xmodzracer |
02-27-2009 11:27 PM |
all this is property of damo and someperson. the whole plan was to make a 50mph radline
Quote:
anyway, here are some basic definitions (abbreviations are in parenthesis) with an equation expressing the idea after the definition:
division: /
multiplication: *
power of x: ^x
Time (s): we'll use seconds for most, if not all, things here
Kilogram (Kg): the base S.I. unit of mass
distance: symbolized by x for horizontal movement, by y for vertical. "m" is the generic symbol for distance in meters. It is used when the direction of the distance is irrelevant.
velocity (v): speed in a certain direction. defined as change in distance in a certain direction divided by change in time.
v=m/s
Acceleration (a): change in velocity over time. defined as change in velocity divided by change in time.
a= m/s^2
Newton (N): the S.I. unit of force. defined as mass (in kilograms) times acceleration (meters per second per second). One Newton is the amount of force required to accelerate a 1 kilogram object by 1 m/s^2
N=Kg*m/s^2
Joule (J): the S.I. unit of work and energy. defined as force (in Newtons) times distance moved in the direction of the force.
J=N*x
^^when force and direction are in same direction
J=N*(cosine of angle made by direction of force and direction of movement)*x
^^this only applies when force and direction are in different directions
Watt (W): the S.I. unit of power. defined as the amount of work per amount of time. One Watt is equal to 1 Joule per second.
W=J/s
In electrical terms, Watts are equal to Volts (V) times Amperes (A), or amps as they're more commonly called
W=V*A
For volts and amperage, think of your wire as like a highway. To my understanding (maybe Damo could clear some of this up for us), the volts are like the speed at which the cars are moving on the highway (higher speed limit=higher volts), and the amps are like how many cars can fit across the highway (more lanes=more amps).
for a more in-depth explanation, check out this website: http://www.microbiologybytes.com/maths/1010-7.html
So the first thing up for question is:
Can you determine speed and torque your motor will put out based on the volts and amperage put into it? As you can see in the multiple definitions of a Watt, Joules per second are related to Volts times Amps like this:
J/s=V*A
By playing around with this equation, we can find that the number of Joules put out by the motor should be equal to Volts times Amps times seconds: J=V*A*s
(Note that this assumes that efficiency is at 100%, which is technically impossible, but for simplicity's sake, I will assume that it is for now)
This means that if you run a 4.8 volt setup (4s ni-mh) with a motor that pulls 1 amp and hold the throttle down for one second, then you'll get 4.8 joules of energy from your motor. Since Joules are equal to Newton-meters, you can then further simplify the equation to determine the amount of torque (in Newtons) your motor produces:
N=(V*A*s)/m
This means that if the setup previously described pushed the car for a total of 2 meters during that second you held the throttle, then the motor put out a force of 2.4 Newtons. It also means that if the car moves 1 meter with this setup, then the motor put out a force of 4.8 meters.
That's where I get a little confused. Basic thinking tells me that the car should go farther if the motor produces more torque, but when I think about it more, torque does not necessarily mean speed. Actually, the faster rpm motors put out less torque than the lower rpm motors. So did I do all the math correctly in those equations, or did I mess up somewhere?
No load chili does 50K rpm at 6V with a 2.3 A draw.
Electrical power (Pe) is VxA = 6 x 2.3 = 13.8 Watts.
Assume 100% conversion of electrical power to mechanical power.
50K rpm is (50,000 x 2 x PI)/60 or 5236 rad/s
Mechanial power (Pm) is torque x angular velocity (in rad/s) so..
torque (in N.m) is Pm / angular velocity = 0.003 N.m (or 0.002 ft.lbs)
This is a very different result.
Yes - but it's power to weight ratio has to be a lot better than any of the 130 or 180 sized motors otherwise people would use these instead of a 370.
In terms of the calculation you were trying to do:
The KE of a rotating object is expressed differently to the typical KE equation:
KErot = 1/2 x I x w^2 where I is the moment of inertia and w is the angular velocity in rads/sec (which is RPM x 2 x Pi / 60).
I = 1/2 x m x r^2 for a cylinder of mass m and radius r.
The Radline modifed arm has a radius of 6.5 mm (0.0065 m) and weighs 10 g (0.010 kg)
So the KErot = 1/2 x [1/2 x 0.01 x 0.0065^2] x [50000 x 2 x PI / 60]^2 = 2.9 Joules
Apologies to xmodracer for the headache he now has.
At a constant 50 mph there are two forces working against the vehicle.
(1) Rolling resistance
(2) Air resistance.
The power required to keep the car moving at 50 mph is given by:
P = (Fr + Fa) * v
P = power required to maintain velocity v. Fr and Fa are the roll and air resistances respectively.
Rolling Resistance (Fr)
Fr = M * Cr * g
M = mass, Cr = coefficient of rolling resistance and g = gravitational constant
Estimates for Radline are M = 350 grams, Cr = 0.015. g = 9.8 m/s^2
Therefore, at 50 mph Fr = approx 0.05 N
Air Resistance
Fa = 0.5 * Cd * A * Rho * v^2
Cd = drag coefficient, A = frontal area of the car, Rho = density of air and V = velocity
Estimates for Radline are Cd = 0.5, A = 0.008 m^2, Rho = 1.2 (kg/m^3)
Therefore, at 50 mph Fair = 1.25 N (notice how much more of the power requirements are due to air resistance?)
Going back to the power required, P = (Fr + Fa) * v the numbers become:
P = (0.05 N + 1.25 N) * 22 m/s = 30 Watts (approx).
Assume that we get 50% efficiency in conversion from electrical power input to the motor into mechanical power out (including drivetrain losses).
Then the total power required is 30 Watts * 2 = 60 Watts (ever notice how hot a 60 watt lightbulb gets?)
Assuming we are running 3S2P at 11.1V the the current draw is 60 / 11.1 = 5.2 amps.
Additionally, we know that at 50 mph, with a 22T pinion and 34T spur the motor will need to be doing 50K rpm.
If the output power required from the motor is 30 Watts, then the torque required is 0.006 N.m.
So, at 50 mph we have the motor rpm and torque needed and the voltage and current draw required to create the aforementioned torque and motor rpm.
These numbers can be used to deduce what motor characteristics (Kt and Kv) are required (later)
Notes
The model does not include power lost at the batteries
Cd and Cr are informed guesses
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