# Tag Info

## Hot answers tagged drag

9

The concept of what you're doing is sound, and as Russell McMahon notes the efficiency gains could be significant enough to justify the change. I'd strongly suggest that you consider adding a ramp to the back edge as well. Drag force is very sensitive to the downstream (rear) end of a body as well You get some positive pressure at the front of the vehicle, ...

6

The air used for the air cushions will come from the air still in the tube pressurized by the pod itself. Thus the system remains closed (all air that is released is sucked out of the environment).

4

Your intuition is correct, those are high. However, you would need to be moving very slowly for wave-making resistance to be negligible. And since it is typically higher than skin-friction I don't think that you can realistically expect to have a significant skin friction and a negligible wave-making resistance. Perhaps a better simplified approach would be ...

4

The examples you mentioned both occur due to turbulent flow of a gas. However in a lot of mechanical systems damping often occurs due visco-elastic deformation or due to shearing of the lubricant inside a bearing, which due to different scales can often be described as a laminar flow. In these cases a good approximation for the "drag forces" would be a ...

3

You do not want to overcome the force of static friction when accelerating because that's when the dynamic friction takes over and your wheels slip. That means that the force applied on the edge of your wheel is limited by the static friction which limits how much effective torque you can apply to accelerate (or maintain speed while overcoming losses due to ...

3

The equation $F_{\textrm{A}} = C_{\textrm{D}}A\rho\dot{x}^2/2$ doesn't tell you that drag is proportional to the square of velocity. In itself, that equation doesn't tell you anything at all about how drag depends on velocity, because that equation is just a definition of the drag coefficient $C_{\textrm{D}}$. What tells you how drag depends on velocity is ...

3

Whether these are sensible Froude numbers or not depends on the length of the vessel in question. For a 100 m ship these are probably high but for a 5 m dinghy these would be quite low numbers. Speed-to-length ratio is the critical factor that determines the importance of skin and wave friction. Skin friction scales with $V^2$, whereas wave drag increases ...

3

Stress on panel mounts needs considering. This can be greatly reduced by a "ramp" that fully protects the panel from air impact in the forward direction. Flat plate drag for a panel will be <= classic drag equation result. $$P_{lost}= D \cdot V$$ = power required to accelerate all opposing air out of the path of the object from rest to velocity $V$ ...

2

I can remember this being discussed during a fluids lecture in my student days a long time ago, when the subject of car roof racks and truck fuel scoops was raised. The panel will increase the drag of the RV and reduce its designed streamlined characteristics. A ramp as you are planning will reduce the drag of the panel but will not restore the original ...

2

"Hull speed" is actually the ratio of speed to the square root of length. To make things even more confusing, length is in feet, and speed is in knots. That's how the constant 1.34 arises. (ProTip: Let's never speak of it again!) Wave resistance ($R_w$) begins its rapid rise at a Froude number (Fr) of about 0.35. Below that Fr, $R_w$ is usually small ...

2

When $C_d$ is an appropriate parameter for discussion, road vehicle aerodynamics uses the planform area of the wing. That's the area formed by the chord (see below) and total width of the wing. You'd use the same area in lift calculations for $C_L$ For complicated wings, or whole vehicles, where defining an area makes little sense, the common practice is ...

2

In the 50's the DARPA designed such a craft but due to aerodynamic instability it was a complete failure and abandoned. The problem you have is maintaining 4 separate intakes which have a greater volumetric capacity than each propeller can handle at maximum power in coldest conditions. You cannot combine the intakes and expect stability, it's just not ...

2

I suggest using the relationship below to derive the equations you are looking for. $F = ma$ $W = mg$ $D = \dfrac{C_d\rho V^2A}{2}$ $C_d$ = Drag Coefficient (Shape dependent) $\rho$ = Atmospheric Density The terminal velocity is reached when $W = D$, $mg = \dfrac{C_d\rho V^2A}{2}$, thus $V_t = \sqrt{\dfrac{2mg}{C_d \rho A}}$ Note, at this stage, $a = \dfrac{... 1 TL;DR: I will expand on DKNguyen answer, which provides all the salient points. i.e. (my interpretation) the downforce is desirable to provide a better interface/grip between the car and the ground.) Comparison of drag and lift coefficients for automobiles. The following graph presents a comparison of drag and lift coefficients. figure (source ... 1 TL;DR: the time required to reach a$p$percentile of the Terminal velocity $$t_p= \sqrt{\frac{2\rho_{sphere}r}{g \cdot C_D \rho_{air}}}arctanh\left(p \right)$$ the distance travelled required to reach a$p$percentile of the Terminal velocity $$x_p= \frac{2\rho_{sphere}r}{C_D \rho_{air}} \log \left(\cosh\left(arctanh\left(p \right)\right )\right )$$ ... 1 The terminal velocity$V_t$will be reached when the drag coefficient is equal to the force of gravity: $$F_{drag} = mg$$ $$C_D\frac{1}{2}\cdot \rho_{air} \cdot A \cdot V_{t}^2= m\cdot g$$ where:$C_D$: is the drag coefficient (for a sphere is 0.5)$\rho_{air}$is density of the liquid the sphere is passing through (if air then 1.225 kg/m3)$A$: cross-... 1 First, pay attention to the fact that the total force acting on the wood is zero. Since your device is turning, it would be much easier to calculate the power as torque times the angular velocity: P = tau * w. When Tau is the total torque around the rotation axis (Nm) and w is the item angular speed (rad/sec). If the drag force is coming from a forced flow ... 1$d C_d/dC_l^2 = dC_d/(2 \cdot C_l \cdot dC_l) = (1/(2\cdot C_l)) \cdot(1/$Slope of$C_d/C_l$curve$)$. Use a few$\alpha$s of interest and read the$C_l$s. Find the slope at those$C_l\$s. The ratio should be consistant.

1

I found some possibly applicable explanations here. The figure referred to in the quote below is The attached plot [Hofman 2000] illustrates the basic principles of added drag due to shallow water. A non-dimensional depth-based speed parameter is on the X axis (usually noted as FNH, but shown here as FL), and a ratio of drag in shallow water to ...

1

The area is named "reference area". As long as it is defined by someone, you are free to choose any arbitrary value for A. To make comparisons easier among different designs for a certain project, engineers define A and they do trade off studies with that reference. This way they can compare the performance of the wings for a certain car/plane model. On ...

1

I believe under Results>Reports>Forces, you can calculate the force on a chosen surface in a particular direction vector (by providing x,y,z coordinates). Does that help?

1

So the formula you need is Density = P /(R*T) and the value of R you have been given is for air. Once you have re-worked your answer then you should calculate the Reynolds number and then you can check your result here : Drag on a sphere by NASA

1

I see how you're thinking: the car is a "piston" and if you heat the air behind this "piston", it will push it forward. Now think how inefficient this "engine" is - you are adding fractions of a degree to the air (working gas) which makes for horrible thermodynamic efficiency. On top of that, you have no "piston walls".

1

The thermic energy output of a car is too small to make a difference. Also the technical implemention would be very difficult. Did you ever hold your hand near to the exhaust. Its really not that hot compared to a camp fire. But this was in idle so when driving it might be different. On the other hand when driving you also have a lot of air mass to be ...

1

So first, there things known as dashpots, for which the force really is proportional to velocity. These are the things that intended to be represented by the damping term in the classical mass-spring-damper systems that you find in the textbooks. But there is a much bigger picture idea here. The mass-spring-damper systems in textbook are just mathematical ...

1

Most race car wings are a lot more complex than an aircraft wing as they are usually designed to manage airflow over the whole body of the car, particularly the wheels and also have to deal with turbulent flow both from preceding aerodynamic surfaces and cars which they are following as well as controlling vortex generation to control airflow over their own ...

1

You are correct that the Froude number (Fr) is very important for wave resistance. The answer given by nivag regarding the speed-length ratio (also known as "hull speed") is not correct. That limit is often quoted, but it is a myth that it is somehow impossible for displacement hulls to surpass it. Ships can travel faster than that ratio implies, but for ...

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