G > V5 THEN 3640

2900 V8 = E4 * COS(-A2 - V6) - E4 * COS(-A2 - ATN(X(3, 1) / Z(3, 1)))

2910 O3 = SQR(2 * Mc * G * V8 / (Q3 + Mc * E4 * E4) + O2 * O2)

2920 K9 = X(3, 1)

2930 K5 = Z(3, 1)

2940 K6 = Z(3, 1) + T3

2950 K7 = E2 - X(3, 1)

2960 K8 = U: DD2 = Q3 + Mc * K6 * K6 + Mc * K7 * K7

2970 O4 = (Q3 + Mc * K5 * K6 - K8 * Mc * K7 * K7 - (1 + K8) * Mc * K9 * K7) * O3 / DD2

2980 F3 = ATN(V0 / (Y(3, 4) - Y(3, 6)))

2990 O5 = O4 * COS(F3)

3000 REM TRANSFORMATION OF THE COORDINATES FROM 3 TO 4

3010 FOR K = 1 TO 7 STEP 1

3020 X(4, K) = X(3, K) * COS(F3) + (Y(3, K) - Y(3, 4)) * SIN(F3)

3030 Y(4, K) = (Y(3, K) - Y(3, 4)) * COS(F3) - X(3, K) * SIN(F3)

3040 Z(4, K) = Z(3, K)

3050 NEXT K

3060 A4 = ATN(TAN(A0) / SQR(1 + (TAN(F2 + F3)) ^ 2 / (COS(A0)) ^ 2))

3070 C3 = ATN(Z(4, 7) / X(4, 7))

3080 C4 = 0

3090 C5 = SQR(X(4, 7) * X(4, 7) + Z(4, 7) * Z(4, 7))

3100 C6 = C4 / C5

3110 C7 = C5 * (Y(4, 6) - Y(4, 1)) / (Y(4, 6) - Y(4, 7))

3120 C8 = C6 * C7

3130 M1 = SQR(X(4, 1) ^ 2 + Z(4, 1) ^ 2)

3140 M2 = ATN(X(4, 1) / Z(4, 1))

3150 REM ROTATION OF THE TRACTOR FROM THE POSITION 4 TO 5

3160 FOR K = 1 TO 7 STEP 1

3170 IF Z(4, K) <> 0 THEN 3200

3180 C9 = -3.14159 / 2

3190 GOTO 3210

3200 C9 = ATN(X(4, K) / Z(4, K))

3210 X(5, K) = SQR(X(4, K) ^ 2 + Z(4, K) ^ 2) * SIN(C9 + C3 + C6)

3220 Y(5, K) = Y(4, K)

3230 Z(5, K) = SQR(X(4, K) ^ 2 + Z(4, K) ^ 2) * COS(C9 + C3 + C6)

3240 NEXT K

3250 Z(5, 7) = 0

3260 Q5 = Q * (COS(F2 + F3)) ^ 2 + 3 * Q * (SIN(F2 + F3)) ^ 2

3270 IF -M2 > A4 THEN 3290

3280 GOTO 3320

3290 M3 = M1 * (1 - COS(-A4 - M2))

3300 M4 = (Q5 + Mc * M1 * M1) * O5 * O5 / 2

3310 IF M3 * Mc * G > M4 THEN 3640

3315 MM1 = M1 * COS(-A4 - ATN(X(5, 1) / Z(5, 1)))

3320 M5 = M1 * COS(-A4 - ATN(X(4, 1) / Z(4, 1))) - MM1

3330 O6 = SQR(2 * Mc * G * M5 / (Q5 + Mc * M1 * M1) + O5 * O5)

3340 M6 = X(5, 1)

3350 M7 = Z(5, 1)

3360 M8 = Z(5, 1) + C8

3370 M9 = C7 - X(5, 1)

3380 N1 = U: DD3 = (Q5 + Mc * M8 * M8 + Mc * M9 * M9)

3390 O7 = (Q5 + Mc * M7 * M8 - N1 * Mc * M9 * M9 - (1 + N1) * Mc * M6 * M9) * O6 / DD3

3400 F5 = ATN(C5 / (Y(5, 6) - Y(5, 7)))

3410 A6 = ATN(TAN(A0) / SQR(1 + (TAN(F2 + F3 + F5)) ^ 2 / (COS(A0)) ^ 2))

3420 REM TRANSFORMATION OF THE COORDINATES FROM THE POSITION 5 TO 6

3430 FOR K = 1 TO 7 STEP 1

3440 X(6, K) = X(5, K) * COS(F5) + (Y(5, K) - Y(5, 6)) * SIN(F5)

3450 Y(6, K) = (Y(5, K) - Y(5, 6)) * COS(F5) - X(5, K) * SIN(F5)

3460 Z(6, K) = Z(5, K)

3470 NEXT K

3480 O8 = O7 * COS(-F5)

3490 N2 = ATN(X(6, 1) / Z(6, 1))

3500 N3 = SQR(X(6, 1) ^ 2 + Z(6, 1) ^ 2)

3510 Q6 = Q * (COS(F2 + F3 + F5)) ^ 2 + 3 * Q * (SIN(F2 + F3 + F5)) ^ 2

3520 IF -N2 > A6 THEN 3540

3530 GOTO 3580

3540 N4 = N3 * (1 - COS(-A6 - N2))

3550 N5 = (Q6 + Mc * N3 * N3) * O8 * O8 / 2

3560 P9 = (N4 * Mc * G - N5) / (N4 * Mc * G)

3570 IF N4 * Mc * G > N5 THEN 3640

3580 IF -N2 < A6 THEN 3610

3590 N6 = -N4

3600 O9 = SQR(2 * Mc * G * N6 / (Q6 + Mc * N3 * N3) + O8 * O8)

3610 GOSUB 3740

3620 GOSUB 4170

3630 GOTO 4330

3640 GOSUB 3740

3650 GOSUB 4240

3660 GOTO 4330

3670 REM

3680 IF Z(3, 7) > -.2 THEN 2060

3685 CLS : PRINT : PRINT : PRINT STRING$(80, 42): LOCATE 24, 30, 0

3690 PRINT "THE ENGINE BONNET TOUCHES THE GROUND BEFORE THE ROPS"

3695 LPRINT STRING$(80, 42)

3700 LPRINT "THE ENGINE BONNET TOUCHES THE GROUND BEFORE THE ROPS"

3710 PRINT : PRINT "METHOD OF CALCULATION NOT FEASIBLE" : GOTO 3720

3715 CLS : PRINT : PRINT "METHOD OF CALCULATION NOT FEASIBLE"

3720 LPRINT "METHOD OF CALCULATION NOT FEASIBLE"

3725 LPRINT STRING$(80, 42)

3730 GOTO 4330

3740 REM ***

3750 CLS : LOCATE 13, 15, 0: PRINT "VELOCITY O0="

3755 LOCATE 13, 31, 0: PRINT USING"#.###"; O0: LOCATE 13, 40, 0: PRINT "rad/s"

3760 LOCATE 14, 15, 0: PRINT "VELOCITY O1="

3765 LOCATE 14, 31, 0: PRINT USING"#.###"; O1

3770 LOCATE 15, 15, 0: PRINT "VELOCITY O2="

3775 LOCATE 15, 31, 0: PRINT USING"#.###"; O2

3780 LOCATE 16, 15, 0: PRINT "VELOCITY O3="

3785 LOCATE 16, 31, 0: PRINT USING"#.###"; O3

3790 LOCATE 17, 15, 0: PRINT "VELOCITY O4="

3795 LOCATE 17, 31, 0: PRINT USING"#.###"; O4

3800 LOCATE 18, 15, 0: PRINT "VELOCITY O5="

3805 LOCATE 18, 31, 0: PRINT USING"#.###"; O5

3810 LOCATE 19, 15, 0: PRINT "VELOCITY O6="

3815 LOCATE 19, 31, 0: PRINT USING"#.###"; O6

3820 LOCATE 20, 15, 0: PRINT "VELOCITY O7="

3825 LOCATE 20, 31, 0: PRINT USING"#.###"; O7

3830 LOCATE 21, 15, 0: PRINT "VELOCITY O8="

3835 LOCATE 21, 31, 0: PRINT USING"#.###"; O8

3840 LOCATE 22, 15, 0: PRINT "VELOCITY O9="

3845 LOCATE 22, 31, 0: PRINT USING"#.###"; O9

3850 LPRINT "VELOCITY O0=";

3860 LPRINT USING"#.###"; O0;

3870 LPRINT " rad/s";

3880 LPRINT TAB(40); "VELOCITY O1=";

3890 LPRINT USING"#.###"; O1;

3900 LPRINT " rad/s"

3910 LPRINT "VELOCITY O2=";

3920 LPRINT USING"#.###"; O2;

3930 LPRINT " rad/s";

3940 LPRINT TAB(40); "VELOCITY O3=";

3950 LPRINT USING"#.###"; O3;

3960 LPRINT " rad/s"

3970 LPRINT "VELOCITY O4=";

3980 LPRINT USING"#.###"; O4;

3990 LPRINT " rad/s";

4000 LPRINT TAB(40); "VELOCITY O5=";

4010 LPRINT USING"#.###"; O5;

4020 LPRINT " rad/s"

4030 LPRINT "VELOCITY O6=";

4040 LPRINT USING"#.###"; O6;

4050 LPRINT " rad/s";

4060 LPRINT TAB(40); "VELOCITY O7=";

4070 LPRINT USING"#.###"; O7;

4080 LPRINT " rad/s"

4090 LPRINT "VELOCITY O8=";

4100 LPRINT USING"#.###"; O8;

4110 LPRINT " rad/s";

4120 LPRINT TAB(40); "VELOCITY O9=";

4130 LPRINT USING"#.###"; O9;

4140 LPRINT " rad/s"

4150 LPRINT

4160 RETURN

4170 PRINT STRING$(80, 42)

4180 LOCATE 24, 30, 0: PRINT "THE TILTING CONTINUES"

4190 PRINT STRING$(80, 42)

4200 LPRINT STRING$(80, 42)

4210 LPRINT TAB(30); "THE TILTING CONTINUES"

4220 LPRINT STRING$(80, 42)

4230 RETURN

4240 PRINT STRING$(80, 42)

4250 LOCATE 24, 30, 0: PRINT "THE ROLLING STOPS"

4260 PRINT STRING$(80, 42)

4270 LPRINT STRING$(80, 42)

4280 LPRINT TAB(30); "THE ROLLING STOPS"

4290 LPRINT STRING$(80, 42)

4300 RETURN

4310 REM ***

4320 REM END OF THE CALCULATION

4330 FOR I = 1 TO 5: LPRINT : NEXT: LPRINT " LOCATION : "; CAMPO$(6): LPRINT

4340 LPRINT " DATE : "; CAMPO$(7): LPRINT

4350 LPRINT ; " ENGINEER : "; CAMPO$(8): LPRINT

4360 FOR I = 1 TO 4: LPRINT : NEXT: PRINT

4370 INPUT " Do you whish to carry out another test ? (Y/N)"; Y$

4380 IF Y$ = "Y" OR Y$ = "y" THEN 190

4390 IF Y$ = "N" OR Y$ = "n" THEN SYSTEM

4400 LOCATE F(NC), C(NC) + L, 1: A$ = INKEY$: IF A$ = "" THEN GOTO 4400

4410 IF LEN(A$) > 1 THEN GOSUB 4570: GOTO 4400

4420 A = ASC(A$)

4430 IF A = 13 THEN L = 0: GOTO 4450

4440 GOTO 4470

4450 IF NC < 8 OR NC > 8 AND NC < 25 THEN NC = NC + 1: GOTO 4400

4460 GOTO 4840

4470 IF A > 31 AND A < 183 THEN GOTO 4490

4480 BEEP: GOTO 4400

4490 IF L = LON(NC) THEN BEEP: GOTO 4400

4500 LOCATE F(NC), C(NC) + L: PRINT A$;

4510 L = L + 1

4520 IF L = 1 THEN B$(NC) = A$: GOTO 4540

4530 B$(NC) = B$(NC) + A$

4540 IF LEN(C$(NC)) > 0 THEN C$(NC) = RIGHT$(CAMPO$(NC), LEN(CAMPO$(NC)) - L)

4550 CAMPO$(NC) = B$(NC) + C$(NC)

4560 GOTO 4400

4570 REM * SLIDE

4580 IF LEN(A$) <> 2 THEN BEEP: RETURN

4590 C = ASC(RIGHT$(A$, 1))

4600 IF C = 8 THEN 4620

4610 GOTO 4650

4620 IF LEN(C$(NC)) > 0 THEN BEEP: RETURN

4630 IF L = 0 THEN BEEP: RETURN

4640 CAMPO$(NC) = LEFT$(CAMPO$(NC), LEN(CAMPO(NC)))

4645 L = L - 1: PRINT A$: RETURN

4650 IF C = 30 THEN 4670

4660 GOTO 4700

4670 IF NC = 1 THEN BEEP: RETURN

4680 NC = NC - 1: L = 0

4690 RETURN

4700 IF C = 31 THEN 4720

4710 GOTO 4760

4720 IF NC <> 8 THEN 4740

4730 BEEP: RETURN

4740 NC = NC + 1: L = 0

4750 RETURN

4760 IF C = 29 THEN 4780

4770 GOTO 4800

4780 IF L = 0 THEN BEEP: RETURN

4790 L = L - 1: C$(NC) = RIGHT$(CAMPO$(NC), LEN(CAMPO$(NC)) - (L + 1))

4795 B$(NC) = LEFT$(CAMPO$(NC), L): LOCATE F(NC), C(NC) + L + 1: PRINT ""

4796 RETURN

4800 IF C = 28 THEN 4820

4810 GOTO 4400

4820 IF C$(NC) = "" THEN BEEP: RETURN

4830 L = L + 1: C$(NC) = RIGHT$(CAMPO$(NC), LEN(CAMPO$(NC)) - (L))

4835 B$(NC) = LEFT$(CAMPO$(NC), L): LOCATE F(NC), C(NC) + L, 1: PRINT ""

4840 RETURN

4850 RETURN

4860 FOR II = 1 TO 7

4870 X(1, II) = 0: X(2, II) = 0: X(3, II) = 0

4875 X(4, II) = 0: X(5, II) = 0: X(6, II) = 0

4880 Y(1, II) = 0: Y(2, II) = 0: Y(3, II) = 0

4885 Y(4, II) = 0: Y(5, II) = 0: Y(6, II) = 0

4890 Z(1, II) = 0: Z(2, II) = 0: Z(3, II) = 0

4895 Z(4, II) = 0: Z(5, II) = 0: Z(6, II) = 0

4900 NEXT II

4910 RETURN

4920 REM * THE SYMBOLS USED HERE ARE THE SAME AS IN THE CODE 6.

For tractor fitted with tracks, the following lines have to be substituted in the original model:

520 LOCATE 12, 40: PRINT ‘HEIGHT OF THE REAR TRACKS D3=’

*540 PRINT ‘HEIGHT OF THE FRT TRACKS D2=’: LOCATE 13, 29: PRINT ‘ ’

650 LOCATE 17, 40: PRINT ‘REAR TRACKS WIDTH B0=’

970 LPRINT TAB(40); ‘HEIGHT OF THE REAR TRACKS D3=’;

*980 LPRINT ‘HEIGHT OF THE FRT TRACKS D2=’;

1160 LPRINT TAB(40); ‘REAR TRACK WIDTH B0=’;

1390 W2 = SQR(H0 * H0 + L0 * L0): S1 = S/2 + B0/2

1530 F2 = 2 * ATN(- L0/D3 + SQR((L0/D3) ^ 2 – (D2/D3) + 1))

1590 X(1, 5) = D3

1660 Y(1, 5) = -L3

• *if applicable

Example 6.1

The tilting continues

TEST NR:

FRONT MOUNTED-OVER PROTECTIVE STRUCTURE OF THE NARROW TRACTOR:

Characteristic units:

LINEAR (m): MASS (kg):

MOMENT OF INERTIA (kgm²): ANGLE (radian)

HEIGHT OF THE COG	H1 = 0.7620	H. DIST. COG-REAR AXLE	L3 = 0.8970
H. DIST. COG - FRONT AXLE	L2 = 1.1490	HEIGHT OF THE REAR TYRES	D3 = 1.2930
HEIGHT OF THE FRT TYRES	D2 = 0.8800	OVERALL HEIGHT( PT IMPACT)	H6 = 2.1000
H. DIST. COG-LEAD PT INTER.	L6 = 0.2800	PROTECTIVE STRUCT. WIDTH	B6 = 0.7780
HEIGHT OF THE ENG. B.	H7 = 1.3370	WIDTH OF THE ENG. B.	B7 = 0.4900
H. DIST. COG-FRT COR. ENG. B.	L7 = 1.6390	HEIGHT FRT AXLE PIVOT PT	H0 = 0.4450
REAR TRACK WIDTH	S = 1.1150	REAR TYRE WIDTH	B0 = 0.1950
FRT AXLE SWING ANGLE	D0 = 0.1570	TRACTOR MASS	Mc = 2565.000
MOMENT OF INERTIA	Q = 295.0000
---	---
VELOCITY O0 = 3.881 rad/s	VELOCITY O1 = 1.078 rad/s
VELOCITY O2 = 1.057 rad/s	VELOCITY O3 = 2.134 rad/s
VELOCITY O4 = 0.731 rad/s	VELOCITY O5 = 0.000 rad/s
VELOCITY O6 = 0.000 rad/s	VELOCITY O7 = 0.000 rad/s
VELOCITY O8 = 0.000 rad/s	VELOCITY O9 = 0.000 rad/s
VELOCITY O0 = 3.881 rad/s	VELOCITY O1 = 1.078 rad/s
VELOCITY O2 = 1.057 rad/s	VELOCITY O3 = 2.134 rad/s
VELOCITY O4 = 1.130 rad/s	VELOCITY O5 = 0.993 rad/s
VELOCITY O6 = 0.810 rad/s	VELOCITY O7 = 0.629 rad/s
VELOCITY O8 = 0.587 rad/s	VELOCITY O9 = 0.219 rad/s

THE TILTING CONTINUES

Location:	Date:	Engineer:

Example 6.2

The rolling stops

TEST NR:

FRONT MOUNTED-OVER PROTECTIVE STRUCTURE OF THE NARROW TRACTOR:

Characteristic units:

LINEAR (m): MASS (kg):

MOMENT OF INERTIA (kgm²): ANGLE (radian)

HEIGHT OF THE COG	H1 = 0.7653	H. DIST. COG-REAR AXLE	L3 = 0.7970
H. DIST. COG - FRONT AXLE	L2 = 1.1490	HEIGHT OF THE REAR TYRES	D3 = 1.4800
HEIGHT OF THE FRT TYRES	D2 = 0.8800	OVERALL HEIGHT( PT IMPACT)	H6 = 2.1100
H. DIST. COG-LEAD PT INTER.	L6 = -0.0500	PROTECTIVE STRUCT. WIDTH	B6 = 0.7000
HEIGHT OF THE ENG. B.	H7 = 1.3700	WIDTH OF THE ENG. B.	B7 = 0.8000
H. DIST. COG-FRT COR. ENG. B.	L7 = 1.6390	HEIGHT FRT AXLE PIVOT PT	H0 = 0.4450
REAR TRACK WIDTH	S = 1.1150	REAR TYRE WIDTH	B0 = 0.1950
FRT AXLE SWING ANGLE	D0 = 0.1570	TRACTOR MASS	Mc = 1800.000
MOMENT OF INERTIA	Q = 250.0000
---	---
VELOCITY O0 = 3.840 rad/s	VELOCITY O1 = 0.281 rad/s
VELOCITY O2 = 0.268 rad/s	VELOCITY O3 = 1.586 rad/s
VELOCITY O4 = 0.672 rad/s	VELOCITY O5 = 0.000 rad/s
VELOCITY O6 = 0.000 rad/s	VELOCITY O7 = 0.000 rad/s
VELOCITY O8 = 0.000 rad/s	VELOCITY O9 = 0.000 rad/s
VELOCITY O0 = 3.840 rad/s	VELOCITY O1 = 0.281 rad/s
VELOCITY O2 = 0.268 rad/s	VELOCITY O3 = 1.586 rad/s
VELOCITY O4 = 0.867 rad/s	VELOCITY O5 = 0.755 rad/s
VELOCITY O6 = 1.218 rad/s	VELOCITY O7= 0.969 rad/s
VELOCITY O8 = 0.898 rad/s	VELOCITY O9 = 0.000 rad/s

THE ROLLING STOPS

Location:	Date:	Engineer:

Example 6.3

The rolling stops

TEST NR:

FRONT MOUNTED-OVER PROTECTIVE STRUCTURE OF THE NARROW TRACTOR:

Characteristic units:

LINEAR (m): MASS (kg):

MOMENT OF INERTIA (kgm²): ANGLE (radian)

HEIGHT OF THE COG	H1 = 0.7180	H. DIST. COG-REAR AXLE	L3 = 0.8000
H. DIST. COG - FRONT AXLE	L2 = 1.1590	HEIGHT OF THE REAR TYRES	D3 = 1.5200
HEIGHT OF THE FRT TYRES	D2 = 0.7020	OVERALL HEIGHT( PT IMPACT)	H6 = 2.0040
H. DIST. COG-LEAD PT INTER.	L6 = -0.2000	PROTECTIVE STRUCT. WIDTH	B6 = 0.6400
HEIGHT OF THE ENG. B.	H7 = 1.2120	WIDTH OF THE ENG. B.	B7 = 0.3600
H. DIST. COG-FRT COR. ENG. B.	L7 = 1.6390	HEIGHT FRT AXLE PIVOT PT	H0 = 0.4400
REAR TRACK WIDTH	S = 0.9000	REAR TYRE WIDTH	B0 = 0.3150
FRT AXLE SWING ANGLE	D0 = 0.1740	TRACTOR MASS	Mc = 1780.000
MOMENT OF INERTIA	Q = 279.8960
---	---
VELOCITY O0 = 3.884 rad/s	VELOCITY O1 = 0.107 rad/s
VELOCITY O2 = 0.098 rad/s	VELOCITY O3 = 0.000 rad/s
VELOCITY O4 = 0.000 rad/s	VELOCITY O5 = 0.000 rad/s
VELOCITY O6 = 0.000 rad/s	VELOCITY O7 = 0.000 rad/s
VELOCITY O8 = 0.000 rad/s	VELOCITY O9 = 0.000 rad/s
VELOCITY O0 = 3.884 rad/s	VELOCITY O1 = 0.107 rad/s
VELOCITY O2 = 0.098 rad/s	VELOCITY O3 = 0.000 rad/s
VELOCITY O4 = 0.000 rad/s	VELOCITY O5 = 0.000 rad/s
VELOCITY O6 = 0.000 rad/s	VELOCITY O7 = 0.000 rad/s
VELOCITY O8 = 0.000 rad/s	VELOCITY O9 = 0.000 rad/s

THE ROLLING STOPS

Location:	Date:	Engineer:

Example 6.4

The rolling stops

TEST NR:

FRONT MOUNTED-OVER PROTECTIVE STRUCTURE OF THE NARROW TRACTOR:

Characteristic units:

LINEAR (m): MASS (kg):

MOMENT OF INERTIA (kgm²): ANGLE (radian)

HEIGHT OF THE COG	H1 = 0.7180	H. DIST. COG-REAR AXLE	L3 = 0.8110
H. DIST. COG - FRONT AXLE	L2 = 1.1590	HEIGHT OF THE REAR TYRES	D3 = 1.2170
HEIGHT OF THE FRT TYRES	D2 = 0.7020	OVERALL HEIGHT( PT IMPACT)	H6 = 2.1900
H. DIST. COG-LEAD PT INTER.	L6 = -0.3790	PROTECTIVE STRUCT. WIDTH	B6 = 0.6400
HEIGHT OF THE ENG. B.	H7 = 1.2120	WIDTH OF THE ENG. B.	B7 = 0.3600
H. DIST. COG-FRT COR. ENG. B.	L7 = 1.6390	HEIGHT FRT AXLE PIVOT PT	H0 = 0.4400
REAR TRACK WIDTH	S = 0.9000	REAR TYRE WIDTH	B0 = 0.3150
FRT AXLE SWING ANGLE	D0 = 0.1740	TRACTOR MASS	Mc = 1780.000
MOMENT OF INERTIA	Q = 279.8960
---	---
VELOCITY O0 = 3.884 rad/s	VELOCITY O1 = 1.540 rad/s
VELOCITY O2 = 1.488 rad/s	VELOCITY O3 = 2.162 rad/s
VELOCITY O4 = 0.405 rad/s	VELOCITY O5 = 0.000 rad/s
VELOCITY O6 = 0.000 rad/s	VELOCITY O7 = 0.000 rad/s
VELOCITY O8 = 0.000 rad/s	VELOCITY O9 = 0.000 rad/s
VELOCITY O0 = 3.884 rad/s	VELOCITY O1 = 1.540 rad/s
VELOCITY O2 = 1.488 rad/s	VELOCITY O3 = 2.162 rad/s
VELOCITY O4 = 0.414 rad/s	VELOCITY O5 = 0.289 rad/s
VELOCITY O6 = 0.000 rad/s	VELOCITY O7= 0.000 rad/s
VELOCITY O8 = 0.000 rad/s	VELOCITY O9 = 0.000 rad/s

THE ROLLING STOPS

Location:	Date:	Engineer:

Example 6.5

The tilting continues

TEST NR:

FRONT MOUNTED-OVER PROTECTIVE STRUCTURE OF THE NARROW TRACTOR:

Characteristic units:

LINEAR (m): MASS (kg):

MOMENT OF INERTIA (kgm²): ANGLE (radian)

HEIGHT OF THE COG	H1 = 0.7660	H. DIST. COG-REAR AXLE	L3 = 0.7970
H. DIST. COG - FRONT AXLE	L2 = 1.1490	HEIGHT OF THE REAR TYRES	D3 = 1.4800
HEIGHT OF THE FRT TYRES	D2 = 0.8800	OVERALL HEIGHT( PT IMPACT)	H6 = 2.1100
H. DIST. COG-LEAD PT INTER.	L6 = -0.2000	PROTECTIVE STRUCT. WIDTH	B6 = 0.7000
HEIGHT OF THE ENG. B.	H7 = 1.3700	WIDTH OF THE ENG. B.	B7 = 0.8000
H. DIST. COG-FRT COR. ENG. B.	L7 = 1.6390	HEIGHT FRT AXLE PIVOT PT	H0 = 0.4450
REAR TRACK WIDTH	S = 1.1150	REAR TYRE WIDTH	B0 = 0.9100
FRT AXLE SWING ANGLE	D0 = 0.1570	TRACTOR MASS	Mc = 1800.000
MOMENT OF INERTIA	Q = 250.0000
---	---
VELOCITY O0 = 2.735 rad/s	VELOCITY O1 = 1.271 rad/s
VELOCITY O2 = 1.212 rad/s	VELOCITY O3 = 2.810 rad/s
VELOCITY O4 = 1.337 rad/s	VELOCITY O5 = 0.000 rad/s
VELOCITY O6 = 0.000 rad/s	VELOCITY O7 = 0.000 rad/s
VELOCITY O8 = 0.000 rad/s	VELOCITY O9 = 0.000 rad/s

THE TILTING CONTINUES

Location:	Date:	Engineer:

Example 6.6

The tilting continues

TEST NR:

FRONT MOUNTED-OVER PROTECTIVE STRUCTURE OF THE NARROW TRACTOR:

Characteristic units:

LINEAR (m): MASS (kg):

MOMENT OF INERTIA (kgm²): ANGLE (radian)

HEIGHT OF THE COG	H1 = 0.7653	H. DIST. COG-REAR AXLE	L3 = 0.7970
H. DIST. COG - FRONT AXLE	L2 = 1.1490	HEIGHT OF THE REAR TYRES	D3 = 1.2930
HEIGHT OF THE FRT TYRES	D2 = 0.8800	OVERALL HEIGHT( PT IMPACT)	H6 = 1.9600
H. DIST. COG-LEAD PT INTER.	L6 = -0.4000	PROTECTIVE STRUCT. WIDTH	B6 = 0.7000
HEIGHT OF THE ENG. B.	H7 = 1.3700	WIDTH OF THE ENG. B.	B7 = 0.8750
H. DIST. COG-FRT COR. ENG. B.	L7 = 1.6390	HEIGHT FRT AXLE PIVOT PT	H0 = 0.4450
REAR TRACK WIDTH	S = 1.1150	REAR TYRE WIDTH	B0 = 0.1950
FRT AXLE SWING ANGLE	D0 = 0.1570	TRACTOR MASS	Mc = 1800.000
MOMENT OF INERTIA	Q = 275.0000
---	---
VELOCITY O0 = 3.815 rad/s	VELOCITY O1 = 1.130 rad/s
VELOCITY O2 = 1.105 rad/s	VELOCITY O3 = 2.196 rad/s
VELOCITY O4 = 0.786 rad/s	VELOCITY O5 = 0.000 rad/s
VELOCITY O6 = 0.000 rad/s	VELOCITY O7 = 0.000 rad/s
VELOCITY O8 = 0.000 rad/s	VELOCITY O9 = 0.000 rad/s
VELOCITY O0 = 3.815 rad/s	VELOCITY O1 = 1.130 rad/s
VELOCITY O2 = 1.105 rad/s	VELOCITY O3 = 2.196 rad/s
VELOCITY O4 = 0.980 rad/s	VELOCITY O5 = 0.675 rad/s
VELOCITY O6 = 0.000 rad/s	VELOCITY O7 = 0.000 rad/s
VELOCITY O8 = 0.000 rad/s	VELOCITY O9 = 0.548 rad/s

THE TILTING CONTINUES

Location:	Date:	Engineer:

Example 6.7

Method of calculation not feasible

TEST NR:

FRONT MOUNTED-OVER PROTECTIVE STRUCTURE OF THE NARROW TRACTOR:

Characteristic units:

LINEAR (m): MASS (kg):

MOMENT OF INERTIA (kgm²): ANGLE (radian)

HEIGHT OF THE COG	H1 = 0.7620	H. DIST. COG-REAR AXLE	L3 = 0.7970
H. DIST. COG - FRONT AXLE	L2 = 1.1490	HEIGHT OF THE REAR TYRES	D3 = 1.5500
HEIGHT OF THE FRT TYRES	D2 = 0.8800	OVERALL HEIGHT( PT IMPACT)	H6 = 2.1000
H. DIST. COG-LEAD PT INTER.	L6 = -0.4780	PROTECTIVE STRUCT. WIDTH	B6 = 0.7780
HEIGHT OF THE ENG. B.	H7 = 1.5500	WIDTH OF THE ENG. B.	B7 = 0.9500
H. DIST. COG-FRT COR. ENG. B.	L7 = 1.6390	HEIGHT FRT AXLE PIVOT PT	H0 = 0.4450
REAR TRACK WIDTH	S = 1.1150	REAR TYRE WIDTH	B0 = 0.1950
FRT AXLE SWING ANGLE	D0 = 0.1570	MOMENT OF INERTIA	Q = 200.0000
TRACTOR MASS	Mc = 1800.000

THE ENGINE BONNET TOUCHES THE GROUND BEFORE THE ROPS

METHOD OF CALCULATION NOT FEASIBLE

Location:	Date:	Engineer:

Example 6.8

The rolling stops

TEST NR:

FRONT MOUNTED-OVER PROTECTIVE STRUCTURE OF THE NARROW TRACTOR:

Characteristic units:

LINEAR (m): MASS (kg):

MOMENT OF INERTIA (kgm²): ANGLE (radian)

HEIGHT OF THE COG	H1 = 0.7180	H. DIST. COG-REAR AXLE	L3 = 0.8110
H. DIST. COG - FRONT AXLE	L2 = 1.1590	HEIGHT OF THE REAR TYRES	D3 = 1.2170
HEIGHT OF THE FRT TYRES	D2 = 0.7020	OVERALL HEIGHT( PT IMPACT)	H6 = 2.0040
H. DIST. COG-LEAD PT INTER.	L6 = -0.3790	PROTECTIVE STRUCT. WIDTH	B6 = 0.6400
HEIGHT OF THE ENG. B.	H7 = 1.2120	WIDTH OF THE ENG. B.	B7 = 0.3600
H. DIST. COG-FRT COR. ENG. B.	L7 = 1.6390	HEIGHT FRT AXLE PIVOT PT	H0 = 0.4400
REAR TRACK WIDTH	S = 0.9000	REAR TYRE WIDTH	B0 = 0.3150
FRT AXLE SWING ANGLE	D0 = 0.1740	TRACTOR MASS	Mc = 1780.000
MOMENT OF INERTIA	Q = 279.8960
---	---
VELOCITY O0 = 3.884 rad/s	VELOCITY O1 = 1.540 rad/s
VELOCITY O2 = 1.488 rad/s	VELOCITY O3 = 2.313 rad/s
VELOCITY O4 = 0.581 rad/s	VELOCITY O5 = 0.000 rad/s
VELOCITY O6 = 0.000 rad/s	VELOCITY O7 = 0.000 rad/s
VELOCITY O8 = 0.000 rad/s	VELOCITY O9 = 0.000 rad/s
VELOCITY O0 = 3.884 rad/s	VELOCITY O1 = 1.540 rad/s
VELOCITY O2 = 1.488 rad/s	VELOCITY O3 = 2.313 rad/s
VELOCITY O4 = 0.633 rad/s	VELOCITY O5 = 0.373 rad/s
VELOCITY O6 = 0.000 rad/s	VELOCITY O7 = 0.000 rad/s
VELOCITY O8 = 0.000 rad/s	VELOCITY O9 = 0.000 rad/s

THE ROLLING STOPS

Location:	Date:	Engineer:

Example 6.9

The tilting continues

TEST NR:

FRONT MOUNTED-OVER PROTECTIVE STRUCTURE OF THE NARROW TRACTOR:

Characteristic units:

LINEAR (m): MASS (kg):

MOMENT OF INERTIA (kgm²): ANGLE (radian)

HEIGHT OF THE COG	H1 = 0.7620	H. DIST. COG-REAR AXLE	L3 = 0.7970
H. DIST. COG - FRONT AXLE	L2 = 1.1490	HEIGHT OF THE REAR TYRES	D3 = 1.2930
HEIGHT OF THE FRT TYRES	D2 = 0.8800	OVERALL HEIGHT( PT IMPACT)	H6 = 1.9670
H. DIST. COG-LEAD PT INTER.	L6 = -0.3000	PROTECTIVE STRUCT. WIDTH	B6 = 0.7700
HEIGHT OF THE ENG. B.	H7 = 1.3500	WIDTH OF THE ENG. B.	B7 = 0.9500
H. DIST. COG-FRT COR. ENG. B.	L7 = 1.6390	HEIGHT FRT AXLE PIVOT PT	H0 = 0.4450
REAR TRACK WIDTH	S = 1.1150	REAR TYRE WIDTH	B0 = 0.1950
FRT AXLE SWING ANGLE	D0 = 0.1570	TRACTOR MASS	Mc = 1800.000
MOMENT OF INERTIA	Q = 300.0000
---	---
VELOCITY O0 = 3.790 rad/s	VELOCITY O1 = 1.159 rad/s
VELOCITY O2 = 1.133 rad/s	VELOCITY O3 = 2.118 rad/s
VELOCITY O4 = 0.801 rad/s	VELOCITY O5 = 0.000 rad/s
VELOCITY O6 = 0.000 rad/s	VELOCITY O7 = 0.000 rad/s
VELOCITY O8 = 0.000 rad/s	VELOCITY O9 = 0.000 rad/s
VELOCITY O0 = 3.790 rad/s	VELOCITY O1 = 1.159 rad/s
VELOCITY O2 = 1.133 rad/s	VELOCITY O3 = 2.118 rad/s
VELOCITY O4 = 0.856 rad/s	VELOCITY O5 = 0.562 rad/s
VELOCITY O6 = 0.000 rad/s	VELOCITY O7 = 0.000 rad/s
VELOCITY O8 = 0.000 rad/s	VELOCITY O9 = 0.205 rad/s

THE TILTING CONTINUES

Location:	Date:	Engineer:

Example 6.10

The tilting continues

TEST NR:

FRONT MOUNTED-OVER PROTECTIVE STRUCTURE OF THE NARROW TRACTOR:

Characteristic units:

LINEAR (m): MASS (kg):

MOMENT OF INERTIA (kgm²): ANGLE (radian)

HEIGHT OF THE COG	H1 = 0.7653	H. DIST. COG-REAR AXLE	L3 = 0.7970
H. DIST. COG - FRONT AXLE	L2 = 1.1490	HEIGHT OF THE REAR TYRES	D3 = 1.3800
HEIGHT OF THE FRT TYRES	D2 = 0.8800	OVERALL HEIGHT( PT IMPACT)	H6 = 1.9600
H. DIST. COG-LEAD PT INTER.	L6 = -0.3000	PROTECTIVE STRUCT. WIDTH	B6 = 0.7000
HEIGHT OF THE ENG. B.	H7 = 1.3700	WIDTH OF THE ENG. B.	B7 = 0.8900
H. DIST. COG-FRT COR. ENG. B.	L7 = 1.6390	HEIGHT FRT AXLE PIVOT PT	H0 = 0.4450
REAR TRACK WIDTH	S = 1.1150	REAR TYRE WIDTH	B0 = 0.1950
FRT AXLE SWING ANGLE	D0 = 0.1570	TRACTOR MASS	Mc = 1800.000
MOMENT OF INERTIA	Q = 275.0000
---	---
VELOCITY O0 = 3.815 rad/s	VELOCITY O1 = 0.748 rad/s
VELOCITY O2 = 0.724 rad/s	VELOCITY O3 = 1.956 rad/s
VELOCITY O4 = 0.808 rad/s	VELOCITY O5 = 0.000 rad/s
VELOCITY O6 = 0.000 rad/s	VELOCITY O7 = 0.000 rad/s
VELOCITY O8 = 0.000 rad/s	VELOCITY O9 = 0.407 rad/s

THE TILTING CONTINUES

Location:	Date:	Engineer:

Example 6.11

The rolling stops

TEST NR:

FRONT MOUNTED-OVER PROTECTIVE STRUCTURE OF THE NARROW TRACTOR:

Characteristic units:

LINEAR (m): MASS (kg):

MOMENT OF INERTIA (kgm²): ANGLE (radian)

EIGHT OF THE COG	H1 = 0.7653	H. DIST. COG-REAR AXLE	L3 = 0.7970
H. DIST. COG - FRONT AXLE	L2 = 1.1490	HEIGHT OF THE REAR TYRES	D3 = 1.4800
HEIGHT OF THE FRT TYRES	D2 = 0.9000	OVERALL HEIGHT( PT IMPACT)	H6 = 1.9600
H. DIST. COG-LEAD PT INTER.	L6 = -0.4000	PROTECTIVE STRUCT. WIDTH	B6 = 0.7000
HEIGHT OF THE ENG. B.	H7 = 1.3700	WIDTH OF THE ENG. B.	B7 = 0.8000
H. DIST. COG-FRT COR. ENG. B.	L7 = 1.6390	HEIGHT FRT AXLE PIVOT PT	H0 = 0.4450
REAR TRACK WIDTH	S = 1.1150	REAR TYRE WIDTH	B0 = 0.1950
FRT AXLE SWING ANGLE	D0 = 0.1570	TRACTOR MASS	Mc = 1800.000
MOMENT OF INERTIA	Q = 250.0000
---	---
VELOCITY O0 = 3.840	VELOCITY O1 = 0.246
VELOCITY O2 = 0.235	VELOCITY O3 = 0.000
VELOCITY O4 = 0.000	VELOCITY O5 = 0.000
VELOCITY O6 = 0.000	VELOCITY O7 = 0.000
VELOCITY O8 = 0.000	VELOCITY O9 = 0.000
VELOCITY O0 = 3.840	VELOCITY O1 = 0.246
VELOCITY O2 = 0.235	VELOCITY O3 = 0.000
VELOCITY O4 = 0.000	VELOCITY O5 = 0.000
VELOCITY O6 = 0.000	VELOCITY O7 = 0.000
VELOCITY O8 = 0.000	VELOCITY O9 = 0.000

THE ROLLING STOPS

Location:	Date:	Engineer:

Explanatory notes to Annex IX:

• ⁽¹⁾ Unless the numbering of sections B2 and B3 which has been harmonised with the whole Annex, the text of the requirements and the numbering set out in point B are identical with the text and numbering of the OECD standard Code for the official testing of front mounted roll-over protective structures on narrow-track agricultural and forestry tractors, OECD standard Code 6, Edition 2017 of February 2017.

• ⁽²⁾ Users are reminded that the seat index point is determined according to ISO 5353:1995 and is a fixed point with respect to the tractor that does not move as the seat is adjusted away from the mid-position. For purposes of determining the clearance zone, the seat shall be placed in the rear and uppermost position.

• ⁽³⁾ The programme and the examples are available on the OECD website.

• ⁽⁴⁾ Permanent + elastic deflection measured at the point when the required energy level is obtained.

ANNEX X

Requirements applying to roll-over protection structures (rear mounted roll-over protective structures on narrow-track tractors)

A. GENERAL PROVISIONS

• 1.The Union requirements applying to roll-over protection structures (rear mounted roll-over protective structures on narrow track tractors) are set out in point B.

• 2.Tests may be performed in accordance with the static or alternatively the dynamic test procedures as set out in sections B1 and B2. The two methods are deemed equivalent.

• 3.In addition to the requirements set out in point 2, the performance requirements for foldable roll-over protective structures (ROPS) set out in section B3 shall be met.

B. REQUIREMENTS APPLYING TO ROLL-OVER PROTECTION STRUCTURES (REAR MOUNTED ROLL-OVER PROTECTIVE STRUCTURES ON NARROW TRACK TRACTORS)⁽¹⁾

1. Definitions

1.1.   [Not applicable]

1.2.   Roll-Over Protective Structure (ROPS)

Roll-over protective structure (protective cab or frame), hereinafter called ‘protective structure’, means the structure on a tractor the essential purpose of which is to avoid or limit risks to the driver resulting from roll-over of the tractor during normal use.

The roll-over protective structure is characterized by the provision of space for a clearance zone large enough to protect the driver when seated either inside the envelope of the structure or within a space bounded by a series of straight lines from the outer edges of the structure to any part of the tractor that might come into contact with flat ground and that is capable of supporting the tractor in that position if the tractor overturns.

1.3.   Track

1.3.1.   Preliminary definition: median plane of the wheel or track.

The median plane of the wheel is equidistant from the two planes containing the periphery of the rims or tracks at their outer edges.

1.3.2.   Definition of track

The vertical plane through the wheel axis intersects its median plane along a straight line which meets the supporting surface at one point. If A and B are the two points thus defined for the wheels on the same axle of the tractor, then the track width is the distance between points A and B. The track may be thus defined for both front and rear wheels. Where there are twin wheels, the track is the distance between two planes each being the median plane of the pairs of wheels. For track-laying tractors, the track is the distance between the median planes of the tracks.

1.3.3.   Additional definition: median plane of the tractor

Take the extreme positions of points A and B for the tractor rear axle, which gives the maximum possible value for the track. The vertical plane at right angles to the line AB at its centre point is the median plane of the tractor.

1.4.   Wheelbase

The distance between the vertical planes passing through the two lines AB as defined above, one for the front wheels and one for the rear-wheels.

1.5.   Determination of seat index point; Seat location and adjustment for test

1.5.1.   Seat index point (SIP)⁽²⁾

The seat index point shall be determined in accordance with ISO 5353:1995

1.5.2.   Seat location and adjustment for test

• 1.5.2.1. where the seat position is adjustable, the seat must be adjusted to its rear uppermost position;

• 1.5.2.2. where the inclination of the backrest is adjustable, it must be adjusted to the mid position;

• 1.5.2.3. where the seat is equipped with suspension, the latter must be blocked at mid-travel, unless this is contrary to the instructions clearly laid down by the seat manufacturer;

• 1.5.2.4. where the position of the seat is adjustable only lengthwise and vertically, the longitudinal axis passing through the Seat Index Point shall be parallel with the vertical longitudinal plane of the tractor passing through the centre of the steering wheel and not more than 100 mm from that plane.

1.6.   Clearance zone

1.6.1.   Reference plane

The clearance zone is illustrated in figures 7.1 and 7.2. The zone is defined in relation to the reference plane and the Seat Index Point. The reference plane is a vertical plane, generally longitudinal to the tractor and passing through the Seat Index Point and the centre of the steering wheel. Normally the reference plane coincides with the longitudinal median plane of the tractor. This reference plane shall be assumed to move horizontally with the seat and steering wheel during loading but to remain perpendicular to the tractor or the floor of the roll-over protective structure. The clearance zone shall be defined on the basis of Sections 1.6.2 and 1.6.3.

1.6.2.   Determination of the clearance zone for tractors with a non-reversible seat

The clearance zone for tractors with a non-reversible seat is defined in 1.6.2.1 to 1.6.2.13 below and is bounded by the following planes, the tractor being on a horizontal surface, the seat adjusted and located as specified in Sections 1.5.2.1 to 1.5.2.4⁽²⁾, and the steering wheel, where adjustable, adjusted to the mid position for seated driving:

• 1.6.2.1. a horizontal plane A1 B1 B2 A2, (810 + a

h-10) mm behind the SIP;

• 1.6.2.2. an inclined plane H1 H2 G2 G1, perpendicular to the reference plane, including both a point 150 mm behind line B1B2 and the rearmost point of the seat backrest;

• 1.6.2.3. a cylindrical surface A1 A2 H2 H1 perpendicular to the reference plane, having a radius of 120 mm, tangential to the planes defined in 1.6.2.1 and 1.6.2.2 above;

• 1.6.2.4. a cylindrical surface B1 C1 C2 B2, perpendicular to the reference plane, having a radius of 900 mm extending forward for 400 mm and tangential to the plane defined in 1.6.2.1 above along line B1B2;

• 1.6.2.5. an inclined plane C1 D1 D2 C2, perpendicular to the reference plane, joining the surface defined in 1.6.2.4 above and passing 40 mm from the forward external edge of the steering wheel. In the case of a high steering wheel position, this plane extends forward from line B1B2 tangentially to the surface defined in 1.6.2.4 above;

• 1.6.2.6. a vertical plane D1 K1 E1 E2 K2 D2 perpendicular to the reference plane 40 mm forward of the external edge of the steering wheel;

• 1.6.2.7. a horizontal plane E1 F1 P1 N1 N2 P2 F2 E2 passing through a point (90-av) mm below the Seat Index Point;

• 1.6.2.8. a surface G1 L1 M1 N1 N2 M2 L2 G2, if necessary curved from the bottom limit of the plane defined in 1.6.2.2 above to the horizontal plane defined in 1.6.2.7 above, perpendicular to the reference plane, and in contact with the seat backrest throughout its length;

• 1.6.2.9. two vertical planes K1 I1 F1 E1 and K2 I2 F2 E2 parallel to the reference plane, 250 mm either side of the reference plane, and bounded at the top 300 mm above the plane defined in 1.6.2.7 above;

• 1.6.2.10. two inclined and parallel planes A1 B1 C1 D1 K1 I1 L1 G1 H1 and A2 B2 C2 D2 K2 I2 L2 G2 H2 starting from the upper edge of the planes defined in 1.6.2.9 above and joining the horizontal plane defined in 1.6.2.1 above at least 100 mm from the reference plane on the side where the loading is applied;

• 1.6.2.11. two portions of vertical planes Q1 P1 N1 M1 and Q2 P2 N2 M2 parallel to the reference plane, 200 mm either side of the reference plane, and bounded towards the top 300 mm above the horizontal plane defined in 1.6.2.7 above

• 1.6.2.12. two portions of I1 Q1 P1 F1 and I2 Q2 P2 F2 of a vertical plane, perpendicular to the reference plane and passing (210-ah) mm in front of the SIP;

• 1.6.2.13. two portions I1 Q1 M1 L1 and I2 Q2 M2 L2 of the horizontal plane passing 300 mm above the plane defined in 1.6.2.7 above.

1.6.3.   Determination of the clearance zone for tractors with a reversible driver’s position

For tractors with a reversible driver’s position (reversible seat and steering wheel), the clearance zone is the envelope of the two clearance zones defined by the two different positions of the steering wheel and the seat.

• 1.6.3.1. If the protective structure is a of a rear two-post type, for each position of the steering wheel and of the seat, the clearance zone shall respectively be defined on the basis of above sections 1.6.1 and 1.6.2 for driver’s position in normal position, and on the basis of sections 1.6.1 and 1.6.2 of Annex IX for driver’s position in reverse position (see figure 7.2.a).

• 1.6.3.2. If the protective structure is of another type, for each position of the steering wheel and of the seat, the clearance zone shall be defined on the basis of sections 1.6.1 and 1.6.2 of this Annex (see figure 7.2.b).

1.6.4.   Optional seats

• 1.6.4.1. In case of tractors that could be fitted with optional seats, the envelope comprising the Seat Index Points of all options offered shall be used during the tests. The protective structure shall not enter the larger clearance zone which takes account of these different Seat Index Points.

• 1.6.4.2. In the case where a new seat option is offered after the test has been performed, a determination shall be made to see whether the clearance zone around the new SIP falls within the envelope previously established. If it does not, a new test must be performed.

• 1.6.4.3. Optional seat does not include a seat for a person in addition to the driver and from where the tractor cannot be controlled. The SIP shall not be determined because the definition of the clearance zone is in relation to the driver seat.

1.7.   Mass

1.7.1.   Unballasted/Unladen Mass

The mass of the tractor excluding optional accessories but including coolant, oils, fuel, tools plus the protective structure. Not included are optional front or rear weights, tyre ballast, mounted implements, mounted equipment or any specialised components;

1.7.2.   Maximum Permissible Mass

The maximum mass of the tractor stated by the manufacturer to be technically permissible and declared on the vehicle’s identification plate and/or in the Operator’s Handbook;

1.7.3.   Reference Mass

The mass, selected by the manufacturer, used in formulae to calculate the height of fall of the pendulum block, the energy inputs and crushing forces to be used in the tests. Must not be less than the unballasted mass and must be sufficient to ensure the Mass Ratio does not exceed 1,75 (see Section 1.7.4);

1.7.4.   Mass Ratio

The ratio of

Image omitted; see official source

This must not be greater than 1,75

1.8.   Permissible measurement tolerances

Linear dimension:	± 3 mm
except for: – tyre deflection:	± 1 mm
–structure deflection during horizontal loadings:	± 1 mm
– height of fall of the pendulum block:	± 1 mm
Masses:	± 0,2 % (of the sensor full scale)
Forces:	± 0,1 % (of the sensor full scale)
Angles:	± 0,1°

1.9.   Symbols

• (mm) Half of the horizontal seat adjustment

• (mm) Half of the vertical seat adjustment

• (mm) Minimum overall width of the tractor;

• (mm) Maximum outer width of the protective structure;

• (mm) Deflection of the structure at the point of impact (dynamic tests) or at the point of, and in line with, the load application (static tests);

• (mm) Deflection of the structure for the calculated energy required;

• (J) Strain energy absorbed at point when load is removed. Area contained within F-D curve;

• (J) Strain energy absorbed. Area under F-D curve;

• (J) Strain energy absorbed after additional loading following a crack or tear;

• (J) Strain energy absorbed in overload test in the event of the load having been removed before starting this overload test. Area under F-D curve;

• (J) Energy input to be absorbed during longitudinal loading;

• (J) Energy input to be absorbed during side loading;

• (N) Static load force;

• (N) Loading force for calculated energy required, corresponding to E’i ;

Force/deflection diagram;

• (N) Maximum static load force occurring during loading, with the exception of the overload;

• (N) Vertical crushing force;

• (mm) Falling height of the pendulum block (dynamic tests);

• (mm) Falling height of the pendulum block for additional test (dynamic tests);

• (kgm²) Tractor reference moment of inertia about the centre line of the rear wheels, whatever the mass of these rear wheels may be;

• (mm) Tractor reference wheelbase;

• (kg) Tractor reference mass during strength tests.

2. Field of application

• 2.1.This Annex shall is applicable to tractors having at least two axles for pneumatic tyred wheels or having tracks instead of wheels and having the following characteristics:

• 2.1.1. ground clearance of not more than 600 mm beneath the lowest points of the front and rear axles, allowing for the differential;

• 2.1.2. fixed or adjustable minimum track width with one of the axles less than 1 150 mm fitted with tyres of a larger size. It is assumed that the axle mounted with the wider tyres is set at a track width of not more than 1 150 mm. It must be possible to set the track width of the other axle in such a way that the outer edges of the narrower tyres do not go beyond the outer edges of the tyres of the other axle. Where the two axles are fitted with rims and tyres of the same size, the fixed or adjustable track width of the two axles must be less than 1 150 mm;

• 2.1.3. mass greater than 400 kg unladen but including the roll-over protective structure and tyres of the largest size recommended by the manufacturer. For tractors with a reversible driver’s position (reversible seat and steering wheel), the unladen mass shall be less than 3 500 kg and the maximum permissible mass shall not exceed 5 250 kg. For all tractors, the Mass Ratio (Maximum Permissible Mass / Reference Mass) must not be greater than 1,75.

• 2.1.4. roll-over protective structure of the rollbar, frame or cab type, mounted partly or entirely behind the Seat Index Point and having a zone of clearance whose upper limit is (810 + av) mm above the seat index point in order to provide a sufficiently large area or unobstructed space for the protection of the driver.

• 2.2.It is recognised that there may be designs of tractors, for example, special forestry machines, such as forwarders and skidders, for which this Annex is not applicable.

B1   STATIC TEST PROCEDURE

3. Rules and directions

3.1.   Conditions for testing the strength of protective structures and of their attachment to tractors

3.1.1.   General requirements

3.1.1.1.   Test purposes

Tests made using special rigs are intended to simulate such loads as are imposed on a protective structure, when the tractor overturns. These tests enable observations to be made on the strength of the protective structure and any brackets attaching it to the tractor and any parts of the tractor which transmit the test load.

3.1.1.2.   Test methods

Tests may be performed in accordance with the static procedure or the dynamic procedure (see Annex II). The two methods are deemed equivalent.

3.1.1.3.   General rules governing preparation for tests

• 3.1.1.3.1.The protective structure must conform to the series production specifications. It shall be attached in accordance with the manufacturer’s recommended method to one of the tractors for which it is designed.

Note: A complete tractor is not required for the static strength test; however, the protective structure and parts of the tractor to which it is attached represent an operating installation, hereinafter referred to as ‘the assembly’.

• 3.1.1.3.2.For both the static test and the dynamic test the tractor as assembled (or the assembly) must be fitted with all series production components which may affect the strength of the protective structure or which may be necessary for the strength test.

• Components which may create a hazard in the clearance zone must also be fitted on the tractor (or the assembly) so that they may be examined to see whether the requirements of the Acceptance Conditions in 3.1.3 have been fulfilled. All components of the tractor or the protective structure including weather protective must be supplied or described on drawings.

• 3.1.1.3.3.For the strength tests, all panels and detachable non-structural components must be removed so that they may not contribute to the strengthening of the protective structure.

• 3.1.1.3.4.The track width must be adjusted so that the protective structure will, as far as possible, not be supported by the tyres or the tracks during the strength tests. If these tests are conducted in accordance with the static procedure, the wheels or tracks may be removed.

3.1.2.   Tests

3.1.2.1.   Sequence of tests according to the Static Procedure

The sequence of tests, without prejudice to the additional tests mentioned in sections 3.2.1.6, and 3.2.1.7 is as follows:

• (1) loading at the rear of the structure (see 3.2.1.1);

• (2) rear crushing test (see 3.2.1.4);

• (3) loading at the front of the structure (see 3.2.1.2);

• (4) loading at the side of the structure (see 3.2.1.3);

• (5) crushing at the front of the structure (see 3.2.1.5).

3.1.2.2.   General requirements

• 3.1.2.2.1.If, during the test, any part of the tractor restraining equipment breaks or moves, the test shall be restarted.

• 3.1.2.2.2.No repairs or adjustments of the tractor or protective structure may be carried out during the tests.

• 3.1.2.2.3.The tractor gear box shall be in neutral and the brakes off during the tests.

• 3.1.2.2.4.If the tractor is fitted with a suspension system between the tractor body and the wheels, it shall be blocked during the tests.

• 3.1.2.2.5.The side chosen for application of the first load on the rear of the structure shall be that which, in the opinion of the testing authorities, will result in the application of the series of loads under the most unfavourable conditions for the structure. The lateral load shall be applied to the opposite side of the median plane of the tractor than the longitudinal load. The front load shall be applied on the same side of the longitudinal median plane of the protective structure as the lateral load.

3.1.3.   Acceptance conditions

• 3.1.3.1. A protective structure is regarded as having satisfied the strength requirements if it fulfils the following conditions:

• 3.1.3.1.1. during static testing, at the point when the energy required is attained in each horizontal load test prescribed or in the overload test the force must be greater than 0,8 F;

• 3.1.3.1.2. if during the test, cracks or tears appear as a result of the application of the crushing force, an additional crushing test as defined in 3.2.1.7 must be performed immediately after the crushing test which caused these cracks or tears to appear;

• 3.1.3.1.3. during the tests other than the overload test, no part of the protective structure must enter the clearance zone as defined in 1.6;

• 3.1.3.1.4. during the tests other than the overload test, all parts of the clearance zone shall be secured by the structure, in accordance with 3.2.2.2;

• 3.1.3.1.5. during the tests the protective structure must not impose any constraints on the seat structure;

• 3.1.3.1.6. the elastic deflection, measured in accordance with 3.2.2.3 shall be less than 250 mm.

• 3.1.3.2. There shall be no accessories presenting a hazard for the driver. There shall be no projecting part or accessory which is liable to injure the driver should the tractor overturn, or any accessory or part which is liable to trap him — for example by the leg or the foot — as a result of the deflections of the structure.

3.1.4.   [Not applicable]

3.1.5.   Test apparatus and equipment

3.1.5.1.   Static testing rig

• 3.1.5.1.1.The static testing rig must be designed in such a way as to permit thrusts or loads to be applied to the protective structure.

• 3.1.5.1.2.Provision must be made so that the load can be uniformly distributed normal to the direction of loading and along a flange having a length of one of the exact multiples of 50 between 250 and 700 mm. The stiff beam shall have a vertical face dimension of 150 mm. The edges of the beam in contact with the protective structure shall be curved with a maximum radius of 50 mm.

• 3.1.5.1.3.The pad shall be capable of being adjusted to any angle in relation to the load direction, in order to be able to follow the angular variations of the structure’s load-bearing surface as the structure deflects.

• 3.1.5.1.4.Direction of the force (deviation from horizontal and vertical):

• — at start of test, under zero load: ± 2°;

• — during test, under load: 10° above and 20° below the horizontal. These variations must be kept to a minimum.

• 3.1.5.1.5.The deflection rate shall be sufficiently slow, less than 5 mm/s so that the load may at all moments be considered as static.

3.1.5.2.   Apparatus for measuring the energy absorbed by the structure

• 3.1.5.2.1.The force versus deflection curve shall be plotted in order to determine the energy absorbed by the structure. There is no need to measure the force and deflection at the point where the load is applied to the structure; however, force and deflection shall be measured simultaneously and co-linearly.

• 3.1.5.2.2.The point of origin of deflection measurements shall be selected so as to take account only of the energy absorbed by the structure and/or by the deflection of certain parts of the tractor. The energy absorbed by the deflection and/or the slipping of the anchoring must be ignored.

3.1.5.3.   Means of anchoring the tractor to the ground

• 3.1.5.3.1.Anchoring rails with the requisite track width and covering the necessary area for anchoring the tractor in all the cases illustrated must be rigidly attached to a non-yielding base near the testing rig.

• 3.1.5.3.2.The tractor must be anchored to the rails by any suitable means (plates, wedges, wire ropes, jacks, etc.) so that it cannot move during the tests. This requirement shall be checked during the test, by means of the usual devices for measuring length.

• If the tractor moves, the entire test shall be repeated, unless the system for measuring the deflections taken into account for plotting the force versus deflection curve is connected to the tractor.

3.1.5.4.   Crushing rig

A rig as shown in figure 7.3 shall be capable of exerting a downward force on a protective structure through a rigid beam approximately 250 mm wide, connected to the load-applying mechanism by means of universal joints. Suitable axle stands must be provided so that the tractor tyres do not bear the crushing force.

3.1.5.5.   Other measuring apparatus

The following measuring devices are also needed:

• 3.1.5.5.1. device for measuring the elastic deflection (the difference between the maximum momentary deflection and the permanent deflection, see figure 7.4).

• 3.1.5.5.2. device for checking that the protective structure has not entered the clearance zone and that the latter has remained within the structure’s protective during the test (section 3.2.2.2).

3.2.   Static test procedure

3.2.1.   Loading and crushing tests

3.2.1.1.   Loading at the rear

• 3.2.1.1.1.The load shall be applied horizontally, in a vertical plane parallel to the tractor’s median plane.

• The load application point shall be that part of the roll-over protective structure likely to hit the ground first in a rearward overturning accident, normally the upper edge. The vertical plane in which the load is applied shall be located at a distance of 1/6 of the width of the top of the protective structure inwards from a vertical plane, parallel to the median plane of the tractor, touching the outside extremity of the top of the protective structure.

• If the structure is curved or protruding at this point, wedges enabling the load to be applied thereon shall be added, without thereby reinforcing the structure.

• 3.2.1.1.2.The assembly shall be lashed to the ground as described in 3.1.6.3.

• 3.2.1.1.3.The energy absorbed by the protective structure during the test shall be at least:

Image omitted; see official source

• or

Image omitted; see official source

• 3.2.1.1.4.For tractors with a reversible driver’s position (reversible seat and steering wheel), the energy shall be whichever is the higher of the formula selected above or the following:

Image omitted; see official source

3.2.1.2.   Loading at the front

• 3.2.1.2.1.The load shall be applied horizontally in a vertical plane parallel to the tractor’s median plane. The point of application shall be that part of the protective structure likely to hit the ground first if the tractor overturns sideways while travelling forward, i.e. normally the upper edge. The point of application of the load shall be 1/6 of the width of the top of the protective structure inwards from a vertical plane parallel to the median plane of the tractor touching the outside extremity of the top of the protective structure.

• If the structure is curved or protruding at this point, wedges enabling the load to be applied thereon shall be added, without thereby reinforcing the structure.

• 3.2.1.2.2.The assembly shall be lashed to the ground as described in 3.1.6.3.

• 3.2.1.2.3.The energy absorbed by the protective structure during the test shall be at least:

Image omitted; see official source

• 3.2.1.2.4.In case of tractors with a reversible driver’s position (reversible seat and steering wheel):

• — if the protective structure is a rear two-post rollbar the preceding formula shall also apply;

• — for other types of protective structures, the energy shall be whichever is the higher of the above or either of the following as selected: Image omitted; see official source or Image omitted; see official source

3.2.1.3.   Loading from the side

• 3.2.1.3.1.The side loading shall be applied horizontally, in a vertical plane perpendicular to the tractor’s median plane passing 60 mm in front of the Seat Index Point, the seat being at the mid position of the longitudinal adjustment. The load application point shall be that part of the roll-over protective structure likely to hit the ground first in a sideways overturning accident, normally the upper edge.

• 3.2.1.3.2.The assembly shall be lashed to the ground as described in 3.1.6.3.

• 3.2.1.3.3.The energy absorbed by the protective structure during the test shall be at least:

Image omitted; see official source

• 3.2.1.3.4.For tractors with a reversible driver’s position (reversible seat and steering wheel), the load application point shall be in the plane at right angles to the median plane and passing at the midpoint of the segment joining the two Seat Index Points defined by joining the two different positions of the seat. For protective structures having a two-post system, the load shall be located on one of the two posts.

• 3.2.1.3.5.In case of tractors with a reversible driver’s position (reversible seat and steering wheel) where the protective structure is a rear two-post rollbar, the energy shall be whichever is higher of the following:

Image omitted; see official source

• or

Image omitted; see official source

3.2.1.4.   Crushing at the rear

The beam shall be positioned over the rear uppermost structural member(s) and the resultant of crushing forces shall be located in the tractor’s median plane A force Fv shall be applied where:

Image omitted; see official source

The force Fv shall be maintained for five seconds after the cessation of any visually detectable movement of the protective structure.

Where the rear part of the protective structure roof will not sustain the full crushing force, the force shall be applied until the roof is deflected to coincide with the plane joining the upper part of the protective structure with that part of the rear of the tractor capable of supporting the tractor when overturned.

The force shall then be removed, and the crushing beam repositioned over that part of the protective structure which would support the tractor when completely overturned. The crushing force Fv shall then be applied again.

3.2.1.5.   Crushing at the front

The beam shall be positioned across the front uppermost structural member(s) and the resultant of crushing forces shall be located in the tractor’s median plane. A force Fv shall be applied where:

Image omitted; see official source

The force Fv shall be maintained for five seconds after the cessation of any visually detectable movement of the protective structure.