Propulsion that does not involve driving wheels

[comphappy]

Originally Posted by EricH

Weight, when directed perpendicularly to a surface such as the ground, is also known as Normal Force for purposes of determining friction between an object and the surface. This increases the frictional force that is available, increasing the traction.

Yes but that does nothing for your acceleration.

m*g*u=m*a
gu=a

So a robot that is 500lbs wheels will spinout like a robot that is 10lbs.
Collisions are a different story, and my thought on that will be left between me and my team for now.

[DonRotolo]

Originally Posted by gorrilla

I dont think FIRST would let a team have a large,SHARP!, spinning object on the robots

I disagree. With proper guards it is no more dangerous than a fan in your house.

Originally Posted by RogerHebert

<R06>

Originally Posted by smurfgirl

this goes against the spirit of the rules

Originally Posted by EricH

For those of you saying that a downforce would be legal

Originally Posted by GUI

increasing normal force will increase traction, which is in violation of R06.

Originally Posted by Ziaholic

Downward force does increase traction which would be illegal.

Originally Posted by professorX

I believe that you are not able to do this.

Originally Posted by EricH

This increases the frictional force that is available, increasing the traction.

I get the feeling that all of you are serious in thinking that increasing the normal force is not legal?

Originally Posted by Kevin Sevcik

I agree with Bongle that using a fans pushing down to increase downforce is a little silly. I don't think it's going to be illegal, however.

Finally, a voice of reason. Thank You Kevin.


With the caveat that my proposal was to use fans horizontally to push air and move the robot and not to push downwards harder (since the effect would be somewhat, as Kevin puts it, silly) I have to say...

If increasing the normal force is not legal, then your robot must weigh nothing, since adding ANY weight at all increases the normal force. Clearly, we are allowed robots that weight something...

Put another way: There is no difference between adding a 20 pound weight atop a 100 pound robot and adding a fan pushing downwards with 20 pounds of force atop a 100 pound robot. Both "increase traction" by modifying the normal force. And if it is legal on that 100 pound robot, then it is perfectly legal on a 120 pound robot, since the rules define (and inspectors measure) a specific robot weight, not a specific downforce.

I hope someone asks Q&A, because I find the whole discussion silly (and will eat crow if wrong), but can also assure you that this is not a consideration for our robot.

What we are considering is wheels aided by fans. After all, what would you do with an extra 50% (or so) maneuverability???

[feilmeier]

Actually, a 500 lb robot will get more frictional force, but on the same account will have a larger inertia, thus retarding it's acceleration. By simulating an increase weight, you will increase the frictional force, but leave the inertia alone, thereby allowing you to accelerate at a faster rate.

120 lb robot will generate roughly 32 Newtons of frictional force, which will then give it an acceleration of .58 m/s^2.

A 120 lb robot with an additional 10 Newtons(120 lb is roughly 534 N) will generate roughly 32.5 Newtons of frictional force, which, when related in the F=MA equation, will give you an acceleration of roughly .6 m/s^2.

As you can see, this is a very small increase, only leading to about 0.3 m/s or so of end velocity at the end of the run of the entire 54 ft/ 16.5 m. To make this effective, a comparable amount of pressure must be induced to allow a robot to perform significantly better. In a collision, the robot that has the higher rate of speed in this game will probably win out, and because of the low coefficient of friction, the collisions will very likely be semi-eliastic.

As to the original post, my personal leaf blower can exert about 10 N when in contact with concrete. I have a feeling that the team with the best manuvability will have a nice system to direct airflow, and power the wheels

[EricH]

Originally Posted by comphappy

Yes but that does nothing for your acceleration.

m*g*u=m*a
gu=a

So a robot that is 500lbs wheels will spinout like a robot that is 10lbs.
Collisions are a different story, and my thought on that will be left between me and my team for now.

F=m*a
Frictional force = mu*N, where N = m*g, where g= 9.8 m/s^2= the gravitational constant acceleration

You're saying that m*a=mu*m*g, right? m cancels out, leaving a=mu*g.

However, the g has been increased by using a fan or something to add downwards force! Guess what? a must increase!

And lbs are NOT a unit of mass, they are a unit of force, which is computed by m*g. Increase g and you increase the force.

Let's look at your scenario: 500 lbs=32f/s^2*x slugs (slug being the English system's unit of mass) 10 lbs =32 f/s^2*y slugs
you are saying that 32f/s^2 * x slugs = 32f/s^2 * y slugs. Cancel out 32f/s^2 and you get x slugs = y slugs. Cancel out the units and x = y. However, because 500/32 = x slugs and 10/32 = y slugs, you get 500/32=10/32, which simplifies to 500=10. This isn't true, is it?

[EricH]

Originally Posted by Don Rotolo

I get the feeling that all of you are serious in thinking that increasing the normal force is not legal?
Finally, a voice of reason. Thank You Kevin.


With the caveat that my proposal was to use fans horizontally to push air and move the robot and not to push downwards harder (since the effect would be somewhat, as Kevin puts it, silly) I have to say...

If increasing the normal force is not legal, then your robot must weigh nothing, since adding ANY weight at all increases the normal force. Clearly, we are allowed robots that weight something...

I am not saying that adding weight is illegal. However, what I am saying is that using a fan or other method to effectively increase your weight beyond the 120# + battery + bumpers is. I have no problem with teams going right up to 120.0, even using a fan. But as soon as you use some of that weight to add extra weight on top of the 120.0, then I have a problem. I would assume that the others who said that would say the same thing.

Again, I have no problem with adding stuff up to the limit; I have a problem with using some of that to go beyond the limit.

[ZakuAce]

Originally Posted by EricH

I am not saying that adding weight is illegal. However, what I am saying is that using a fan or other method to effectively increase your weight beyond the 120# + battery + bumpers is. I have no problem with teams going right up to 120.0, even using a fan. But as soon as you use some of that weight to add extra weight on top of the 120.0, then I have a problem. I would assume that the others who said that would say the same thing.

Again, I have no problem with adding stuff up to the limit; I have a problem with using some of that to go beyond the limit

Spot on. This is exactly the way I feel. Now, can we please get back on track to the fans as propulsion idea?

I really hate this idea. I don't see how it is practical. Also, I feel like acting against a hit would be a problem. If you get hit while you are maneuvering into position to score, with wheel power you have at least some chance of repositioning yourself. You are using the small amount of traction you can get to your advantage, which is not something I would throw away. With a hovercraft like propulsion, I think the ability to counter-act this is much harder. Not to mention the power requirements, though like Don said, the batteries can run the CIMS at full power for a while.

HOWEVER, I think it could be a viable design if you are focusing your robot on only keeping your robot moving and keeping the opposing alliance from scoring, since being hit will send you flying away.

[wizardofoz]

You can't use fans on the moon since there is almost no atmosphere on the moon. If the goal of the game is to simulate a moon environment, wouldn't it be going against the theme of the competition to use fans?

[comphappy]

Originally Posted by EricH

F=m*a
Frictional force = mu*N, where N = m*g, where g= 9.8 m/s^2= the gravitational constant acceleration

You're saying that m*a=mu*m*g, right? m cancels out, leaving a=mu*g.

However, the g has been increased by using a fan or something to add downwards force! Guess what? a must increase!

And lbs are NOT a unit of mass, they are a unit of force, which is computed by m*g. Increase g and you increase the force.

Let's look at your scenario: 500 lbs=32f/s^2*x slugs (slug being the English system's unit of mass) 10 lbs =32 f/s^2*y slugs
you are saying that 32f/s^2 * x slugs = 32f/s^2 * y slugs. Cancel out 32f/s^2 and you get x slugs = y slugs. Cancel out the units and x = y. However, because 500/32 = x slugs and 10/32 = y slugs, you get 500/32=10/32, which simplifies to 500=10. This isn't true, is it?

What you said is very WRONG check your equations before correcting me!!!



m*a=u*m*g
becomes
a=u*g

therefore mass does not matter, now if you are useing downdraft gas then the "mass" part of Fn will not equal the mass of the robot. But that is not what I was talking about. Watch your equations.

[Justin Stiltner]

As a long time participant in FIRST (10 years) I think a few of you are "reverse lawyering" the rule about traction. To me it is pretty clear that the intent of that rule is just what most think, the rover wheels are the ONLY thing that can act against the floor to provide motive force. They actually mention legs, other wheels, etc. I think that a robot with a fan would be a great idea, and something I would look into the feasibility of. I am sure this will be settled soon via the Q and A. What you are talking about doing is done in all motor sports (cars anyway) the car uses down force generated from the aerodynamics of the vehicle to provide extra normal force, and the end result is more traction. If it were properly designed, and guarded I would have no problems with a robot that had normal force increasing devices, but be ready for a lengthy inspection, and have your engineering numbers ready if you designed it yourself, or the data sheet of the fan from the mfg.

To go ahead and comment on various other posts ideas, and take these as you will...

WORN WHEELS
If a team came in with worn in wheels I dont see a problem, It is the same as a team that has run in a previous regional and not changed their wheels, If you are going to say that you cannot have worn wheels then you would need to make everyone change their wheels every few matches. However the manual does state that the profile of the wheel may not be modified. But one thing to ponder, if we assume these to be ideal surfaces, then the surface area of the contact patch doesn't matter.

FANS FOR PROPULSION
Dont think of a fan which is normally moving a relatively low volume of air and relatively low velocity. Think of a helicopter rotor, by using this you can get an off the shelf product, that has been tested to certain rpm limits, and the best part, instead of reversing the rotation of your "fan" you only have to change the collective pitch of the rotor blades, (done with a servo on model helicopters). I have not tested the numbers myself, however there are reports that a heli with a 27.6" rotor diameter was lifting itsself as well as generating 5lbs of upward force, this while consuming about 420W via a brushless motor. However you would need casters to turn if this was all you had, which could result in a wild ride when you had a collision.

FAN FOR DOWNFORCE
To me, a fan blowing up alone would help, however you would most likely see more of an effect if you were to use your fan to generate a low pressure area under the robot. The force in this case being pressure differential multiplied by the area of the robot under this low pressure. Assuming you could affect the entire underside of a full legal size robot with a lower pressure you would have 1064in^2 of area, with something this large, the differential would not have to be much to have a rather large increase in force. For instance a random fan mfg. I looked at quoted 50cfm at 2inches of water (.072PSI), this doesn't sound like much, but over the full area of the bottom of the a fore mentioned maximum dimension robot you would achieve an extra 76lb of downward force, assuming a 120lb robot and a coefF of .05 that would be 6lb of linear force before the blower, and another
3.8 if you could establish -2IWG of pressure under the robot meaning you get 63% more force if you can establish the low pressure.

Take all of this with a grain of salt, plenty of off the cuff calculations there, and food for thought. I also think a good traction control system would be worth its weight in gold!!!

[EricH]

Originally Posted by comphappy

What you said is very WRONG check your equations before correcting me!!!



m*a=u*m*g
becomes
a=u*g

therefore mass does not matter, now if you are useing downdraft gas then the "mass" part of Fn will not equal the mass of the robot. But that is not what I was talking about. Watch your equations.

Frictional force (static) = Ffs
Max frictional force = μ*N, N = the normal force. So far so good, right?

N = m*g, where m = the mass of the object (robot) and g = either 9.8 m/s^2 or 32 f/s^2 depending on your system of measurement. Am I not correct?

For the condition where nothing is slipping or about to slip, Ffs <= μ*N. When the object is about to slip, Ffs = μ*N. When the object slips, you get into Ffk, or the force of kinetic friction. I'm still correct in this, right?

By your equations, you are assuming that the wheel is about to slip. That is, F=m*a (the standard force equation) == μ*N. But wait! Where, oh where, does Ffs come in? It is the force exerted by the static friction, so it is m*a also, I'm assuming. Please correct me if I am wrong here.

Now, on to business.

Ffs = m*a = μ*m*g, assuming a flat plane. I think this is quite reasonable considering the application and that you're about to slip your wheels.

So, m cancels. a = μ*g. You are correct. The mass does not matter.

HOWEVER: you add a fan using your available mass (which doesn't matter) which adds additional force going downwards, correct? mu is constant due to the rules, so we can set that aside. That leaves g and a. a = g. g is a component of the normal force, under N = m*g. The normal force is equal and opposite to the weight m*Ag , such that W (weight) + N = 0. So far so good, right?

When you add the force Ffan going downwards (or upwards), you now have W + N + Ffan = 0. Ffan would be added to either W or N if the fan were pushing down, and you have W+Ffan = -N or N + Ffan = -W.

Your full new equation, therefore, would read: m*a = μ*m*g + Ffan. (Frictional Force = Normal force times mu plus Fan force) Correct me if I am wrong here.

Fan force is dependent on area and speed, is it not? There is not a mass component there, because it is already accounted for. So you can no longer divide out mass. (basic algebra)

As for your other comment, YOU also need to watch your equations. Unless you can point out what is wrong with my response to your example of a 500 lb and a 10 lb robot, that stands. I could also run it in metric if I wanted to.

[GameSlammer7]

I noticed that the question was raised as to whether the 26 N was for each wheel or all combined. That's a physics answer I'll nab: it does not matter. Increased surface area does not affect the frictional force. Only normal force and the coefficient of friction affects. So, more wheels I believe will produce same friction output. I believe.

[feilmeier]

Let us get away from the mass of the robot times the gravity, and consider just the weight, since it comes in force measurements, and is much easier to calculate with.

Maximum Static Friction = Coefficient of Friction times the normal force/weight

If you increase the normal force, either by adding more mass, or by applying force in the form of a downdraft caused by a fan, then the Maximum Static Friction becomes larger. That's Algebra. 0.06 * 10 < 0.06 * 20

Now wheels that roll without slipping should ideally roll without friction, but as we all know, that's false. Rolling friction equals coefficient of friction times the normal force. This coefficient can also be the same as the one used in the static friction. As the tread touching the ground is at rest relative to the ground this makes it true.

If you can refute this, please point me to where you found this. I found all of my equations inside of Physics for Scientists and Engineers 5th edition, Extended Version page 118-127

[comphappy]

Originally Posted by EricH

Frictional force (static) = Ffs
Max frictional force = mu*N, N = the normal force. So far so good, right?

NO
Lets lay down some facts
is u of static friction which FIRST has at 0.06 for static inline.
the Fn is going to be mg asumming the robot is flat on the ground.
120 lb = 54.43 kg
Fn = 54.43 * g
Fn = 533.41 N
now lets calculate the maximum frictional force that can be exerted by wheels before they slip on the surface (static friction as the wheel does not "leave the surface")
Fu = Fn * u
Fu = 533.41N * 0.06 = 32N
now lets find out the maximum acceleration because we must be equal to or less then Fu.
Fu = ma where a is the acceleration parallel to the surface
32 = 54.43 * a
a = 0.58 m/s/s

low lets try that with a robot at 10kg

Fn = 10 * g =98N
Fu = 98N * 0.06 = 5.88 N
5.8 = 10 * a
a = 0.58

Now Lets try it with the equation that I derived
a=u*g
a = 0.06 * 9.8 = 0.58 m/s/s
So with no wind propulsion which I said in my first response. mass does not mater in terms of acceleration.

N = m*g, where m = the mass of the object (robot) and g = either 9.8 m/s^2 or 32 f/s^2 depending on your system of measurement. Am I not correct?

For the condition where nothing is slipping or about to slip, Ffs <= mu*N. When the object is about to slip, Ffs = mu*N. When the object slips, you get into Ffk, or the force of kinetic friction. I'm still correct in this, right?

No the m is N you are putting that in twice, which is wrong, see the image that I slipped in my last post.

By your equations, you are assuming that the wheel is about to slip. That is, F=m*a (the standard force equation) == mu*N. But wait! Where, oh where, does Ffs come in? It is the force exerted by the static friction, so it is m*a also, I'm assuming. Please correct me if I am wrong here.

This is not right, see above.

Now, on to business.

Ffs = m*a = mu*m*g, assuming a flat plane. I think this is quite reasonable considering the application and that you're about to slip your wheels.

So, m cancels. a = mu*g. You are correct. The mass does not matter.

mass is not in there twice take it out.

HOWEVER: you add a fan using your available mass (which doesn't matter) which adds additional force going downwards, correct? mu is constant due to the rules, so we can set that aside. That leaves g and a. a = g. g is a component of the normal force, under N = m*g. The normal force is equal and opposite to the weight m*Ag , such that W (weight) + N = 0. So far so good, right?

When you add the force Ffan going downwards (or upwards), you now have W + N + Ffan = 0. Ffan would be added to either W or N if the fan were pushing down, and you have W+Ffan = -N or N + Ffan = -W.

Your full new equation, therefore, would read: m*a = mu*m*g + Ffan. (Frictional Force = Normal force times mu plus Fan force) Correct me if I am wrong here.

Fan force is dependent on area and speed, is it not? There is not a mass component there, because it is already accounted for. So you can no longer divide out mass. (basic algebra)

As for your other comment, YOU also need to watch your equations. Unless you can point out what is wrong with my response to your example of a 500 lb and a 10 lb robot, that stands. I could also run it in metric if I wanted to.

I have pointed it out, in metric (not sure why that mattered). As for above with a fan you can accelerate a little more. although the equations above for the fan suffer from the same error as the wheels.

This is important to make clear as intuition for most people is wrong here.

[comphappy]

Originally Posted by feilmeier

Let us get away from the mass of the robot times the gravity, and consider just the weight, since it comes in force measurements, and is much easier to calculate with.

Maximum Static Friction = Coefficient of Friction times the normal force/weight

If you increase the normal force, either by adding more mass, or by applying force in the form of a downdraft caused by a fan, then the Maximum Static Friction becomes larger. That's Algebra. 0.06 * 10 < 0.06 * 20

Now wheels that roll without slipping should ideally roll without friction, but as we all know, that's false. Rolling friction equals coefficient of friction times the normal force. This coefficient can also be the same as the one used in the static friction. As the tread touching the ground is at rest relative to the ground this makes it true.

If you can refute this, please point me to where you found this. I found all of my equations inside of Physics for Scientists and Engineers 5th edition, Extended Version page 118-127

Ah my friend you are oh so close, they do have friction the static friction. And yes the frictional force increases with mass, however it is harder to get a more massive object to move. As in my proof above you will see friction force down is proportional to the force to push the object (with out moving into dynamic friction ie slipping) across. All the forces are greater but only by the mass, they cancel. Its all the same until a collision. then its just p=mv with some lost to heat and deformation.

[EricH]

Comphappy, I see where we really differ, now that you have explained where you and I differ. What you call "u" (the coefficient of friction) is normally hand-written as "μ", which is the Greek letter "mu". I used "mu" to designate this letter, while you use "u". I then used a "*" to designate multiplication.

I have edited my last post to reflect hand-written usage. Please go through again and tell me if I am still wrong.

As for why metric matters, I am simply much more familiar with the mass/weight units in metric. I couldn't even tell you what the units for slugs (the English system version of kilograms) are.