UNIVERSITY OF ORADEA

19

AIR-TECH

UNIVERSITY OF ORADEA

Mechatronics

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………………..2015…………….

……………………………………………

aláírás

A jármű bemutatása

Vehicle introduction:

„AETHER” is a compressed air powered experimental vehicle, designed to participate at the „PNEUMOBIL 2014” race, organized by the BOSCH-REXROTH Company. The vehicle was designed by the Air-Tech team students, supported by the coordinating teacher. The vehicle can operate in two modes: the „energy saving mode”, and the „maximum speed mode”. It can switch fully automated from one mode to the other. The control of the vehicle engine is made with a PLC and electro-pneumatic valves.

The design of the vehicle is presented in figure 1. It must be said that this design can undergo some modification in the following versions regarding the motor positioning and the drive chain.

Figure.1.

Méretek és súly

Mass and dimensions

a) The overall mass of the vehicle is estimated to be around 86 kg without the driver.

The vehicle has the overall dimensions of:

– length: 2.344 m

– width: 1.426 m

– height: 1.100 m

-clearance between the bottom of the vehicle and ground surface is 0.100 m

b) driver’s shoulder-level measured from the ground is 0.700 m

The compressed air bottle is placed at the back of the engine and protected with an aluminum sheet.

Figure 2. Dimensions are given in mm.

On the figure above the vehicle’s various dimensions are shown to be within regulations, and the centre of gravity (G) is represented as determined by the CAD software.

Váz és karosszéria

Framework, car-body

The frame of the vehicle is made of aluminum tubes of 35 mm diameter and a wall thickness of 2 mm and is designed to support all the other parts and the driver. The vehicle’s frame is also designed to protect the driver in case of accidents.

The vehicle’s framework is shown in figure 3.

Figure 3.

In figure 4. it is shown below that the car-body fulfilles other technical specifications than protection, required by the autovehicle type Pneumobiles: it holds the entire drive chain, the compressed air bottle and the driven wheel. The bottle is also not suspended under the car-body directly above ground.

Figure 4.

Vezető biztonsága

Safety prescriptions:

The driver is protected by the following elements:

the structure has protective bars over the head of the driver (figure 1);

the driver will have a safety belt fixed in 4 points .

the driver will wear a protective helmet during the race.

The pneumatic circuit and the electronic circuit have a safety release buttons as required: the large red buttons next to the steering wheel. The battery and the PLC are also safely housed under the driver’s seat, in an enclosed structure. (figure 20)

The air tank is well fixed with metal straps on the vehicle and is protected by aluminum housing. The air tank will have an additional Emergency –stop switch

Figure 20.

Sűrített levegő tartály

Compressed air bottle

The place of the air tank is shown below . The air tank will be fixed with metal straps and secured with screws like it is presented in figure 19.

Figure 19.

Puffer tartály

Input of external energy

For heating up the compressed air which is cooled by the reduction of pressure from 200 bars to 10 bars, the air will pass through a heat exchanger. The only additional energy used is the energy gained from the surrounding air and 2 buffers for a bigger airflow. These buffers will be mounted between the 4 cylinders, over the transmission. The buffers are made of alluminium and these can resist up 12-14 bars.

Motor

Engine

The engine of the vehicle is composed of 4 Rexroth cylinders (2 pieces of Ø 63×200 and 2 pieces of Ø100×200) driven by electric valves (24 VDC) controlled by PLC. Transmission of motion from the engine axle to the driven wheels axle is made with a chain and sprockets. The motor is designed to be a single structural unit.

For engine torque calculations the geometrical notations are shown in figure 5 and the construction of the engine is shown in figure 6 for “energy saving mode” operation. In order to choose the proper cylinder diameters we have developed computer programs for each operation mode. The MATLAB source codes of these programs are shown in APPENDIX 1.

As it is shown in figure 6, for the “energy saving mode” the small cylinder can be filled with compressed air in a part of the stroke (h0) and then the air can expand from p0 (10 bars) having a variable pressure px. The large cylinder is powered by the air which is exhausted from the small cylinder and will expand in the two chambers having a variable, decreasing pressure px1. Representing the mean torque of the engine as function of h0 and the rate of large cylinder and small cylinder area we obtain the diagram in figure 7. In the “maximum speed mode” we power both cylinders with 10 bars for a part of the stroke h0. The diagram in this case is shown in figure 8.

Figure 5.

Figure 6.

Figure 7.

Figure 8.

In the simulations we assumed a polytropic coefficient of 1.2, a crank radius r=0.122 and a small cylinder radius of 0.0315 m (63 mm). In these diagrams the friction forces within the cylinders are included.

As it can be observed from diagram in figure 7, the best option is to use a radius rate of approximately 1.6. So the large cylinder radius will be of 0.050 m (100 mm).

The pneumatic diagram of the engine is shown in figure 9. In the upper right corner the control logic is presented. In the lower part of the figure we present the phases of the engine operation. Concentric arcs are representing cams which are switching the sensors I1 and I2. In the figure 9 is presented the car’s pneumatic scheme

Figure 9.

Vezérlőrendszer

The whole system is made from pneumatic distributor (5/3 and 3/2).The variable resistance , the flow valve, the flow and presure sensor are commanded by PLC through buttons.

Alkatrész lista

Rexroth elements:

The demanded elements are specified in the attached XLS file

Hajtás lánc

Drive chain

The power is transmitted from motor shaft to shifting gear to the differential which powers the wheels.The shifting gear will be automated and it will be powered by two pneumatic pistons which are commanded by the PLC.

Table 2.

Gear shift

In the next image presentet the automatic

gear shift’s pneumatic scheme which will be

commanded by the PLC

Kerék felfügesztés

Suspension

The wheels are mounted on the frame like it is shown in figures 10. and 11. for the front wheels and in figure 12 and 13 for the rear wheel. For the damping of shocks the vehicle relies on the rubber tires of the wheels.

Figure 10. Top view of the right wheel’s mounting

Figure 11. Front view of the right wheel’s mouing

Figure 12. Left side view of the rear wheel’s mouting

Figure 13. Showing the rear wheel’s mounting from the back, with the bearings highlighted in green

Kormánymű

Steering

The design of vehicle direction is shown in figure 14., 15., and 16.

Figure 14.

The car turns around in a six meter radius and the agle of pin is about 10 degrees

Figure 15.

Figure 16

Fékek

Brake

Figure 17

In figure 17 is presented the handbrake disk which is monted on the rear axel on the differential.

The vehicle is equipped with two separate braking sistems. The first system is the foot brake disk, the second sistem is the hand brake system.

Figure 18

In figure 18 is presented the foot brake disk. When the foot brake is pressed all 4 wheels are braked.

The foot brake system is composed of four brake disks with king pads operated through a cable system connected to a pedal. The king system has one pedal and two cables (one cable for the front brakes and one cable for the rear brakes).

The hand brake will be a separate redundant system, having a leaver for the brake. The braking pads for the hand brake will be driven using a cable system completely separate from the foot braked .By operating the hand brake it will disable the rear axle, the king disk for the hand brake being mounted on the rear axle, which has the power transmission system.

Computing of the king distance

Friction force in given by:

When inertia is missing (in static state or uniform movement)

where:

m = 160 kg –the mass of the vehicle including the pilot

g = 9,81 m/s2– gravitational acceleration

L = 1.734 m –distance between front and rear wheels

l2 = 1.180 m –distance between font wheels and center of mass

l1 = 0.554 m –distance between rear wheels and center of mass

h = 0.540 m –distance from road surface to center of mass

Tf – friction force

N -reaction force normal to the running surface

F – friction coefficient

During king, due to a moment of inertia, the load on the rear wheel will decrease:

1569.6=N1+N2

1569.6 *1180=N1 * 1.734

N1 =-P1=1569.6 *1180 / 1734 = 1068.12[N]

1569.6 *0.540= N2 * 1734

N2= -P2=1569.6 *0.540 /1.734 = 488.8[N]

L,l1,c, are constants which depend on vehicle structural characteristics. In our case these are:

c= h/l2=0.660 / 1.180 = 0.559

N1=1068.12 *(1- 0.559 *a/g)

N2=488.8 *(1 + 0.559 *a/g)

The friction coefficient depends on the quality of the road, pressure in the tires and the vehicle speed. The friction coefficient is decreasing with the speed.

For a speed of 40 km/h and dry bitumen road the friction coefficient is: µf=0.72

In order to have a king without slipping:

The deceleration value will be:

a<g / 1.180/1.734(1+0.72 * 0.559) -0.540/1.734(1- 0.72 *0.559) *0.72

a < 6.1 m/s2

Considering the kinetic energy of the vehicle at a speed of 40 km/h

“e” representing the king distance (v is given in m/s and V is given in km/h):

In case we consider the theoretical maximal value of the deceleration for the computing of minimal king distance:

In the following table the king distance is given for different speeds

Függelék

APPENDIX 1.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% MATLAB SOURCE CODE FOR TORQUE DIAGRAM COMPUTATION %

% SPEED MODE %

% AETHER PNEUMOBILE 2014 %

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

r = 0.122; % crank radius [m]

l0 = 0.2; % cylinder fixture [m]

h = 2*r; % cylinder stroke [m]

l = h + r + l0; % OA distance [m]

rp = 0.0315; % small cylinder radius [m]

A = pi*rp^2; % small cylinder area [m^2]

p0 = 1000000; % initial pressure [Pa]

pa = 100000; % aprox. atm. pressure [Pa]

gama = 1.2;

%%%%%%%%%%%%%%%%%%%%% Operation

% Pneumatic cylinders

Ffric3 = 100; % friction force of 1 cyl. is 100 N

nr_grade = 180;

k = p0/pa;

% h0 – fill lemgth [m]

% k1 – (fill length)/(total length) – small cylinder

% kv – (fill length)/(current displacement) – small cylinder

% ka – A1/A – (large cyl. area)/(small cyl. area)

for ka = 1:10

for j = 1:10

h0(j) = 0.02*j;

k1 = (h0(j)/h)^gama;

for i = 1:nr_grade

alfa(i) = i*pi/180;

c(i) = sqrt(l^2 – 2*l*r*cos(alfa(i)) + r^2);

a(i) = l*r*sin(alfa(i))/c(i);

x(i) = c(i) – (h + l0);

if x(i)< h0(j)

kv(i) = 1;

end

if x(i)>= h0(j)

kv(i) = (h0(j)/x(i))^gama;

end

kv1(i) = (h/(h – x(i) + ka*x(i)))^gama;

end

% M = Torque 2 cyl

px = pa.*k.*kv;

M = A.*((px – pa).*a + (px – pa).*ka.*a);

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

Mfric3 = 2*Ffric3*a;

M = M – Mfric3;

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

Mm1(j,:) = M;

end

% 10 rotations

Mm2 = Mm1;

nr_rot = 10;

for i_rot = 1:2*nr_rot

Mm2 = [Mm2 Mm1];

end

[lin,col] = size(Mm2);

Mm3 = [zeros(10,90) Mm2(:,1:col-90)];

Mm4 = Mm2 + Mm3;

Mm5(:,ka) = mean(Mm4,2);

end

figure;

[C,h] = contour(Mm5);

set(h,'ShowText','on','LevelStep',50);

ka_scale = round(sqrt(1:1:10).*100)./100;

set(gca,'XTicklabel',ka_scale)

set(gca,'YTicklabel',h0)

xlabel('Large cyl. radius per Small cyl. radius')

ylabel('h0')

title('Motor Mean Torque [Nm]')

grid on

% Needed Force and Torque on the wheel

r_wheel = 0.0254*20/2;

% We want the vehicle to accelerate from 0 to v0 km/h in s meters

g = 9.81; % [m/(sec^2)]

m = 160; % penumobile + driver mass[kg]

s = 20; % [m]

v0 = 45*1000/3600; % transform km/h in m/sec

acc = v0^2/(2*s); % [m/(sec^2)]

% Estimated Friction Torques

% 1. Front wheels

% 1.a. Front wheel bearings

miu1a = 0.03; % bearing friction coef.

Ffric1a = m*2*0.25*g*miu1a; % 2 wheels = 0.25 of weight each

Mfric1a = Ffric1a*0.01; % radius of bearing 0.01 [m]

% 1.b. Front wheel rolling resistance

miu1b = 0.01; % rolling friction coef.

Ffric1b = m*2*0.25*g*miu1a;

Mfric1b = Ffric1b*r_wheel;

% 2. Rear wheel

% 2.a. Rear wheel bearings

miu2a = 0.05; % bearing friction coef.

Ffric2a = m*0.5*g*miu2a; % 1 wheel = 0.5 of weight

Mfric2a = Ffric2a*0.01; % radius of bearing 0.01 [m]

% 2.b. Rear wheel rolling resistance

miu2b = 0.02; % rolling friction coef.

Ffric2b = m*0.5*g*miu2b;

Mfric2b = Ffric2b*r_wheel;

% Total Friction torque Nm

M_fric = Mfric1a + Mfric1b + Mfric2a + Mfric2b;

% Inertial force

F_inertia = m*acc; % N

% Needed Torque during acceleration

M_accel = F_inertia*r_wheel + M_fric % Nm

% Needed Torque during steady run

M_st_run = M_fric % Nm

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% MATLAB SOURCE CODE FOR TORQUE DIAGRAM COMPUTATION %

% ENERGY SAVING MODE %

% AETHER PNEUMOBILE 2011 %

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

r = 0.122; % crank radius [m]

l0 = 0.2; % cylinder fixture [m]

h = 2*r; % cylinder stroke [m]

l = h + r + l0; % OA distance [m]

rp = 0.0315; % small cylinder radius [m]

A = pi*rp^2; % small cylinder area [m^2]

p0 = 1000000; % initial pressure [Pa]

pa = 100000; % aprox. atm. pressure [Pa]

gama = 1.2;

%%%%%%%%%%%%%%%%%%%%% Operation

% Pneumatic cylinders

Ffric3 = 100; % friction force of 1 cyl. is 30 N

nr_grade = 180;

k = p0/pa;

% h0 – fill lemgth [m]

% k1 – (fill length)/(total length) – small cylinder

% kv – (fill length)/(current displacement) – small cylinder

% ka – A1/A – (large cyl. area)/(small cyl. area)

for ka = 1:10

for j = 1:10

h0(j) = 0.02*j;

k1 = (h0(j)/h)^gama;

for i = 1:nr_grade

alfa(i) = i*pi/180;

c(i) = sqrt(l^2 – 2*l*r*cos(alfa(i)) + r^2);

a(i) = l*r*sin(alfa(i))/c(i);

x(i) = c(i) – (h + l0);

if x(i)< h0(j)

kv(i) = 1;

end

if x(i)>= h0(j)

kv(i) = (h0(j)/x(i))^gama;

end

kv1(i) = (h/(h – x(i) + ka*x(i)))^gama;

end

% M = Torque 2 cyl

px = pa.*k.*kv;

px1 = pa.*k.*k1.*kv1;

M = A.*((px – px1).*a + (px1 – pa).*ka.*a);

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

Mfric3 = 2*Ffric3*a;

M = M – Mfric3;

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

Mm1(j,:) = M;

end

% 10 rotations

Mm2 = Mm1;

nr_rot = 10;

for i_rot = 1:2*nr_rot

Mm2 = [Mm2 Mm1];

end

[lin,col] = size(Mm2);

Mm3 = [zeros(10,90) Mm2(:,1:col-90)];

Mm4 = Mm2 + Mm3;

Mm5(:,ka) = mean(Mm4,2);

end

figure;

[C,h] = contour(Mm5);

set(h,'ShowText','on','LevelStep',10);

ka_scale = round(sqrt(1:1:10).*100)./100;

set(gca,'XTicklabel',ka_scale)

set(gca,'YTicklabel',h0)

xlabel('Large cyl. radius per Small cyl. radius')

ylabel('h0')

title('Motor Mean Torque [Nm]')

grid on

% Needed Force and Torque on the wheel

r_wheel = 0.0254*20/2;

% We want the vehicle to accelerate from 0 to v0 km/h in s meters

g = 9.81; % [m/(sec^2)]

m = 160; % penumobile + driver mass[kg]

s = 50; % [m]

v0 = 15*1000/3600; % transform km/h in m/sec

acc = v0^2/(2*s); % [m/(sec^2)]

% Estimated Friction Torques

% 1. Front wheels

% 1.a. Front wheel bearings

miu1a = 0.03; % bearing friction coef.

Ffric1a = m*2*0.25*g*miu1a; % 2 wheels = 0.25 of weight each

Mfric1a = Ffric1a*0.01; % radius of bearing 0.01 [m]

% 1.b. Front wheel rolling resistance

miu1b = 0.01; % rolling friction coef.

Ffric1b = m*2*0.25*g*miu1a;

Mfric1b = Ffric1b*r_wheel;

% 2. Rear wheel

% 2.a. Rear wheel bearings

miu2a = 0.05; % bearing friction coef.

Ffric2a = m*0.5*g*miu2a; % 1 wheel = 0.5 of weight

Mfric2a = Ffric2a*0.01; % radius of bearing 0.01 [m]

% 2.b. Rear wheel rolling resistance

miu2b = 0.02; % rolling friction coef.

Ffric2b = m*0.5*g*miu2b;

Mfric2b = Ffric2b*r_wheel;

% Total Friction torque Nm

M_fric = Mfric1a + Mfric1b + Mfric2a + Mfric2b;

% Inertial force

F_inertia = m*acc; % N

% Needed Torque during acceleration

M_accel = F_inertia*r_wheel + M_fric % Nm

% Needed Torque during steady run

M_st_run = M_fric % Nm

APPENDIX 2.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% %

% CHARACTERISTICS %

% %

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

h = 0.15; % piston stroke length [m]

r_wheel = 0.0254*20/2; % wheel diameter [m]

perimeter = 2*pi*r_wheel; % wheel perimeter [m]

volume_c1 = 0.0315^2*pi*h*1000; % volume of small cylinder [dm^3]

volume_c2 = 0.050^2*pi*h*1000; % volume of large cylinder [dm^3]

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% %

% ENERGY SAVING MODE %

% (15 km/h) %

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

tr = 8; % transmission rate

dist = perimeter*tr; % distance made by the vehicle

% for one engine rotation [m]

vs = 15*1000/3600; % vehicle speed [m/s]

ds = vs/dist; % piston double strokes per sec

fr_c1 = 2*volume_c1*ds*60; % flow rate for one

% small cylinder in (dm^3)/min

om(1,1) = 2*fr_c1; % flow rate for 2 small

% cylinders in (dm^3)/min

om(2,1) = ds*60; % engine RPM

om(3,1) = om(2,1)*tr; % wheel RPM

om(4,1) = ds*2*h; % piston speed

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% %

% SPEED MODE %

% (45 km/h) %

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

tr = 4; % transmission rate

dist = perimeter*tr; % distance made by the vehicle

% for one engine rotation [m]

vs = 45*1000/3600; % vehicle speed [m/s]

ds = vs/dist; % piston double strokes per sec

fr_c1 = 2*volume_c1*ds*60; % flow rate for one

% small cylinder in (dm^3)/min

fr_c2 = 2*volume_c2*ds*60; % flow rate for one

% large cylinder in (dm^3)/min

om(1,2) = 2*(fr_c1 + fr_c2); % flow rate for 2 small + 2 large

% cylinders in (dm^3)/min

om(2,2) = ds*60; % engine RPM

om(3,2) = om(2,2)*tr; % wheel RPM

om(4,2) = ds*2*h; % piston speed

disp(' Flow rate [l/min] Engine RPM Wheel RPM Piston Speed [m/s]')

disp(om')