Simple Aerodynamics question

murphey

murphey

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murphey
I'm sitting in the aerodynamics for pilots course as the local college (I don't get a grade but then, I didn't pay tuition, either). I was expecting a bit more depth, but it's for students who plan to fly big iron. Here's the question - I'm sure I'm overthinking it.

Weight of airplane 4300 pounds = 133 slugs
Move the airplane at a rate of 3.4 feet per second (fps), level runway, ignore friction.
So, what's the Force to move the plane? I get 454 pounds. Afterall, Force = mass & acceleration

So the next question is given the same details, and the calculated force, how much time does it take to get the airplane to 148 fps? I'm using the time = speed/force, hence I end up with less than 1 second, which is absurd! So, where am I going wrong?
 
Force = Mass x Acceleration, not Mass x Velocity. Looks to me like you’re plugging a velocity (3.4 fps) into the formula.
 
Is the plane on a treadmill? :)
 
F = Mass * acceleration. You have Mass and speed (3.4 ft/sec) . You need acceleration, or, if the acceleration is zero (steady state speed), you need to know something about friction / rolling resistance. if you neglect friction, the force is zero to maintain a steady speed.

time = speed/force should be time = speed / acceleration. or time = speed * mass/force
 
Your problem doesn't make sense. If it is frictionless, any force will accelerate the airplane.
 
Checkout_my_Six said:
you are discounting the momentum affects due to mass.....
Click to expand...
No I am not. 1) An airplane not moving has zero momentum; 2) A=F/M, doesn't matter how large a value you have for M, any positive F gives you acceleration.
 
acceleration is a given.....3.4 feet per second (fps)...it's an algebra problem.
 
Maybe... I'm guessing they mean "from zero to 3.4 fps", which would give you an acceleration *if* they also told you *either* a) how much runway distance it took, or b) how much time it took to get up to that speed. Was there additional information not in your original post?
If this is the case, then part B of the question would also make more sense.

Otherwise, yeah, the question makes no sense. It takes zero Newtons of force to move something at constant speed with no friction or drag.
 
murphey said:
I'm sitting in the aerodynamics for pilots course as the local college (I don't get a grade but then, I didn't pay tuition, either). I was expecting a bit more depth, but it's for students who plan to fly big iron. Here's the question - I'm sure I'm overthinking it.

Weight of airplane 4300 pounds = 133 slugs
Move the airplane at a rate of 3.4 feet per second (fps), level runway, ignore friction.
So, what's the Force to move the plane? I get 454 pounds. Afterall, Force = mass & acceleration

So the next question is given the same details, and the calculated force, how much time does it take to get the airplane to 148 fps? I'm using the time = speed/force, hence I end up with less than 1 second, which is absurd! So, where am I going wrong?
Click to expand...

Seems like you are using force and work and velocity and acceleration interchangeably in the question, so it's not valid.
 
Velocity (fps) is a constant speed. It takes an acceleration (fps/s) to get a mass moving, with a corresponding force. Once that mass reaches a velocity, the force has to stop or it will be accelerated faster. Momentum will keep it moving at a constant speed, ignoring friction.

The problem here is that any acceleration can get a mass to move at a specified velocity, it's just going to take more time at a lower acceleration, and corresponding force. That is why the question asked doesn't really make sense.

If, however, you wanted to determine the time involved to accelerate a mass to a certain speed, then it is an algebra equation which must be solved in order to determine the time element from the acceleration.

In the real world application of an airplane being accelerated, you will have your force given to you. It will be the thrust of the engines. It is probably informative to know the time (and distance) required to to accelerate that airplane to takeoff speed given the thrust force pushing it.

Also in the real world we like to use pound-mass (lbm) vs. slug in English unit calculations involving mass and acceleration.
 
This is a physics question, not aerodynamics. But yes, the problem as stated in the OP is incomplete or incorrect, no answer possible with the given information.
 
And by the way, calculating the time required to accelerate the plane is actually stupid simple. You have your mass. You have your force (thrust). Calculate what the acceleration is. Divide your target speed by the acceleration then whammo, you get the number of seconds needed to reach your target velocity.

You can send the consulting fee to my Paypal account.
 
Dana said:
This is a physics question, not aerodynamics. But yes, the problem as stated in the OP is incomplete or incorrect, no answer possible with the given information.
Click to expand...
As noted by CheckoutmySix, this is a trivial algebra problem....but it's driving me wacko because it shouldn't be this difficult to resolve.
 
kath said:
Maybe... I'm guessing they mean "from zero to 3.4 fps", which would give you an acceleration *if* they also told you *either* a) how much runway distance it took, or b) how much time it took to get up to that speed. Was there additional information not in your original post?
If this is the case, then part B of the question would also make more sense.

Otherwise, yeah, the question makes no sense. It takes zero Newtons of force to move something at constant speed with no friction or drag.
Click to expand...
I'm thinking this is the real question...."from zero to 3.4 fps"

Here's the exact question:
"What is the minimum force to accelerate 4300# airplane 3.4 fps?"

Or should I be thinking of this as a kinetic energy problem?

How embarassing...I have 2 degrees in math w/ minor in physics, but all were decades ago.
 
murphey said:
I'm thinking this is the real question...."from zero to 3.4 fps"

Here's the exact question:
"What is the minimum force to accelerate 4300# airplane 3.4 fps?"
Click to expand...

If you have enough time, a nano newton will do it. On the other hand, you'd have to push with a force about 10.6% of the airplane's weight to accelerate it to 3.4 FPS in a second. 3.4/32 x 4300 = 456 lbs.
 
Typo in problem statement, they must mean 3.4ft/sec•sec. Then...

Then F=m•a = 4300•lbs•3.4•ft/(sec•sec) = 14,620•ft•lb/(sec•sec)

The time to velocity is: V = Vo + (a•t) but Vo = 0 so

v = a•t or t=v/a t=148/3.4 = 43.5•sec

Or maybe I have it all wrong?
 
kyleb said:
If you have enough time, a nano newton will do it. On the other hand, you'd have to push with a force about 10.6% of the airplane's weight to accelerate it to 3.4 FPS in a second. 3.4/32 x 4300 = 456 lbs.
Click to expand...
Which is what I got. But using 456 as the force and 3.4 fps for V, the distance is V^2/2a or 3221 ft, which is reasonable. But the time makes no sense....t = V/a
 
Sinistar said:
Typo in problem statement, they must mean 3.4ft/sec•sec. Then...

Then F=m•a = 4300•lbs•3.4•ft/(sec•sec) = 14,620•ft•lb/(sec•sec)

The time to velocity is: V = Vo + (a•t) but Vo = 0 so

v = a•t or t=v/a t=148/3.4 = 43.5•sec

Or maybe I have it all wrong?
Click to expand...
Hope not, just sent an email asking for clarification.
 
Sinistar said:
Typo in problem statement, they must mean 3.4ft/sec•sec. Then...

Then F=m•a = 4300•lbs•3.4•ft/(sec•sec) = 14,620•ft•lb/(sec•sec)

The time to velocity is: V = Vo + (a•t) but Vo = 0 so

v = a•t or t=v/a t=148/3.4 = 43.5•sec

Or maybe I have it all wrong?
Click to expand...
Nope, but the calculation for Force is in slugs for this class.
 
Many thanks to all....obviously the 3.4 fps is wrong units, should be seconds squared. I'm dithering over the calculation in slugs or pounds. Using slugs, the force is 456 pounds, using pounds, it's 14,620 pounds. Now, time for the "does it make sense?" review. For a 4300# airplane, which makes more sense, 456# or 14.6K#? I'm not sure. I know moving my 2400# cherokee, I certainly can't exert 14K ft pounds! Or am I not understanding the concepts again?
 
Time @ 3.4 ft/sec/sec = 138 fps/3.4ft/s/s =40.6 seconds. D = 138 fps/2 x 40.6 s = 4450'.

Or maybe I've forgotten my fizix.

You should have seen me trying to solve a high school trig/geometry/algebra problem this morning. I hadn't done that stuff in >30 years.
 
Update....just got a response from the instructor....yes, a typo (everything is due tomorrow am and I'm the only one who noticed the typo) and should be 3.4 fps^2, so that is the acceleration.

*whew*
 
murphey said:
As noted by CheckoutmySix, this is a trivial algebra problem....but it's driving me wacko because it shouldn't be this difficult to resolve.
Click to expand...

I disagree. It is a non-trivial problem that requires calculus with a bunch of simplifying assumptions to turn it into what appears to be a trivial (but flawed) algebra problem.

All the discussion above assumes no drag. Thrust may be constant, but the net force is not, since drag increases with V^2 as the plane accelerates. You may simplify it by saying there is no drag, but then I think that means there’s no lift either, in which case what’s the point? Another unstated assumption is electric airplane, because otherwise mass is not constant (although effects of changing mass are probably negligible over the short time/distance).

It looks like it was intended to be a simple F=ma problem that is poorly worded, probably due to some copy and paste error.

Edit: just saw Murphey’s update. Took too long composing my response.
 
kyleb said:
Time @ 3.4 ft/sec/sec = 138 fps/3.4ft/s/s =40.6 seconds. D = 138 fps/2 x 40.6 s = 4450'.

Or maybe I've forgotten my fizix.
Click to expand...
Looks good to me.
Knowing a = delta-v/delta-t, find the time by rearranging it as you have. (Assuming the plane starts from rest.)
Then: delta-x = (average v)*time = [(v_final + v_initial)/2] * time. (Again, we're assuming v_initial = 0, that it starts from rest and accelerates smoothly.)

We need to know the mass (in slugs) to use it as the "m" in F=ma, to find the force necessary to accelerate the plane at this rate.
Remember, "pounds" is not a unit of mass, it's a unit of force.

Of course, this only works for a spherical airplane in a vacuum. :)
 
kyleb said:
Time @ 3.4 ft/sec/sec = 138 fps/3.4ft/s/s =40.6 seconds. D = 138 fps/2 x 40.6 s = 4450'.

Or maybe I've forgotten my fizix.

You should have seen me trying to solve a high school trig/geometry/algebra problem this morning. I hadn't done that stuff in >30 years.
Click to expand...
I think you have a simple error, it's 148fps . Not 138fps.

I don't think they asked for this distance but yours is right except the velocity typo.
 
murphey said:
Many thanks to all....obviously the 3.4 fps is wrong units, should be seconds squared. I'm dithering over the calculation in slugs or pounds. Using slugs, the force is 456 pounds, using pounds, it's 14,620 pounds. Now, time for the "does it make sense?" review. For a 4300# airplane, which makes more sense, 456# or 14.6K#? I'm not sure. I know moving my 2400# cherokee, I certainly can't exert 14K ft pounds! Or am I not understanding the concepts again?
Click to expand...
Maybe I am missing something but slugs are not pounds. It's no different than if they gave you pounds or kg.

The 14,600 ft lb./ (sec x sec) just is what it is since that's the numbers you were given and no friction or other counter forces. It's mainly a seemingly large number because of the mass. The acceleration iyou were given was only around 1/10 of earth gravity acceleration. Plus it takes over 40 seconds to reach that speed.
 
murphey said:
Many thanks to all....obviously the 3.4 fps is wrong units, should be seconds squared. I'm dithering over the calculation in slugs or pounds. Using slugs, the force is 456 pounds, using pounds, it's 14,620 pounds.
Click to expand...
Slugs or pounds doesn't make a difference - you get the same force.

force pounds = mass slugs * acceleration ft/sec^2 , 133*3.4 = 454 lbf

force pounds = (mass pounds/32.2) * acceleration feet/sec^2 = 454 lbf

Common units:
upload_2019-9-17_6-16-8.png

 
Last edited:
murphey said:
I'm thinking this is the real question...."from zero to 3.4 fps"

Here's the exact question:
"What is the minimum force to accelerate 4300# airplane 3.4 fps?"

Or should I be thinking of this as a kinetic energy problem?

How embarassing...I have 2 degrees in math w/ minor in physics, but all were decades ago.
Click to expand...

As stated, the question only makes sense if you consider aerodynamic drag. The "minimum" force will be that required to sustain the airplane's drag at 3.4 fps.

But, you've established that it was a typo, and it should have been an acceleration of 3.4 fps/s. But again, as worded, the problem still only makes sense if you consider aerodynamic drag. In a vacuum, there won't be a "minimum" force required to achieve that acceleration, there will be a specific force. In the atmosphere, That will be the force required to initially get it to move, but after that additional force will need to be applied to overcome drag.
 
Sac Arrow said:
As stated, the question only makes sense if you consider aerodynamic drag. The "minimum" force will be that required to sustain the airplane's drag at 3.4 fps.

But, you've established that it was a typo, and it should have been an acceleration of 3.4 fps/s. But again, as worded, the problem still only makes sense if you consider aerodynamic drag. In a vacuum, there won't be a "minimum" force required to achieve that acceleration, there will be a specific force. In the atmosphere, That will be the force required to initially get it to move, but after that additional force will need to be applied to overcome drag.
Click to expand...
Very true but.....remember the name of this course....Aerodynamics for Pilot....the school has 3 levels of physics courses: Physics for Physics majors (the one I took, but not at this school), Physics for Liberal Arts (no, I'm not kidding), and then Physics of Aviation, which was the prereq for this course. A few years ago the prereq was removed. In theory, everyone has college algebra as a required course in the school, not just the dept. But, the school has "Integrated X" courses, which are really "how would you used this topic in the real world, altho we know you'll never remember any of this, but it keeps the school certified...."
 
To all...many thanks, feel free to continue the discussion, but I just submitted the homework. I tend to be very literal on exams and such, your confirmation the original assumption of the typo was correct, everything fell into place.

As I mentioned before, I was hoping for a bit more depth in the course, but it's turned into "here's the formulas, here is a bunch of data, what's the answer?" type of class. Rote and drill. For me, 8th grade algebra. But every once in a while, there's something that's new and interesting to me. The dept has a very expensive wind tunnel, but it hasn't been used in years because the software no longer exists or the software doesn't work on the computer used with it. So I sent an email to the lab manager offering to make it work...haven't heard back, and don't expect to. NIH syndrome.
 
murphey said:
Very true but.....remember the name of this course....Aerodynamics for Pilot....the school has 3 levels of physics courses: Physics for Physics majors (the one I took, but not at this school), Physics for Liberal Arts (no, I'm not kidding), and then Physics of Aviation, which was the prereq for this course. A few years ago the prereq was removed. In theory, everyone has college algebra as a required course in the school, not just the dept. But, the school has "Integrated X" courses, which are really "how would you used this topic in the real world, altho we know you'll never remember any of this, but it keeps the school certified...."
Click to expand...
Same premise is given and gives you all the numbers but the question is...

How have you disenfranchised the airplane by assaulting it with force?
 
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