Transcript Basic Motion calculations.

Basic Motion calculations.
Forces needed to move the Vehicle
Prepared for the Florida EAA
October 2008
David Kerzel
Basic Information about the Vehicle
•
•
•
•
•
Weight of the car
Maximum load it can carry
Tire diameter
Final drive ratio
Gear ratio for each transmission speed.
Linear Inertia
Inertia needs to be over come to accelerate
It all begins with Newton’s Second Law
F=Ma
F in Lbf, M in Lb, a in ft/sec2
Most people think acceleration in mph/sec
To convert MPH/sec
divide MPH/sec by 21.95 => ft/sec2
M Mass is the weight of the vehicle in Lb after
conversion with batteries and occupants.
Linear Inertia
For a good starting point use the curb weight and
the maximum load weight, mine is 3,800 Lb
F=Ma
My driving is at 35-45 MPH and it takes about 10
seconds for me to get to 40 MPH.
That leads to an acceleration of 4 MPH/sec.
Acceleration needs to be ft/sec2 so divide 4
MPH/sec by 21.95 to get 0.1822 ft/sec2.
Fal=Ma = 3800 X .1822 = 692Lbf
Rotating Inertia
All rotating parts also need to be accelerated. The wheels
and ties are obvious, but the axels and shafts. Motor
armature and fly wheel all must be considered. These
calculations are impossible without weights and
dimensions of the items. This rotational inertia is typical
5 to 20% of the linear inertia for production vehicles.
Whether you are moving the car with the motor or the car is
causing the motor to rotate this rotational inertia is the
same so we can look at this like a force and make an
assumption it will be 10%.
Far = 0.10 X Fal = 69Lbf
Far = 0.10 X 692 = 69Lbf
Aerodynamic Drag
As the car moves through the air there is more
force needed.
Fdrag = (Cd X A X v2)/391
Cd is the drag coefficient A is frontal area of the
car in ft2 v is speed in MPH
391 fixes the
units
For my donor car it is Cd = 0.34 and A is 20 ft2
Speed
10
20
30
40
50
60
70
80
Fdrag
2
7
16
28
43
63
85
111
Wind Drag
Wind causes more aerodynamic drag.
Basically a head wind is like moving faster.
So driving at 40 MPH with a 10 MPH head
wind give the aerodynamic drag of 50
MPH. similarly a tail wind reduces
aerodynamic drag. I am going to ignore it
because it is a small factor in all this.
Inclines
Even here in South Florida there are some minor
inclines and a few steeper ones. Some Rail
road crossings are steep, entrance ramps for I95, where I work has a 25% grade from street to
parking lot.
Fclimb = W X sin (θ)
θ = ArcTan (rise/run)
For 10% we get a angle of 8˚32’
The sin of 8˚32’ is 0.0996
3800 X 0.096 = 378Lbf
I would estimate 5% would be reasonable
Inclines
Grade
5%
10%
20%
30%
40%
sin (θ)
0.050
0.100
0.196
0.287
0.371
Weight
3800
3800
3800
3800
3800
A 5% incline needs more force than the
aerodynamic drag at 80 MPH
Fdrag
190
378
746
1092
1411
Rolling Drag
Tires have rolling resistance that ends up being a
form of drag. Different tire designs have higher
or lower rolling resistance. Higher air pressure
reduces rolling resistance, there are tradeoffs.
Frolling = Cr X W X cos (θ)
The angle in the formula reduces the rolling
resistance as the road incline angle increases.
Low rolling resistance tires have Cr of about 0.01
and normal tires are 0.02.
On a flat surface:
Frolling = 0.02 X 3800 X 1 = 76Lbf
Total Acceleration Force
All of the above forces need to be added up for the total
force.
Speed
0
10
20
30
40
50
60
70
80
Fal
Far
Fdrag
Fclimb
Frolling
Ftotal
692
692
692
692
692
692
692
692
692
69
69
69
69
69
69
69
69
69
0
2
9
20
35
54
78
107
139
190
190
190
190
190
190
190
190
190
76
76
76
76
76
76
76
76
76
1027
1029
1036
1047
1062
1081
1105
1134
1166
These are the 4 MPH/sec acceleration forces. As long as
force can be provided the acceleration will continue. No
power source has unlimited force and at some point the
required force cannot be provided.
Constant Velocity Force
Once cruising speed is reached the acceleration
forces no longer apply so the totals reduce.
I have also eliminated the climbing factor.
Speed
10
20
30
40
50
60
70
80
Fdrag
Frolling
Ftotal
2
9
20
35
54
78
107
139
76
76
76
76
76
76
76
76
78
85
96
111
130
154
183
215
These cruising forces show how traveling at 70
MPH requires about two times the force at 20
MPH.
It shows how acceleration requires 5 to 10 times the
force that cursing at constant speed does.
Torque at the wheels
Regardless of the number of driven wheels the same
amount of torque needs to be delivered to the
wheels.
Torque is measured in Ft-Lb which is a weight of I Lb at
the end of a 1 foot lever.
In a vehicle on wheels the length of the lever arm is the
radius, half the diameter of the tire, and the weight is
the required force.
My tires have a radius or 12.45 inches or 1.0375 ft.
If I combine that with the 1027Lbf from the acceleration
calculations I get
1027 X 1.0375 = 1065 ft-Lb of torque for starting and
accelerating at 4MPH/sec.
Torque at the motor Acceleration
The gears in the final drive and transmission trade
torque for shaft speeds. As the speed is
decreased the torque increases at the same
ratio.
I need 1065 Ff-Lb at the wheels for 4 MPH/sec
acceleration. My final drive is 3.87:1 ratio so
the input torque is 1065 / 3.87 = 275 Ft-Lb to the
final drive.
There is still the transmission between the motor
and the final drive.
The final drive and transmission are not 100%
efficient but this will be ignored for simplicity.
Torque at the motor Acceleration
There is still the transmission between the motor
and the final drive
Output Torque
275
275
275
275
275
Gear
1
2
3
4
5
Ratio
3.58
1.95
1.38
1.03
0.77
Input Torque
77
141
200
267
358
Torque at the motor Acceleration
A 9 inch Advance DC motor has 240 Ft-Lb peak
torque at low speed. It would give the initial
acceleration well in 3nd gear. In 2nd gear more
than target acceleration could be reached and
maintained as speed increases.
Output Torque
275
275
275
275
275
Gear
1
2
3
4
5
Ratio
3.58
1.95
1.38
1.03
0.77
Input Torque
77
141
200
267
358
2nd gear will actually provide 6.8 MPH/Sec Acceleration to 10 MPH
Motor Torque Curve
Torque at the motor Acceleration
The AC motor and controller I plan to use at 150% continuous torque
has 90 Ft-Lb output so I will need to start in first gear. The AC motor
does not have the incredibly high low speed torque that a series
wound DC motor has. My motor is capable of higher torque but
getting a controller is the issue today. The AC advantage is the
torque stays constant over the full rated speed range. A Series
wound DC motor starts with very high torque but that torque is
reduced as RPM increases.
Output Torque
275
275
275
275
275
Gear
1
2
3
4
5
Ratio
3.58
1.95
1.38
1.03
0.77
Input Torque
77
141
200
267
358
Power & Acceleration
Power is Torque X RPM / 5252
Power in in HP, Torque is in Ft-Lb
If the acceleration was reduced to 2MPH/sec the Fal and
Far inertia forces would be half of the original levels.
Everything else would be the same.
It would take twice as long to get to speed.
With a electric vehicle it takes the same amount of power to
get to a speed. There is no penalty for aggressive
acceleration in electric vehicles.
Actually rapid acceleration may be slightly more efficient
since less time is spent in acceleration.
Power at Constant Speed
More speed still requires more power.
Speed
10
20
30
40
50
60
70
80
Fdrag
Frolling
Ftotal
2
9
20
35
54
78
107
139
76
76
76
76
76
76
76
76
78
135
2
2
85
270
5
3
96
405
8
6
111
540
12
9
130
675
17
13
154
810
25
18
183
945
34
25
215
1081
46
34
Wheel RPM
Power (HP)
Power (KW)
Vehicle Speed
Vehicle speed was in a few calculations but has not been
looked at.
For this calculation we start at the motor, change that speed
with the transmission and final drive. The vehicle speed /
motor speed is locked by these ratios.
Motor RPM
500.0
1000.0
2000.0
3000.0
4000.0
5000.0
6000.0
7000.0
1st Gear MPH
2.8
5.6
11.2
16.9
22.5
28.1
33.7
39.4
2nd Gear MPH
5.2
10.3
20.7
31.0
41.3
51.6
62.0
72.3
3nd Gear MPH
7.3
14.6
29.2
43.8
58.4
73.0
87.5
4th Gear MPH 5th Gear MPH
9.8
13.1
19.5
26.2
39.1
52.3
58.6
78.5
78.2
104.6
97.7
Vehicle Speed
If we look at the 9 inch Advance DC motor running in
second gear we see a reasonable speed range for a
commuter car with no shifting. This motor can reach
6,000 RPM so this is a good fit.
The 9 inch Advance DC motor torque drops to about 50 FtLb at 3,000 RPM. Torque continues to drop as speed
increases. That reducing torque at increasing speed
reduces acceleration as speed increases. At 30 MPH
the 6.8 MPH/sec acceleration we started with has
dropped to 2 MPH/sec acceleration . At some point the
speed and torque will be in balance and there will be no
more acceleration.
Vehicle Speed
For the AC drive and 90 Ft-Lb to 4,000 RPM, I
need to change gears. In first gear I will get my
expected 4 MPH/sec acceleration, but I won’t
have it in second gear. In second gear I will get
about 2.5 MPH/sec acceleration and in 3rd gear
it will be 1.8 MPH/sec acceleration. I doubt I will
ever need the higher gears.
Torque at the motor Constant Velocity
From the constant velocity note 123 Ft-Lb is
needed for 45 MPH at the wheels.
My final drive is 3.87:1 ratio so the input torque is
123 / 3.87 = 32 Ft-Lb.
Output Torque
32
32
32
32
32
Gear
1
2
3
4
5
Ratio
3.58
1.95
1.38
1.03
0.77
Input Torque
9
16
23
31
41
Torque at the motor Constant Velocity
A 9 inch Advance DC motor has 240 Ft-Lb peak torque at
low speed. As the RPM increase the torque available
decreases.
These input torques are well within the torque produced by
this motor at higher speeds for driving at 45 MPH
Output Torque
32
32
32
32
32
Gear
1
2
3
4
5
Ratio
3.58
1.95
1.38
1.03
0.77
Motor torque must exceed the Input Torque
Input Torque
9
16
23
31
41
Motor RPM
8040
4350
3085
2300
1720
Motor Torque
25
49
110
150
Torque at the motor Constant Velocity
If the vehicle speed increase to 60 MPH more torque and
more motor speed is needed.
The required torque needed to maintain these speeds is not
available from the motor in 2nd or 3rd gears.
Even the DC motor powered vehicle could benefit from at
transmission at higher speeds.
Output Torque
40
40
40
40
40
Gear
1
2
3
4
5
Ratio
3.58
1.95
1.38
1.03
0.77
Input Torque
11
21
29
39
52
Motor RPM
10700
5760
4110
3072
2300
Motor Torque
11
25
49
110
Torque at the motor Constant Velocity
The AC motor has a continuous output of 60 Ft-Lb
to 4,000 RPM and drops to 30 Ft-Lb at 8,000
RPM.
Based on the previous torques the AC motor
could power the vehicle at 60 MPH in 2nd, 3rd or
4th gear.
Your Project
• The information presented here is basic physics and uses
information from my conversion project collected from the
internet. It may not exactly apply to your project.
• You need to gather the data for you project and work through
it all.
• Do calculations at different speeds.
• Do not ignore any of the data.
• These calculation should help avoid a conversion that has
poor acceleration, can’t go up a hill, or can’t keep up with
traffic.
• There are still a multitude of things that can rob performance
but this will take care of the elementary mechanics.
• You are in charge and responsible for what you design and
build. Please do it responsibly.