Energy Dissipation Of A Sphere Passing Through a Medium

[Barrow]

Hello,

Sorry if what I'm about to write is TMI to some people...

I agree with the approach presented, but I think I disagree with a couple
of the assumptions:

>So we'll use an approximate solution that is valid for spheres which
>move slowly through a medium.
>The drag force on such a sphere is: f = 1/2Cpav**2

2 comments:

1: I think this is the approximation for high velocity drag, not low velocity
drag, i.e. Reynolds number > ~1000 (from Wikipedia http://en.wikipedia.org/wiki/Drag_(physics) .
For low velocity the drag is proportional to the velocity,
not the velocity squared (same wiki link).

2: Regardless of comment 1, as the projectile slows down the drag force will
decrease. The drag force is not a constant. It
relates to the instantaneous velocity, not the initial velocity. Assuming constant drag
force might be a good assumption at high velocities, but at low velocities
this assumption makes a big difference.

If the high velocity drag term is used, I think this is the resulting differential
equation (dropping constants):

d2x/d2t = (dx/dt)^2 , which would yield a natural logarithm solution x(t) = ln(c1 + t) +c2

But anyway, I think the low velocity drag is better, which is proportional to velocity,
not velocity squared, and gives an exponential solution (dropping constants):

v(t) = 1 - exp(-t)


Happy Thanksgiving...

[Marching Thru Georgia]

Born2see and Con20or live for TMI. So now let's fill their plate. :laugh:

No, I disagree. Stokes Law is valid when the Reynolds number, R, is less than 1. Actually R<=0.1 is more precise. That represents true laminar flow, which in our case is a sphere moving <1cm/sec.

Now if we were discussing how air drag affects the trajectory of a bullet, then I would endorse using Stokes Law, since it greatly simplifies the calculation. Although Raleigh's Law would still be more appropriate, the solution is not nearly so elegant. And physics is all about elegance.

Just to be clear, by low velocity, I meant subsonic speeds. Although Raleigh's Law can be used to determine the drag coefficient, C, Eulers equation is more appropriate in the supersonic realm.

Now let's give Born2see and Con20or some desert to chew on. We all know that it takes more gas to go the same distance at 60MPH versus 50MPH. It is more than can be explained by the velocity squared. Why? Hint: Road friction is not the answer.

[Barrow]

>The drag force on such a sphere is: f = 1/2Cpav**2

In the current situation the projectile is coming to a complete stop.
The drag force is not a constant in this case, regardless of which
approximation is used. I'm disagreeing with the step

Fd=1/2mv^2, where a constant force is used to solve for d.

Assuming a constant force works well in a situation where the
projectile's deceleration is much less than its velocity, but
not as it's about to stop moving.

[Marching Thru Georgia]

You are quite right. To be precise, we would need to take into account the transition from one regime to the other. If I was approving semester problems for a student to work on in a mechanics class, this would be a good one. But in this forum, with this audience, I think trying to keep things reasonably approximate and simple is best.

[Group Captain Mandrake]

Now let's give Born2see and Con20or some desert to chew on. We all know that it takes more gas to go the same distance at 60MPH versus 50MPH. It is more than can be explained by the velocity squared. Why? Hint: Road friction is not the answer.

I am going to venture the extra fuel consumption beyond that can be ascribed to increased aerodynamic drag and road friction is explained by the higher engine revolutions required. Almost all modern cars will be geared such that 60 mph requires more revs that 50 mph. More revs equals more work expended compressing the gas in the cylinders and and reversing the direction of cyinder motion and fan speed and even perhaps some contribution from more internal friction. There may also be some contribution from driver error if he is trying to eat an egg sandwich at 60 mph.

Also note: Fuel consumption is not solely a function of engine revs. Try going up hill in a BMW and watch the fuel cosnmption gage.

So I bought this game. How on earth do you control anything larger than a brigade?

Also, I have this question. Does the AI give intself any fire advantages at the default level?

[Marching Thru Georgia]

Engine and rolling friction are only minor components. There is one big contributor to the gas consumption. Still waiting to hear from the physics aficionados Born2see and Con20or. :laugh:

[con20or]

Engine and rolling friction are only minor components. There is one big contributor to the gas consumption. Still waiting to hear from the physics aficionados Born2see and Con20or. :laugh:

Keep waiting - you don't get to examine the examiner!

[Group Captain Mandrake]

Engine and rolling friction are only minor components. There is one big contributor to the gas consumption. Still waiting to hear from the physics aficionados Born2see and Con20or. :laugh:

Ah, I get it. This is like a Physics question not an automotive engineering question. From an energy balance standpoint most of the energy expended in the operation of an internal combustion engine in a moving vehicle is lost to heat (radiative and hot exhaust gasses).

The heat losses are inadvertent "waste" resulting from the imperfect conversion of the chemical energy in the fuel. Not sure what the ratio of useful mechanical energy to heat loss is but I doubt it is much better than 50%.

My point about the mechanical efficiency still applies. Most cars will have declining fuel efficiency between 50 and 60 because of drag and gear ratios.

[Hancock the Superb]

Air drag / wind resistance would be the answer you are looking for born2see.

In addition, the solution to the original problem may need to account for rotational energy of the projectile, depending upon whether it is a smoothbore or rifled cannon.

[Marching Thru Georgia]

Hancock The Superb wrote:

In addition, the solution to the original problem may need to account for rotational energy of the projectile, depending upon whether it is a smoothbore or rifled cannon.

Please show your work to justify this statement.