Question

Question 3 - Straight line acceleartion (10%) Table: Question 3 Row Parameter Value Unit 1 - Kg 2 m 4 Mass Tire size (radius) Torque Loss in drive train Peak engine torque First gear ratio (0-30 mph) Second gear ratio (30-60 mph) Final drive 900 0.35 20% 350 5:1 3:1 4:1 Nm | 5 6 Determine the 0-60 mph time using 15 mph increment

Answer 1

Given: Straight line acceleration of a car with given parameters.

Question 3 - Straight line acceleartion (10%) Table: Question 3 Row Parameter Value Unit 1 - Kg 2 m 4 Mass Tire size (radius) Torque Loss in drive train Peak engine torque First gear ratio (0-30 mph) Second gear ratio (30-60 mph) Final drive 900 0.35 20% 350 5:1 3:1 4:1 Nm | 5 6 Determine the 0-60 mph time using 15 mph increment

Solution:

Assumptions:

• The car will be accelerated continuously without any delays.
• The engine will be delivering peak torque for the complete duration of the motion.
• Equal torque is distributed in each wheel and the total mass of the vehicle is also distributed among each wheel.
• Load transfer due to acceleration is neglected.
• Each wheel is assumed to be a disc of mass (900/4 = 225kg.) as no information about the mass of the wheels or the position of the CG is given.

Analysis: We will distribute our solution into 4 segments:

1. I segment where the car accelerates from 0 to 15 mph.
2. II segment where the car accelerates from 15 to 30 mph.
3. III segment where car accelerates from 30 to 45 mph.
4. IV segment where car accelerates from 45 to 60 mph.

I segment where the car accelerates from 0 to 15 mph( 15 x 0.447 = 6.705 m/s)

Torque developed by engine = 350 Nm.

Torque transmitted to drive shaft after gear box reduction = 350 x (5/1) = 1750 Nm.

Torque transmitted after reduction due to losses : 1750 x 0.8 = 1400 Nm.

Torque transferred to each wheel = 1400/4 = 350 Nm.

Now the mass of each wheel = 0.5xmxr2 = 0.5 x 225 x (0.35*0.35) = 13.78 kg-m2.

So, net angular acceleration of each wheel is = Toque/MOI = 350/13.78 = 25.39 rad/sec2

Now this equals to linear acceleration of = r x ang acc = 0.35 x 25.39 = 9.904 m/sec2

Now using Equation of motion we have, time taken to accelerate from u=0, to v=6.705, with a=9.904, t1=0.676 sec.

II segment where the car accelerates from 15 to 30 mph

Similarily solving we get, t2 = 0.676 sec.

III segment where car accelerates from 30 to 45 mph.

Torque developed by engine = 350 Nm.

Torque transmitted to drive shaft after gear box reduction = 350 x (3/1) = 1050 Nm.

Torque transmitted after reduction due to losses : 1050 x 0.8 = 840 Nm.

Torque transferred to each wheel = 840/4 = 210 Nm.

Now mass of each wheel = 0.5xmxr2 = 0.5 x 225 x (0.35*0.35) = 13.78 kg-m2.

So net angular acceleration of each wheel is = Toque/MOI = 210/13.78 = 15.239 rad/sec2

Now this equals to linear acceleration of = r x ang acc = 0.35 x 12.24 = 5.338 m/sec2

Now using Equation of motion we have, time taken to accelerate from u=13.4m/s, to v=20.1 m/s, with a=5.388 m/sec2, t3 = 1.24 sec

IV segment where car accelerates from 45 to 60 mph.

Similarily solving we get, t4 = 1.24 sec

Total time taken = t1 + t2 + t3 + t4 = 3.832 secs

Notes:

• A lot of assumptions have been made to solve this problem as not much specifications were provided.
• For any doubts please feel free to ask in comments and hit like.

All the best!