Why does coefficient of drag increase with velocity?

[Camoman1]

I’ve been reading this article by AB
http://appliedballisticsllc.com/Articles/ABDOC130_CDM.pdf

I can’t get why cd increases with velocity, it stated that it’s important but does not explain why it happens?

 

[hogpatrol]

Do you have that backwards? And it is explained.

From the text: "Before we talk anymore about the drag curves, we have to address the elephant in the room: Why does the drag coefficient (CD) go down as speed increases!?!?"

 

[rkittine]

https://www.physicsforums.com/threads/how-to-calculate-drag-coefficient.667709/

 

[Laurie]

It's all to do with the shock waves generated just ahead of the bullet tip which fan out like an inverted V with the bullet inside.

At very high supersonic speeds, the shock waves have a greater amount of separation from the bullet reducing turbulence and drag along its edges. As speeds drop, the amount reduces, turbulence around the bullet and drag increase peaking just above the speed of sound. Transonic speeds are a related but separate issue that sees a large increase in bullet drag and turbulence around it as some parts of the airflow around the bullet are supersonic and others sub as the air follows the bullet body's shape. Once on the speed of sound, the supersonic shock wave disappears entirely, likewise most turbulence around the bullet and drag is very much less as a result.

That's why the typical drag curve as seen for example in Bryan Litz's / Applied Ballistics books shows a curve that starts low (at the right which is the high velocity side), rises steadily as you go left with reducing velocity, more steeply through the transonic zone and peaks just above the speed of sound with an even faster rapid fall to the left (subsonic) side of the 1.0 MACH line.

http://bulletin.accurateshooter.com/?s=bullet+drag+&submit=Search

That's my layman's understanding and there are now doubt many wrinkles and subtleties in the precise nature of what affects drag at different speeds. For instance, we now know that a higher degree of gyroscopic stabilisation is needed with Sg values of 1.5 and above to reduce various flight behaviours that produce extra drag with modern very long for calibre streamlined bullets.

 

[JPeelen]

The drag force on the bullet does not really go down.
It basically increases with the -square- of the velocity. Only after this is computed, the drag coefficient is applied to cater for the variation from pure quadratic drag increase.
Take for example a bullet that has a cd of 0.660 at 1600 fps and 0.607 at 2200 fps (cd goes down).
Taking into account the quadratic law only, an increase from 1600 to 2200 fps (speed increase by 1.38) equals to a drag increase of 1.89. That is the relation of 2200 squared to 1600 squared.
So assuming drag at 1600 fps as 100 percent, at 2200 fps it would grow to 189 percent. Only then the cd is applied:
at 1600 fps it is 100 % (drag of a flat plate) times 0.660 cd of the bullet = 66 percent
at 2200 fps it is 189 % (drag of a flat plate) times 0.607 cd of the bullet = 115 percent

Therefore, don't let the cd curve mislead you. The net drag gets -larger- with speed. The drag coefficient only describes the variation compared to a flat plate at the same conditions. (I simplify a little.)
In the example shown, the drag force that slows down the bullet grows from 66 to 115 percent (compared to a flat plate of the same cross section), although cd drops from 0.660 to 0.607 at the two velocities.

 

[geraldgee]

Net drag increases as velocity increases. I like to think of it in terms of wind resistance - when you put your hand out the car window at 65 mph versus 25 mph. Married folks might relate it better to the resistance from your spouse when purchasing a $7K bench rifle compared to a $500 Tikka. Its all relative.

LOL

 

[swadiver]

as stated, drag is a v squared function

 

[hogpatrol]

All this mathematics and computations. My eyeballs are starting to ache. Anyway, here's another lesson on drag.

(Might as well send this one off the rails. It's been a long winter.)

 

[geraldgee]

HAHAHAHAHAHAHAHAHAHA

 

[Camoman1]

It's all to do with the shock waves generated just ahead of the bullet tip which fan out like an inverted V with the bullet inside.

At very high supersonic speeds, the shock waves have a greater amount of separation from the bullet reducing turbulence and drag along its edges. As speeds drop, the amount reduces, turbulence around the bullet and drag increase peaking just above the speed of sound. Transonic speeds are a related but separate issue that sees a large increase in bullet drag and turbulence around it as some parts of the airflow around the bullet are supersonic and others sub as the air follows the bullet body's shape. Once on the speed of sound, the supersonic shock wave disappears entirely, likewise most turbulence around the bullet and drag is very much less as a result.

That's why the typical drag curve as seen for example in Bryan Litz's / Applied Ballistics books shows a curve that starts low (at the right which is the high velocity side), rises steadily as you go left with reducing velocity, more steeply through the transonic zone and peaks just above the speed of sound with an even faster rapid fall to the left (subsonic) side of the 1.0 MACH line.

http://bulletin.accurateshooter.com/?s=bullet+drag+&submit=Search

That's my layman's understanding and there are now doubt many wrinkles and subtleties in the precise nature of what affects drag at different speeds. For instance, we now know that a higher degree of gyroscopic stabilisation is needed with Sg values of 1.5 and above to reduce various flight behaviours that produce extra drag with modern very long for calibre streamlined bullets.

This makes sense thank you

 

[Camoman1]

The drag force on the bullet does not really go down.
It basically increases with the -square- of the velocity. Only after this is computed, the drag coefficient is applied to cater for the variation from pure quadratic drag increase.
Take for example a bullet that has a cd of 0.660 at 1600 fps and 0.607 at 2200 fps (cd goes down).
Taking into account the quadratic law only, an increase from 1600 to 2200 fps (speed increase by 1.38) equals to a drag increase of 1.89. That is the relation of 2200 squared to 1600 squared.
So assuming drag at 1600 fps as 100 percent, at 2200 fps it would grow to 189 percent. Only then the cd is applied:
at 1600 fps it is 100 % (drag of a flat plate) times 0.660 cd of the bullet = 66 percent
at 2200 fps it is 189 % (drag of a flat plate) times 0.607 cd of the bullet = 115 percent

Therefore, don't let the cd curve mislead you. The net drag gets -larger- with speed. The drag coefficient only describes the variation compared to a flat plate at the same conditions. (I simplify a little.)
In the example shown, the drag force that slows down the bullet grows from 66 to 115 percent (compared to a flat plate of the same cross section), although cd drops from 0.660 to 0.607 at the two velocities.

This clears up some more questions thanks!
Im going down the rabbit hole... My OCD is kicking in and I aint happy until I know everything lol

 

[Laurie]

The drag force on the bullet does not really go down. ........... Therefore, don't let the cd curve mislead you. The net drag gets -larger- with speed. The drag coefficient only describes the variation compared to a flat plate at the same conditions. (I simplify a little.)
In the example shown, the drag force that slows down the bullet grows from 66 to 115 percent (compared to a flat plate of the same cross section), although cd drops from 0.660 to 0.607 at the two velocities.

Ah, thanks for that. These damned ballistic laws; every time you think you're getting a bit of a handle on things ....................!!

 

[243Mendoza]

I’ve been reading this article by AB
http://appliedballisticsllc.com/Articles/ABDOC130_CDM.pdf

I can’t get why cd increases with velocity, it stated that it’s important but does not explain why it happens?

Generally Cd is a function of Mach number and Reynold's Number. For small caliber ballistics, CD is mainly a function of Mach number where Mach number is velocity divided by the speed of sound.

Drag = Cd*density of air*velocity squared*cross-sectional area/2.

The high Cd at transonic Mach numbers is due to normal shocks appearing on the bullet surface causing flow separation.

 

[dcali]

Shell Oil a long while ago created a series of videos that explain drag from subsonic to supersonic better than I’ve ever seen without math. Google “ shell oil supersonic flight” and carve out 45 min or so for an education.

The short answer is “shock waves”. The long answer is much more interesting.

 

[243Mendoza]

Shell Oil a long while ago created a series of videos that explain drag from subsonic to supersonic better than I’ve ever seen without math. Google “ shell oil supersonic flight” and carve out 45 min or so for an education.

The short answer is “shock waves”. The long answer is much more interesting.

good video

 

[243Mendoza]

Generally Cd is a function of Mach number and Reynold's Number. For small caliber ballistics, CD is mainly a function of Mach number where Mach number is velocity divided by the speed of sound.

Drag = Cd*density of air*velocity squared*cross-sectional area/2.

Should add that Cd is also a function of angle of attack at larger angles.

The high Cd at transonic Mach numbers is due to normal shocks appearing on the bullet surface causing flow separation.

 

[Delfuego]

That movie was awesome! Thanks.

 

[wholman]

Hog that is some ugly ass drag. Hope my bullets don't have that mucth drag.