a physics question

"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. /ttiforum/images/graemlins/smile.gif

something must be wrong with me. i find stuff like this fun. /ttiforum/images/graemlins/smile.gif
 
Hi ruthless,
1500lbs (T)x 100mph (v)= 150,000hp? it seems to me like i'd have to know other dimensions
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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.
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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?
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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.
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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.
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Oh there is something wrong with you alright... /ttiforum/images/graemlins/smile.gif 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
 
"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. /ttiforum/images/graemlins/smile.gif
but i do have to say, im feelin purdy durned stupid atm lol.
 
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.. ;-)
 
Hi ruthless,

First, a "mea culpa"... even professors can make mistakes! We just prefer to point them out and correct them as soon as we see we made these mistakes. I gave you an incomplete formula for drag. Here is the correct formula, with an extra term that I forgot in the original:

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

The bolded item is what I forgot. This is some characteristic area that relates the overall SCALE of the vehicle. In aircraft we select the Reference Area as being the wing planform area (the area of the wing that you observe as you look down on the wing from above...from one wingtip, thru the fuselage, to the other wingtip). Now on to your questions:

drag coefficient0.487x(1/2)0.243x1.2 kg/m3 x301ft./sec.2= drag?
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Not bad for a first try. But let's take a step back and first try to understand each and every piece of this calculation. I've already explained Reference Area. So let's handle that by me just giving you the wing reference area for a big jumbo jet that I helped design the autopilot for... the MD-11. The MD-11 wing reference area is = 3648 square feet.

Airspeed Velocity - This is the speed the airplane is moving WITH RESPECT TO THE ATMOSPHERE (the fluid...air!). If we assume the air is calm (no winds, no turbulence) then the airspeed velocity is identically equal to the groundspeed velocity. How about a reasonable number for the MD-11's airspeed velocity? Well, when the MD-11 is cruising across the Atlantic or Pacific Ocean at about 35,000 feet above sea level, it can actuall fly as high as Mach 0.82 (82% of the speed of sound). This would translate into an airspeed velocity of about 473 Knots (Nautical Miles Per Hour). But we need to make sure our units are consistent in the equation, so we need FEET per SECOND, not Nautical Miles per Hour! If we make the conversion, we end up with an airspeed velocity of about 798 Feet per Second. (NOTE that when we compute drag we have to SQUARE the airspeed velocity in this equation!)

Air Density - As I mentioned above, large commercial aircraft usually cruise at fairly high altitudes. 35,000 feet is a common cruise altitude. As it turns out, the fluid dynamic physics of our atmosphere allows us to model the air's density as it changes from sea level up to altitudes of 35,000 feet and higher. In general, the density of air DECREASES with altitude (which is why it gets very hard to breathe above 14,000 feet or more...less air molecules per breath you inhale). My lessons in modeling the atmosphere last for an entire 2 hour lecture...not gonna do that here. So trust me when I tell you that the air density at 35,000 feet on a "standard day" is = .000736108 Slugs per Cubic Foot (a Slug is a unit of mass in the British system of units. When you multiply a mass in Slugs times the acceleration of gravity in feet/second^2, you get the equivalent of 1 pound). These units for air density are consistent with the others above and they will ensure our answer for Drag comes out to be in units of pounds.

Drag Coefficient - Perhaps the most mysterious of them all! And again, it would take me a LONG time to try to explain this to you. But basically this is a NON-DIMENSIONAL measure that is related to the geometric shape of the aircraft itself. The more aerodynamically sleek the airplane is, the lower its drag coefficient will be. The number you chose is actually very, VERY high for a drag coefficient...even for an automobile. The drag coefficient is what an aerodynamics engineer measures when he places a geometric model of an airplane in the wind tunnel and runs wind tunnel tests. For the MD-11, operating at is cruise flight condition, let's just say that its drag coefficient is = 0.0034 (with NO UNITS...it is non-dimensional).

So now let's look at how the units work themselves out:

Density = Slug per Cubic Foot (Slug/Feet^3)
Airspeed Velocity = Feet per Second SQUARED (Feet^2/Second^2)
Reference Area = Feet Squared (Feet^2)

If we combine all these units when we multiply them, what we end up with is the following units:

Drag = Slug*Feet/Second^2 (And as I mentioned above, this is the precise definition of the force we call a pound). So now, when you do the calculations in your calculator, you can be assured that the number you calculate will come out in units of Pounds!

So now your mission, rutheless, should you decide to accept it:

HOMEWORK: Calculate the drag on the MD-11, cruising at an altitude of 35,000 feet and an airspeed velocity of 798 feet per second.

Report back your drag, in pounds, and I will tell you if you got it right! /ttiforum/images/graemlins/smile.gif

And again I will say: If you truly do think this stuff is fun, ruthless, then you should VERY SOON look into where you could possibly start taking classes in aerodynamics!!! Many community colleges offer basic concepts such as those I am explaining to you here.

RMT
 
rmt, your the coolest. ty so much for taking the time to help me with this. it means the world to me. /ttiforum/images/graemlins/smile.gif


AV1596 X RA7296 X DC0.0034 X AD0.002208324= D 87.4297872838656LBS.

did i get it right? huh? huh? huh?! i may have added the ^2 and ^3 stuff wrong, i was never taught how to do that. :oops:
 
rmt, your the coolest. ty so much for taking the time to help me with this. it means the world to me.
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"A person who shows the earnest will to come to understand that which they do not yet comprehend should always be given that opportunity." (Ray Hudson - teaching ARO 101 in 2000) /ttiforum/images/graemlins/smile.gif My pleasure, ruthless!

Now let me construct the calculations to show you how you get the answer. And for the record, the only power calculation you need to do is the squaring of the airspeed. This is nothing more than multiplying the airspeed by itself. So first:

Airspeed^2 = Airspeed*Airspeed = (798 feet/sec)*(798 feet/sec) = 636804 feet^2/sec^2

Now let's multiply this result by the wing reference area (3648 feet^2):

(636804 feet^2/sec^2)*(3648 feet^2) = 2,323,060,992 feet^4/sec^2

Now let's multiply this huge number times the air density (0.000736108 slugs/feet^3):

(2,323,060,992 feet^4/sec^2)*(0.000736108 slugs/feet^3) = 1710023 slugs-feet/sec^2.

Now let's wrap up this calculation by multiplying this result by the drag coefficient and also multiplying it by the (1/2):

(1710023 slugs-feet/sec^2)*(0.0034)*(1/2) = 2907 slugs-feet/sec^2 = 2907 pounds

Did you follow that? If not, and you are proficient at programming Excel spreadsheets, I would encourage you to build a spreadsheet which accepts all these inputs and performs this calculation until you get the answer that I have given here (within 2% and you've done well!)

RMT
 
so what do i need to do if a # has ^2? just multiply?



"Airspeed^2 = Airspeed*Airspeed = (798 feet/sec)*(798 feet/sec) = 636804 feet^2/sec^2"


how did it become feet^2/sec^2? and how do you know when to make it that way?

i imagine these are pretty stupid questions, but i dont know em, so i ask. /ttiforum/images/graemlins/smile.gif
 
so what do i need to do if a # has ^2? just multiply?
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This mathematical operation is called "raising the base number to a power". If you put your Windows calculator into "scientific" mode via the "View" menu, you will see one of the buttons labeled as "X^Y". If I wish to raise the number 798 to the power of 2 (squared, as in our problem) I would first enter "798" then hit the "X^Y" button and then enter "2". When you hit "=" you will see you get the number I stated above for "Airspeed^2". You can also raise base numbers to non-integer powers (a popular one is to raise a number to the power of 0.5, or one half...this is the definition of the square root function).

If you know how to use Excel (and all good engineers are expert Excel programmers), you can enter the formula in a cell like this: "=798^2" (just the stuff inside the quotes). When you hit return you will see the cell has the same answer in it that we calculated above. Again I would suggest you learn how to use Excel. It is a powerful tool for engineering calculations and even simulations of physical phenomenon. In my ARO 202 Fundamentals of Aircraft Peformance course, each student turns in a course project with 7 different spreadsheets in a single Excel workbook. Those spreadsheets accept design parameters for an entire airplane, and calculate ALL of the relevant flight performance for all phases of flight (Taxi, Takeoff, Climb, Cruise, Descent, Loiter, Approach, and Landing). Many students are amazed at what they have accomplished by the end of the term... and they walk out of the class with a powerful aircraft simulation they can use for design studies.

"Airspeed^2 = Airspeed*Airspeed = (798 feet/sec)*(798 feet/sec) = 636804 feet^2/sec^2"

how did it become feet^2/sec^2? and how do you know when to make it that way?
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Note that EACH "copy" of Airspeed is measured in Length per Time (Feet/Sec). Since we are squaring the airspeed, then this means we are multiplying "Feet*Feet" in the numerators and "Sec*Sec" in the denominators of the physical units. The mathematical way to express "Feet*Feet" is just "Feet^2". If we were to CUBE the airspeed (which is what we do when we calculate aerodynamic power...we raise airspeed to the power of 3) then we would be performing "Feet*Feet*Feet" and "Sec*Sec*Sec" and so it would result in "Feet^3/Sec^3".

Do you get it? Does that make sense? Once again, these concepts of physical units are VERY IMPORTANT, and getting mixed units (i.e. incompatible units mixed together) accounts for well over 70% of the errors that I see students make on their midterm and final exams.

i imagine these are pretty stupid questions, but i dont know em, so i ask.
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You might be surprised how many students I have in my classes who are "afraid to ask" because they think they will look stupid to others. But by not asking, they may not LOOK stupid to others, but they are certainly remaining ignorant!!! :eek:

So...you have until mid September to ask questions. After that time I will be heading back to school to teach the next batch of freshman aerospace engineering students... that and my normal "super-secret" job reverse-engineering UFOs with alien consultants! /ttiforum/images/graemlins/smile.gif

RMT
 
ok, could you give me another problem to figure out the drag? i would like to try again, i think i may understand. and i will try to figure excel out. /ttiforum/images/graemlins/smile.gif
 
and this question you dont have to answer if you dont have the time to bother. but im just curious, what all would a person need to take your classes?
 
ok, could you give me another problem to figure out the drag? i would like to try again, i think i may understand.
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OK then. Let's learn the next important thing that an aerodynamicist needs to understand... how the force of drag changes with respect to the airspeed that the airplane is flying at. We will keep the MD-11 as the airplane we wish to analyze, and we will even keep it flying at 35,000 feet so the air density remains the same. But let's see how much the drag goes down if we slow the airplane so that it is flying just above its stall speed: 290 Knots (which is Nautical Miles per Hour). I want you to calculate the drag at this new flight condition. (HINT: Remember you need consistent units, so you will have to convert Knots into Feet/Sec before you do the final calculation!)

and this question you dont have to answer if you dont have the time to bother. but im just curious, what all would a person need to take your classes?
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This depends on what you want to get out of the class. If you want college credit towards a degree then you need to be accepted at the university, and have the prerequisite courses in math and science that most high school graduates have when they start college. If all you wish to do is "audit" the course to gain new knowledge, most universities will often allow you to do this (for a fee, of course) even if you are not enrolled or do not have the prerequisites. I have had several "audit" students over the years. You should look into junior colleges and accredited colleges in your area to see if you could audit a class or two. What is the closest major university to where you live?

RMT
 
"What is the closest major university to where you live?"

mississippi state and alabama are very close to where i live. i live 30 mins from msu, and an hour from alabama.
 
Mississippi State University has a fine aerospace engineering program...

http://ae.msstate.edu/site/pages/department/course_descriptions.php

I would bet you could be permitted to audit their ASE 1013 course, which is similar to the ARO 101 course I teach in the fall quarter here. It seems to cover the same things, except my course also begins to introduce flight mechanics concepts that it seems is covered in their ASE 1023 course. I think you should look into it, and see if you can audit that course.

The first aerospace coure ALWAYS introduces the atmosphere to students and shows them how to model it. This makes sense, of course, since the atmosphere is the medium that aircraft operate within, so it has a large impact on the design and performance of an aircraft.

RMT
 
"You might be surprised how many students I have in my classes who are "afraid to ask" because they think they will look stupid to others. But by not asking, they may not LOOK stupid to others, but they are certainly remaining ignorant!!! "

i never had that problem growing up. i would ask the teacher a question, everyone would laugh. then i would ask everyone "how many knows the answer?" none would raise their hands. then i would ask " who all wanted to ask the same question, but was afraid of looking stupid?" again no hands. then i'd laugh and sit in my seat. :D
 
"But let's see how much the drag goes down if we slow the airplane so that it is flying just above its stall speed: 290 Knots "

drag@290knots=1093 slug-feet/sec^2 =1093lbs.

once i figured it out, i kinda started understanding some things. sorry it took me so long, but i wanted to make sure i got it right, and im pretty sure i did. /ttiforum/images/graemlins/smile.gif
 
If you put your Windows calculator into "scientific" mode
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Just a quick off topic question for RMT. In your opinion, if you use closed source tools during the scientific process does that invalidate your efforts? I know it's no big deal when you are starting out, and calculator isn't a good example but it got me wondering, when you present a paper, is it "acceptable" to use closed source tools or have any papers been rejected? One reason I could think of is without open source software that you compile yourself, you aren't actually repeating the experiment. Have you ever heard of this, or is closed source vs open source never an issue?
 
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