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a physics question


ruthless
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ok, this ones for all you physics gurus.

 

is there any way to turn drag into propulsion? and i dont mean a wind turbine and a generator type of deal. i mean is there any way to make a body design that turns drag into propulsion?

 

in other words, if you were to make a car, and you wanted it to propel itself after a certain speed with no other help other than drag, could it ever be possible?

 

just a thought.

 

 

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ruthless,

 

is there any way to turn drag into propulsion? and i dont mean a wind turbine and a generator type of deal. i mean is there any way to make a body design that turns drag into propulsion?

in other words, if you were to make a car, and you wanted it to propel itself after a certain speed with no other help other than drag, could it ever be possible?

Answer: No. Drag is the aerodynamic force vector that always acts 180 degrees opposite to the velocity vector of any vehicle operating in a fluid. IOW, it is the result of motion itself due to the viscosity of the fluid you are displacing. Part of drag is skin friction drag. Think of this component as the air analogy of the rolling friction on your tires that oppose forward motion. The other component of drag is pressure drag, which results from the pressure behind your vehicle's wake being lower that the pressure in front, thus creating a suction-like force that operates in the opposing direction to your velocity. There is a third component that occurs when you reach sonic and supersonic speeds, and that is called wave drag. But rest assured all forms of drag directly oppose the direction of the velocity vector.RMT

 

 

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so mankind can never overcome drag... what a drag.

i wonder if tt will ever be possible without overcoming drag.

"Eat more clams...it minimizes your drag!" ;)Seriously, though, the idea of OVERCOMING drag is quite a bit different from what you originally asked(turning drag into a propulsive thrust). The simplest way to "overcome drag" is to simply go somewhere that no fluid (like air) exists! There is no aerodynamic drag force in outer space, for the simple reason that there is no air in outer space. This is why a space probe can emit a constant thrust force (even a very small one) and continue to accelerate to higher and higher velocities... because there is no countering force of drag.

 

Time traveling has to overcome a much stronger "barrier" than simple aerodynamic drag. In order to time travel you have to overcome the stiffness of spacetime. Now if you'll excuse me I think I would like some clam chowder for dinner! :yum:

 

RMT

 

 

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ruthless,

 

Listen to Rainman on this one. He's the aero engineer. Drag is a "force" that, as he has stated, acts opposite to the force that propels a body through the air (or any other fluid).

 

And Titorite is also on the right track. Drag is a fact of real physical bodies moving through a fluid of whatever type. You will always have some sort of friction acting on the surfaces of the body that work against the force that propells it. Energy is conserved. If the drag could be turned 180 degrees and assist the force of propulsion then you have an "over unity" situation. You get more power output than the energy input - a perpetual motion machine. For 2,000 years people have tried to invent a perpetual motion machine. So far there has not been a single instance where such a machine has been shown to exist - nor is it expected that any such machine could possibly exist. There is no free lunch.

 

 

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R.I.P. clamorian 2050-2007

poor lil guy... he had no chance against RMT's super intillect.

I think my intellect had little to do with the defeat of this time traveling clam. I think what he couldn't stand up to were my appetite, my incisors, and most of all my digestive enzymes! :) :yum:And thanks to Darby for his always informative discussions of proper physics. May I contine? ;) Darby correctly points out that the hopes and dreams of the "overunity" crowd are based on some magical process which provides greater power output than it uses as input. But in an even more frustrating treatment of reality, they won't even be able to get a "unity" device, much less "overunity". The "unity" that is discussed relates to the physical efficiency of any device or system which provides useable power. We use the Greek letter "eta" as the variable for physical power efficiency, and this is one of the first fundamental concepts I teach in my ARO 103 Intro to Propulsion Systems course. The simplest formulation of "eta" is:

 

Efficiency = eta = Power Output/Power Input

 

The laws of thermodynamics (specifically the 2nd Law) tell us that whenever one changes the FORM of energy there is always some energy lost due to the process of that conversion. For example, a propeller-driven airplane takes energy inherent to a fuel and air mixture (chemical energy) and converts it into mechanical energy (shaft torque and speed, in RPM). Any Internal Combustion Engine performs this type of energy conversion. And even the best ICE's cannot achieve even an efficiency of 0.90 (or 90%). Heat is the largest component of "energy loss due to conversion" in the process of going from chemical to mechanical energy. But now we must add the propeller's efficiency. The propeller accepts mechanical power as an input (torque & shaft speed), and it converts that into aerodynamic (pressure) power (thrust and airspeed). Because there is a lot less heat produced in this conversion, propellers can achieve fairly high efficiencies, but even the most efficient propeller cannot do better than about eta=0.94, or 94% efficiency.

 

So, before anyone claims they have achieved "overunity", one should first look for someone to claim they have actually achieved unity efficiency (eta=1.0). Someone who really understands physics and engineering will also understand that, to be an honest broker of advanced technology, they should first show unity efficiency. If they cannot show that, any claims at "overunity" are likely from people who simply have not accounted for a particular form of energy loss as a result of power conversion.

 

RMT

 

 

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Sure there is something that can overcome drag. It is called -- free-falling, and that is everything that goes around in circles in this Universe. The Earth is free-falling around the sun. So get the car or vehicle up to a certain height and watch it over come drag on the way down because of gravity. Splat!

 

 

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thanks rmt, i understand now. may i ask how you measure efficiency? or would that be asking too much :)

 

and if an internal combustion engine is 90% efficient, then i wonder how efficient nuclear material is.

 

and btw....

 

"I think my intellect had little to do with the defeat of this time traveling clam. I think what he couldn't stand up to were my appetite, my incisors, and most of all my digestive enzymes!"

 

ewwww! :)

 

 

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In a way, you could. Although not directly, it is a result of the drag (or resistance if you prefer) of the forces. If you were to design a function that utilises heat from friction, then some of that energy lost to resistance could be recovered.

 

I mean, if we took all the heat from a space shuttle re-entering the atmosphere, and used it to convert water into steam, which was converted by a suitable rotor function, it could in turn be used to generate electricity, or hot-water features for an on-board spa. Not that it would be actually practical for the few minutes of re-entering (or safe, for that matter, hot-tap = massive third degree burns).

 

 

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Hi TNot,

 

Sure there is something that can overcome drag. It is called -- free-falling, and that is everything that goes around in circles in this Universe. The Earth is free-falling around the sun. So get the car or vehicle up to a certain height and watch it over come drag on the way down because of gravity. Splat!

Actually, no. This will not overcome drag. I post a correction not to insult you, but to make sure others don't get the wrong idea. An object that is free-falling in the atmosphere (the fluid) still experiences drag. The proof that it experiences drag is in the fact that it will not continually accelerate its velocity higher and higher. Rather, any object in free-fall within the atmosphere reaches what we call Terminal Velocity. This is the point where the aerodynamic drag force (which scales with respect to Velocity^2) exactly equals (but in the opposing direction) the force due to gravity (which scales with G*m1*m2/r). So even for a body in free-fall (when immersed in a fluid, such as air), drag still exists.RMT

 

 

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Hi Rusty,

 

If you were to design a function that utilises heat from friction, then some of that energy lost to resistance could be recovered.

Yes. But the operative word here is "some". In fact, the function of a heater in your car does exactly this: It utilizes the energy of the heat created by the engine, and puts it to a usable operational need (i.e. to keep our bodies warm). But the reality is that we cannot reclaim all energy lost due to power conversion. We can never achieve 100% efficiency... but the goal of engineering is precisely to achieve as high an efficiency as possible, within cost constraints.RMT

 

 

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Hi again ruthless,

 

thanks rmt, i understand now. may i ask how you measure efficiency? or would that be asking too much

Not at all! In fact, this little engineering lesson can also serve to illustrate a major problem with Titor's story in that he never defined the basis for "timeline divergence" which he claimed was measured in a percentage. Efficiency is also measured as a percentage, and the equation I gave you earlier actually defines that percentage and its basis metric (measurement).

 

Efficiency = eta = Power Output/Power Input

Note that efficiency is a dimensionless quantity..us engineers really love these kinds of measures, cuz they are perfectly relative! :) This keeps us in the good graces of Uncle Al (Einstein). ;) So now let's look at an example problem for efficiency... one that comes directly from one of my ARO 103 midterms:A propeller-driven aircraft is powered by a Lycoming IO-360-A internal combustion engine that delivers 200 horsepower to the engine output shaft at an efficiency of 88%. Calculate the total chemical horspepower that the combustion process (Otto Cycle) must produce to deliver this amount of shaft horsepower.

 

We use the equation above to solve this problem. They tell us that eta=88%=0.88 and that the total power output (to the shaft) is 200 Hp. Using a little algebra we solve for:

 

Power Input = Power Output/eta = 200 (Hp)/0.88 = 227.27 (Hp)

 

Does this explain the measure of efficiency a bit better? The basis of efficiency, as a percentage relative measure, is how much power we put into a power conversion process. When compared to Titor's story, he was never able to define the basis measurement which corresponded to "timeline divergence". His excuse that "it is an empirical measure" also doesn't wash, as we just don't define a percentage measure without knowing precisely what its basis measurement is! :eek: This is but one case that classifies Titor's story as "bad science".

 

RMT

 

 

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"Power Input = Power Output/eta = 200 (Hp)/0.88 = 227.27 (Hp)"

 

so in other words, it takes 227hp to turn the crankshaft, and it turns with 200hp at the propeller?

 

meaning that it has 88% efficiency?

 

so all i need to know is the input and output to calculate efficiency? if ive got that right, thats purdy durned easy :)

 

 

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so all i need to know is the input and output to calculate efficiency? if ive got that right, thats purdy durned easy

Excellent. You got that right, ruthless! woot! :) Yes, it is an easy concept when you begin at this level. You are already showing a natural tendency towards science and engineering if you can understand at this level... and then move to a more complex understanding. That next level of understanding comes when we learn how to mathematically model both the Input Power and Output Power terms that make up the calculation of Efficiency. This is exactly how I go about teaching these concepts in my ARO course.An aerospace engineer will always first try to describe the total amount of Power Required for any given aircraft to achieve (and maintain) an airborne state. To do so we need to take the really simple form of Efficiency (eta) and make it just a bit more complex. That added complexity will help us more deeply understand Efficiency overall. For any given airplane flying at any given speed through the air, we can write an equation for the total Power Required to keep the airplane flying at this condition and airspeed:

 

Power Required (Aerodynamic) = Net Thrust Force(T) x Aircraft Airspeed(V) = T * V

 

We can analyze many different flight conditions with this equation, some of them quite complex. But if we make a simplifying assumption for CRUISE FLIGHT (constant altitude and velocity) of an airplane, we can say that Thrust = Drag and so at this specific condition we can rewrite the above as:

 

Power Required (Aerodynamic-Cruise) = Net Drag Force(D) x Aircraft Airspeed(V) = D * V

 

This simple equation allows us to specify exactly how much propulsive power we will need for this airplane. It allows us to design the propeller(s) and also select which engines will drive them.

 

Perhaps in this small extension of the idea of efficiency, you can see how deep a solid engineering and scientific analysis can go in many different phenomena.

 

RMT

 

 

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"Power Required (Aerodynamic) = Net Thrust Force(T) x Aircraft Airspeed(V) = T * V

 

We can analyze many different flight conditions with this equation, some of them quite complex. But if we make a simplifying assumption for CRUISE FLIGHT (constant altitude and velocity) of an airplane, we can say that Thrust = Drag and so at this specific condition we can rewrite the above as:

 

Power Required (Aerodynamic-Cruise) = Net Drag Force(D) x Aircraft Airspeed(V) = D * V"

 

so, lemme take a crack at this...

 

1500lbs (T)x 100mph (v)= 150,000hp? it seems to me like i'd have to know other dimensions, like lift vs speed, and weight of the airplaine. but maybe thats why its called NET thrust force? to be honest, i find this one a little confusing, but i'll figure it out.

 

so, i guess my next question is, how do i calculate net thrust force and net drag force?

 

btw, if you dont have the time, or the patience to put on another class, i understand. :)

 

something must be wrong with me. i find stuff like this fun. :)

 

 

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Hi ruthless,

 

1500lbs (T)x 100mph (v)= 150,000hp? it seems to me like i'd have to know other dimensions

Let's stop right there, for you have hit on an important issue that will help you understand why your calculation of hp is not correct. Dimensions. Keeping consistent dimensions and units is one of the most fundamental errors that engineering students commit. So it is no surprise you would have this problem. In fact, it is helpful to have this problem, for that is how human minds learn. Let us first begin with a definition of 1 horsepower (1 Hp). If you go to:http://www.onlineconversion.com/power.htm

 

And enter "1" at the top, and select "horsepower [international]" on the left hand side and select "foot pound-force/second" on the right hand side you will see that in terms of the units we call foot-pound/sec, 1 Hp is equal to 550 foot-pound/second. So now let's look at your equation above and see if we have consistent units. 1500 lbs will work, we do not have to convert it further. However "mph" is miles/hour. We want to convert that into feet/second. When we do this conversion we find that 100 miles/hour = 146.67 feet/second. So now if we use the "feet/sec" value for velocity in the equation we get:

 

Power = T * V = 1500 lbs * 146.67 feet/sec = 220,005 foot-pounds/second.

 

And now we can use the foot-pound/second to Hp conversion above to conver to Hp. We simply divide the value of power in foot-pound/second by 550 to get the value in Hp:

 

Power = (220,005 foot-pound/sec)/(550 foot-pound/second per Hp)

 

Power = 400 Hp

 

like lift vs speed, and weight of the airplaine. but maybe thats why its called NET thrust force? to be honest, i find this one a little confusing, but i'll figure it out.

It is confusing because you are learning something new. I assure you that what we are working thru here are some of the most simple calculations for any engineering student who makes it to their junior year. Don't make it more complex than you can handle for right now. Yes, weight and lift and other quantities (like dynamic pressure) come into play. But a student has to move from simple to more complex. Master what I have shown you above, and then we can move on. I can explain a lot more about aerodynamic drag and how it relates to airspeed, but that will come later.

 

so, i guess my next question is, how do i calculate net thrust force and net drag force?

Let me just give you an equation. You may not understand it all right now, but I assure you it works. I will explain it in words:Drag = Drag Coefficient * (1/2) * air density * airspeed velocity^2

 

"Drag = Cd*one-half*rho*Vee-squared" is how an aero engineer would state it.

 

btw, if you dont have the time, or the patience to put on another class, i understand.

Being a teacher, I always have time for those who wish to learn. But being a practicing engineer (among other activities) I assure you that I will only respond when I can. If I don't reply for a day, or maybe even a week, it is not because I don't wish to answer or help you learn. It is because I have other commitments which are of higher priority. I'm sure you understand.

 

something must be wrong with me. i find stuff like this fun.

Oh there is something wrong with you alright... :) You have engineeringitis! If you find this sort of thing fun (figuring out physical problems and how to solve them), you would do quite well to pursue a degree and career in engineering. It is never too late. ;) RMT
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"Power = T * V = 1500 lbs * 146.67 feet/sec = 220,005 foot-pounds/second.

 

And now we can use the foot-pound/second to Hp conversion above to conver to Hp. We simply divide the value of power in foot-pound/second by 550 to get the value in Hp:

 

Power = (220,005 foot-pound/sec)/(550 foot-pound/second per Hp)

 

Power = 400 Hp"

 

ok, makes more sense now. lemme try again,

 

1000lbs (t)x 100ft/sec. (V)=100,000 ft-lbs./sec. (p)=181.8 hp?

 

"Let me just give you an equation. You may not understand it all right now, but I assure you it works. I will explain it in words:

 

Drag = Drag Coefficient * (1/2) * air density * airspeed velocity^2

 

"Drag = Cd*one-half*rho*Vee-squared" is how an aero engineer would state it."

 

ok... im gonna take a crack at it...nervously...

 

drag coefficient0.487x(1/2)0.243x1.2 kg/m3 x301ft./sec.2= drag? btw, i have no idea how to add this up, so im not even going to try. :D

 

and i'm sure i got it wrong, let me know where.

 

all in all i'd say not bad for a 6th grade drop out. :)

 

but i do have to say, im feelin purdy durned stupid atm lol.

 

 

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Anything is possible, but like rainman said, the meaning of drag negates your hypothesis.

 

But I see your point and the basic question you ask. Here's my quick idea after pondering this for a few seconds. ... You would have to use the principals behind lightning and thunder. The electric charge splits the air, creating a vacuum, then after the atmospheric pressure collapses unto the vacuated space, the air molecules slap together and make thunder. If you created a string of strong enough electric arcs ahead of your craft/car/vehicle then your vehicle would be entering into a vacuum with minimal or no air friction. The slapping of the atmosphere around the vacuated space could PUSH (or really sqeeze) the craft forward into the arced vacuum, thus going faster and faster. You would basically use atmospheric pressure to propel the craft. Thats my idea. You heard it here first. Of course it would take a F-tonnes of electrical energy to do this.

 

The important thing to remember though, is nothing is impossible.. ;-)

 

 

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