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# Question:the images might no appear dont know why but it...

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The Images might no appear, dont know why. But it is just a representation of what is written, I think the problem can still be solved withouth them...

A model vehicle is required to demonstrate a ramp jump operation as shown in Figure 1.  The jump setting has 2 identical tables (same length 1 m each, same width not shown in the elevation, same height) placed with a gap of 200 mm in between.  A ramp is fixed at the edge of Table 1.  The vehicle is placed at the starting location (where the red arrow is).  Power is then switched on to drive the vehicle forward (towards right).

Figure 1: Layout of ramp jump system demonstration

With the help of the ramp, the vehicle jumps over the gap to Table 2.  For safety reason, the vehicle should land on Table 2 with an edge tolerance of at least 100 mm.  After landing, the braking system on the vehicle is activated such that the wheels are not rotating any more.  The vehicle comes to a stop at the stop location by friction (where the green arrow is).  Again, for safety reason, the vehicle should stop before an edge tolerance that is kept at the far edge of Table 2.

The motor used in this vehicle has a gear box such that its gear ratio and power output can be selected as shown in Figure 2.

 Gear Setting Torque (g-cm) 870RPM Power in W 1 161 338 1.43894 2 415 132 1.440995 3 1032 132 1.39943 4 2400 51.3 1.264813

Figure 2: Gear setting available

The following information is given:

• Mass of vehicle is 0.5 kg.
• Coefficient of friction on Table 2 surface is 0.4.

Which set of system design parameters would you use?

Formulae applicable to this question:

The basic laws of mechanics including Newton's laws of motion apply to this question.

Apart from the labelled variables in Figure 1, other system design parameters to be considered in this question:

• rw = wheel radius of the vehicle in m
• h = height of centre of gravity of the vehicle in m
• P = power delivered by the motor and gear box assembly = $\tau \omega$
• $\tau$ = torque delivered by the gear box assembly in Nm
• $\omega$ = angular velocity of the wheel in radian per second

l0 = Runway distance = 0.83 m

l1 = Base length of ramp = 0.05 m

$\varphi$$\varphi$$\varphi$$\varphi$ = Ramp angle = 40 degree

rw = Vehicle wheel diameter = 30 mm

h = Height of centre of gravity of vehicle = 0.03 m

Gear setting no. = 3

l0 = Runway distance = 0.42 m

l1 = Base length of ramp = 0.05 m

$\varphi$$\varphi$$\varphi$$\varphi$ = Ramp angle = 40 degree

rw = Vehicle wheel diameter = 150 mm

h = Height of centre of gravity of vehicle = 0.04 m

Gear setting no. = 4

l0 = Runway distance = 0.42 m

l1 = Base length of ramp = 0.05 m

$\varphi$$\varphi$$\varphi$ = Ramp angle = 20 degree

rw = Vehicle wheel diameter = 150 mm

h = Height of centre of gravity of vehicle = 0.04 m

Gear setting no. = 2

l0 = Runway distance = 1.00 m

l1 = Base length of ramp = 0.05 m

$\varphi$$\varphi$$\varphi$ = Ramp angle = 30 degree

rw = Vehicle wheel diameter = 80 mm

h = Height of centre of gravity of vehicle = 0.04 m

Gear setting no. = 1

l0 = Runway distance = 0.83 m

l1 = Base length of ramp = 0.05 m

$\varphi$$\varphi$$\varphi$$\varphi$ = Ramp angle = 30 degree

rw = Vehicle wheel diameter = 30 mm

h = Height of centre of gravity of vehicle = 0.03 m

Gear setting no. = 3

l0 = Runway distance = 0.51 m

l1 = Base length of ramp = 0.05 m

$\varphi$$\varphi$$\varphi$$\varphi$ = Ramp angle = 35 degree

rw = Vehicle wheel diameter = 150 mm

h = Height of centre of gravity of vehicle = 0.04 m

Gear setting no. = 4

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