How do you calculate the drag of a simple object

[Mat]

How do you calculate the drag of a simple object?

Hi I am wondering how you find the drag of a triangle shape and am wondering if a 12in long and 2in high and 2in wide or an 8in long and 2 in high and 2 in wide would have more drag

[Karl B²]

Need more information.

You don't specify in which direction the object is moving relative to the flow surrounding it. Am I to assume that the sharp point faces the flow, with the short side aft, like a wedge?

For a streamlined, teardrop shaped section (rounded leading edge, tapered trailing edge) with a fixed width, such as a symmetrical airfoil or the shape of a fish, a length to width ratio of about 4 to 1 usually gives the least drag. Surface drag will increase with length, and [I forget the exact term:] dynamic drag will increase as length decreases.

There is likely to be a similar optimal length to width ratio for your triangle if the stated assumed flow arrangement is correct, but I have neither the experience nor the mathematics to model such a flow. For a fixed width, a shorter length will increase the angle of the sides against the airflow and increase dynamic drag, while parasitic drag will increase with length. In this case, some real-world testing would probably be required to optimise for minimal drag.

Karl

[Mat]

The sharp point would be at the front in motion. It make a right angle.

[Karl B²]

Can you sketch what you're considering, such as with a simple graphic program like MS Paint, then post it here? It's hard to visualize from your description.

[Mat]

This picture isnt very good. But here's what I mean.
Would vortex drag effect it alot?
Is there a simple formula to just put in a plane at an angle and find that resistance and not the vortex drag easily?
The picture was too large for me to upload so I put it on my stupid website
scrol down to see it.
http://www.geocities.com/a_awsome_milkman/

[Karl B²]

Got it now. Is this to be mounted to a flat surface, such as a fuselage? I'm not a fluid dynamics engineer, but I play one on TV. :)

Dynamic drag occurs when the airflow is forced to move in a different direction. At the two slopes you're considering (8 & 12" length), the slope itself won't be the largest drag contributor. As the flow moves up the slope it will be compressed slightly, which will also tend to make it move towards the outer edges. This is the same effect seen when raindrops on a car windshield stream not only upwards but outwards.

When this airflow encounters a sharp edge, it suddenly encounters lower pressure which draws the flow. Your airflow will be moving towards the sides of the slope as well as to the apex of the wedge at the back. The flow across these boundaries will likely form turbulent vortexes as they pass the sudden breaks, and vortexes are big contributors to drag.

It should be possible to calculate the drag coefficient to determine how much drag there is at a given speed, but again, that's way beyond my abilities. This thread would benefit from more exposure to the better minds that haunt the Modeling Science forum.

Karl

[Mat]

It is to be a very simle rocket experiment and it is that shape because of how the engine will be mountted to the launcher.

Is Drag porportional to velocity?

[Ham-fisted-oaff]

It is to be a very simle rocket experiment and it is that shape because of how the engine will be mountted to the launcher.

Is Drag porportional to velocity?

Nope. Velocity squared:

Fd = 0.5 * Cd * A * density * U^^2

Cd = drag coefficient. See comment below about proper Cd values.

For drag coefficient go to a library and look at technical books on fluid dynamics or aerodynamics. Text books often have tables of Cd for various shapes in the back.

A = projected area. That's the largest cross section perpendicular to the direction of flow.

density = density of the fluid flowing past the object. For liquids this is constant, but for gases it varies with pressure.

U^^2 = velocity squared.

For metric calculations the units are:

Fd = Newtons

velocity - meters/second

density - kilgrammes per cubic meter

A - meters squared.

Cd is dimensionless

For compressible fluids (gases), the density will change with pressure and temperature, but for modelling level engineering constant values can be assumed.

The drag coefficient will change according to the velocity of the object (or of the gas flowing past), or more scientifically the Reynolds' number of the flow. A Reynolds' number is a "dimensionless group" and measures the criticality of the flow. We engineers can use Reynolds numbers to test flows with one fluid and predict how the flow would be altered for a different fluid, e.g. test with water, pump oil. Most Cd tables will give drag coefficients for ranges of Reynolds' Numbers.

To calculate the Reynolds' Number:

Re = characteristic length * velocity * density / dynamic (absolute) viscosity.

The characteristic length is, in your case, found by (2 x width + 2 x height)/4

For air at 20 Celsius (~70 Fahrenheit)

density = 1.205 kgm^^3

dynamic viscosity = 1.709 * 10^-6 kg * sec / meter

For fresh water. the corresponding numbers are 1000 kg*m^^-3 and 9.82 * 10^-4 kg * sec /meter.

I don't have my texts with me today, but with that shape I'd expect a Cd of around 1.0 to 2.0.

[kallend]

Nope. Velocity squared:

Fd = 0.5 * Cd * A * density * U^^2

Cd = drag coefficient. See comment below about proper Cd values.

For drag coefficient go to a library and look at technical books on fluid dynamics or aerodynamics. Text books often have tables of Cd for various shapes in the back.

A = projected area. That's the largest cross section perpendicular to the direction of flow.

density = density of the fluid flowing past the object. For liquids this is constant, but for gases it varies with pressure.

U^^2 = velocity squared.

For metric calculations the units are:

Fd = Newtons

velocity - meters/second

density - kilgrammes per cubic meter

A - meters squared.

Cd is dimensionless

For compressible fluids (gases), the density will change with pressure and temperature, but for modelling level engineering constant values can be assumed.

The drag coefficient will change according to the velocity of the object (or of the gas flowing past), or more scientifically the Reynolds' number of the flow. A Reynolds' number is a "dimensionless group" and measures the criticality of the flow. We engineers can use Reynolds numbers to test flows with one fluid and predict how the flow would be altered for a different fluid, e.g. test with water, pump oil. Most Cd tables will give drag coefficients for ranges of Reynolds' Numbers.

To calculate the Reynolds' Number:

Re = characteristic length * velocity * density / dynamic (absolute) viscosity.

The characteristic length is, in your case, found by (2 x width + 2 x height)/4

For air at 20 Celsius (~70 Fahrenheit)

density = 1.205 kgm^^3

dynamic viscosity = 1.709 * 10^-6 kg * sec / meter

For fresh water. the corresponding numbers are 1000 kg*m^^-3 and 9.82 * 10^-4 kg * sec /meter.

I don't have my texts with me today, but with that shape I'd expect a Cd of around 1.0 to 2.0.

No, it wouldn't be that high. Even a parachute (traditional round one) has a Cd of only 1.3 according to Hoerner's book "Fluid Dynamic Drag". I'd expect it to be closer to 0.6 - 0.8

[Ham-fisted-oaff]

No, it wouldn't be that high. Even a parachute (traditional round one) has a Cd of only 1.3 according to Hoerner's book "Fluid Dynamic Drag". I'd expect it to be closer to 0.6 - 0.8

But parachutes are made from semi-porous cloth, not a solid object? Therefore effective drag reduced by air flowing through the parachute as much as around it?

Cd of 1.0 to 2.0 may be a little high, but for conservative estimates I wouldn't assume less than 0.7 (cylinder in cross flow?)

[kallend]

OK, according to Hoerner (Figure 3-34) the Cd of a triangular wedge going point first into the airflow is very sensitive to the "half vertex angle", which for your 12" by 2" wedge will be around 5 degrees. Hoerner gives the Cd in this case as around 0.75 (it's on a graph and difficult to get an exact value). As the "half vertex angle" increases, the Cd goes up rapidly.

BTW the porosity of parachute fabric is minimal. (I am a skydiver with over 1500 jumps).

[matttay]

Cd of 1.0 to 2.0 may be a little high, but for conservative estimates I wouldn't assume less than 0.7 (cylinder in cross flow?)

There's an analysis out on the web that examines the torque induced on a servo due to aileron/rudder/elevator positioning, model speed, arm lengths, etc.

In their analysis I think they also assume a cd of 1.0 to 2.0 for flaps. Obviously the frontal area is a function flap deflection. So a flap might have Cd of 2, but without any deflection its frontal area is nearly 0, so there's nearly 0 drag.