Subject
- Posted on
help calculating velocity up a ramp
- 11-18-2007
November 18, 2007, 5:26 pm
C.R.C,
I'm trying to estimate the speed of a LEGO robot traveling as fast as possible
up a ramp of a given slope.... here is what I have:
Given data:
The robot has these characteristics:
mass: 0.55 kilograms
ramp angle a = 20 degrees = 0.349 radians (there will be 2 other ramp
angles as well, but this will show the math)
wheel radius = 0.04445 meters
gear ratio: 25:1 (there will be 2 other gear ratios as well, but this
will show the math)
1 drive motor
The motor has these specs:
RPM: 340 = 35.6 Radians per second
Noload current: 0.009 Amps
Stall Torque: 0.055 Newton-Meters
Stall Current: 0.34 Amps
Here is the procedure I have tried thus far, but I get bogged down in the final
Step...
Step 1: Calculate the power available at the wheel
Applying the gear ratio:
RPM: 35.6 / 25 = 1.424 Radians per second
Stall Torque: 0.055 * 25 = 1.375 Newton-Meters
Applying the wheel radius:
Force at wheel = stall Torque / wheel radius
= 1.375 / 0.04445
= 30.933 Newtons
Is that right?
>From Inspiration to Implementation, page 217 I can calculate the
theoretical maximum possible power:
Pm,max = 1/4 * (Stall Torque) * (radians per second)
= 1/4 * (1.375 ) * (1.424)
= 0.4895 Watts
So, Pm,max tells me how much power I have available at the wheel... 0.4895
Watts?
Step 2: Calculate how much force is needed on the ramp
Assuming constant velocity up the ramp, such that acceleration is ZERO,
F = m * a
= 0.55 * 0
= 0
So the net force must be zero, and the applied force "Framp" must be the sum
of the frictional force "Ff" plus the force of the weight of the robot
acting down "Fw"
Framp = Ff + Fw
Given that Ff = u * Fn where u = coefficient of friction, g = the normal
gravitational force and (m*g) is just the weight of
the robot, so that
Ff = u * (m * g) * cos(ramp angle)
= u * 0.55 * cos(20deg)
= u * 0.55 * 0.9396
= u * 0.5167
Assuming the coefficient of friction u = 0.3 (from same book), then
Ff = 0.3 * 0.5167
= 0.155
So we now have
Framp = 0.155 + Fw
Given that Fw = m * g * sin(ramp angle)
Fw = 0.55 * 9.8 * sin(20deg)
= 0.55 * 9.8 * 0.3420
= 1.8434
So finally, the applied force is...
Framp = 0.155 + 1.8434
= 1.9984
So, Framp tells me how much force is needed on a ramp of 20 degrees... 1.9984
(what are the units? Newtons or Newton-meters or Watts
or kilograms?
Step 3: Calculate the speed up the ramp
Here is where I get uncertain. I need to apply these two formulas?
1) Power = Fapp * Velocity
(Is Power for the available power.... 0.4895 Watts?)
2) angular Speed (a.k.a. rotational velocity) = Velocity / wheelRadius
(Is rotational velocity just the RPMs at the wheel?)
I need to solve for Velocity, so rearranging them yields
1) Velocity = Power / Fapp
2) Velocity = (rotational velocity ) * wheelRadius
And here I am stuck, since I don't know:
a) units for Framp
b) is Power in equation 1 above the same as Pm,max?
c) how to use equation 2 above...
Thanks for your help!
I'm trying to estimate the speed of a LEGO robot traveling as fast as possible
up a ramp of a given slope.... here is what I have:
Given data:
The robot has these characteristics:
mass: 0.55 kilograms
ramp angle a = 20 degrees = 0.349 radians (there will be 2 other ramp
angles as well, but this will show the math)
wheel radius = 0.04445 meters
gear ratio: 25:1 (there will be 2 other gear ratios as well, but this
will show the math)
1 drive motor
The motor has these specs:
RPM: 340 = 35.6 Radians per second
Noload current: 0.009 Amps
Stall Torque: 0.055 Newton-Meters
Stall Current: 0.34 Amps
Here is the procedure I have tried thus far, but I get bogged down in the final
Step...
Step 1: Calculate the power available at the wheel
Applying the gear ratio:
RPM: 35.6 / 25 = 1.424 Radians per second
Stall Torque: 0.055 * 25 = 1.375 Newton-Meters
Applying the wheel radius:
Force at wheel = stall Torque / wheel radius
= 1.375 / 0.04445
= 30.933 Newtons
Is that right?
>From Inspiration to Implementation, page 217 I can calculate the
theoretical maximum possible power:
Pm,max = 1/4 * (Stall Torque) * (radians per second)
= 1/4 * (1.375 ) * (1.424)
= 0.4895 Watts
So, Pm,max tells me how much power I have available at the wheel... 0.4895
Watts?
Step 2: Calculate how much force is needed on the ramp
Assuming constant velocity up the ramp, such that acceleration is ZERO,
F = m * a
= 0.55 * 0
= 0
So the net force must be zero, and the applied force "Framp" must be the sum
of the frictional force "Ff" plus the force of the weight of the robot
acting down "Fw"
Framp = Ff + Fw
Given that Ff = u * Fn where u = coefficient of friction, g = the normal
gravitational force and (m*g) is just the weight of
the robot, so that
Ff = u * (m * g) * cos(ramp angle)
= u * 0.55 * cos(20deg)
= u * 0.55 * 0.9396
= u * 0.5167
Assuming the coefficient of friction u = 0.3 (from same book), then
Ff = 0.3 * 0.5167
= 0.155
So we now have
Framp = 0.155 + Fw
Given that Fw = m * g * sin(ramp angle)
Fw = 0.55 * 9.8 * sin(20deg)
= 0.55 * 9.8 * 0.3420
= 1.8434
So finally, the applied force is...
Framp = 0.155 + 1.8434
= 1.9984
So, Framp tells me how much force is needed on a ramp of 20 degrees... 1.9984
(what are the units? Newtons or Newton-meters or Watts
or kilograms?
Step 3: Calculate the speed up the ramp
Here is where I get uncertain. I need to apply these two formulas?
1) Power = Fapp * Velocity
(Is Power for the available power.... 0.4895 Watts?)
2) angular Speed (a.k.a. rotational velocity) = Velocity / wheelRadius
(Is rotational velocity just the RPMs at the wheel?)
I need to solve for Velocity, so rearranging them yields
1) Velocity = Power / Fapp
2) Velocity = (rotational velocity ) * wheelRadius
And here I am stuck, since I don't know:
a) units for Framp
b) is Power in equation 1 above the same as Pm,max?
c) how to use equation 2 above...
Thanks for your help!
Re: help calculating velocity up a ramp
When implementation is expensive, heavy design is a good idea. The
whole purpose of legos is to make experimentation easy. I don't think
they're built to that tight of tolerances... and don't get caught up
in overdesign; it happens to the best of us.
http://en.wikipedia.org/wiki/Leonhard_Euler#Berlin
Units propagate; trig functions are unitless... Given 1g=9.8 m/s^2,
1g*cos(60 deg) = 4.9 m/s^2.
Yep. If your wheel has a radius of 1, then it has a circumference of
6.28; for every full rotation, you move 6.28 units... angular
velocity * 2 * pi * radius = linear velocity.
If you want to calculate the max speed, it is something like that.
Tweaks/comments on your equations:
<power> = <friction power> + <conversion to potential energy>
Power is a change in mechanical energy per unit time. You are
interested in a steady-state velocity (no acceleration), hence we can
ignore the power needed to gain kinetic energy (from a standstill).
Friction power is lost to heat.
Potential energy is gained as you go up the slope.
note that the friction coefficient is *lost* energy; the wheel/ground
interface may have a very high friction coefficient, but that doesn't
matter since the wheel is rolling, not sliding. Major sources of
friction in wheeled vehicles are
- geartrain friction
- axle bearing friction
- wheels squishing
- (at high speeds) air friction
<conversion to potential energy> ~= <mass> * <gravity> * <slope> *
<velocity>
Potential energy is ~ m*g*height; <slope>*<velocity> gives a change in
height/unit time. e.g. a 60 degree slope results in m*g*v/2
(cos(60)=1/2).
When you substitute <power>=<max power>, solving should result in
<velocity>=<max velocity>.
- Daniel
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