The drag force on the free balloon is proportional to the square of its airspeed, that is, its speed relative to the air mass. Its airspeed is thus the difference between its ground speed and the wind speed.
When the balloon is launched, it has no ground speed, and so its air speed is equal to the wind speed (but opposite in sign). The drag force on it is high, and it accelerates quickly. As it accelerates, its ground speed increases. This causes the air speed to decrease, and thus the drag force on it to decrease. Thus, it continues to gain speed, but more slowly. If the wind remained constant, its ground speed would approach, more and more closely, the wind speed. A gust could cause it to temporarily travel faster than the "normal" wind speed, in which case the drag force would cause it to decelerate.
While only one horizontal force (the wind) acts on free balloon, an airship is subject to such forces, the wind and the propulsive force of its propellers or jets. For the purpose of this article, we assume axial propulsion, that is the engine drives the airship forward.
It's time to talk about velocity. To a physicist, velocity isn't the same as speed, velocity is a vector which has both a magnitude (the speed) and a direction.
The basic equation of airship motion is:
velocity (ship relative to ground) = velocity (ship relative to air mass) + velocity (air mass relative to ground).
This equation may be rearranged to solve for the velocity (ship relative to air mass).
The ground velocity is defined by a ground speed (ship relative to ground) and a course (the geographic direction toward which the ship is moving). The air velocity is defined by an air speed (ship relative to air mass) and a heading (the geographic direction toward which the nose is pointing). And the air mass (true wind) velocity is defined by a wind speed and direction.
Course, heading and wind direction are all defined so that north is zero degrees, and the angle increases clockwise. (This is not, by the way, the same convention that mathematicians used to express angles, so some mathematical conversions are necessary in order to apply trigonometric functions properly.)
A further complication is that for the vector mathematics to work, the wind must be expressed as the direction the air is moving toward, whereas meteorologists define the wind direction as the direction the wind is coming from. If I refer to the direction of the wind vector, I mean the TO direction.
The drag force is based on the apparent wind, that is, the velocity of the air mass relative to the ship. That's the opposite of the velocity of the ship relative to the air mass. The ship's heading is chosen so that this apparent wind is coming over the nose (headwind) or over the tail (tail), i.e., so that there's no crosswind. (The ship, aided by its fins, acts like a weathervane, turning into or away from the wind to make this happen).
If the ship is unable to quite make this heading (because the wind keeps shifting faster then the ship can turn), then it will feel an apparent crosswind, creating a side force that causes it to "crab" or "sideslip," a movement sidewise in the direction the apparent wind is blowing (the equivalent of leeway for a watership). For the purpose of this article, we will be ignoring sideslip and side drag.
If there's no wind, the air speed equals the ground speed. A wind that's a headwind (directly opposing movement down-course) increases the airspeed (and thus the drag), and a tailwind decreases it, by the amount of the wind speed.
Vector mathematics is necessary to calculate the effects of in-between winds. To add (or subtract) vectors, we "resolve" the vector into two mutually perpendicular components, for example, a north-south and an east-west component. If you are heading 20 mph northwest, that resolves to 14.1 mph north and 14.1 mph west.
If vectors are to be added, we separately add up their north-south components, and their east-west components, and then recombine the components to get the combined vector (resultant). For example, motion 40 mph west and 30 mph north corresponds to movement of 50 mph in a direction about 37o north of west. Trigonometry, which is known to the down-timers, is needed to make these calculations, but vector mathematics is new to them.
Sometimes, it's informative to resolve a vector into components other than geographic. For example, if we resolve the wind into a component in the direction ("down") the ground course and one perpendicular ("cross") to it, then we can readily see how much a favorable wind is helping us along and how much it's trying to blow us off course.
The CWV angle in the table below is the unsigned angle between the course (C) and the true wind vector (WV). Thus, if your course is due west, and the wind is from the northeast, the wind vector is to the southwest and the CWV angle is 45o. The CWV angle is 0 for a down-course (tail) wind and 180 for an up-course (head) wind.
The underlying equation, if you're wondering, is rather simple:
(AS/GS)=sqrt((WS/GS)^2 -2*(WS/GS)*cos(CWVang)+1) [equation 1].
It shouldn't be surprising that even an oblique headwind increases airspeed as a percentage of ground speed.
However, notice that even a pure crosswind is bad. Why? because to keep the crosswind from pushing you off course, you have to point the nose a little bit upwind to compensate, and then you have to increase power so you maintain the required ground speed.
If you're familiar with sailing ships, that may seem strange. Sailing ships do quite well with a wind off the beam. However, sailing ships capture the wind mostly on their sails, not their superstructure. The sails can be braced about to face the wind more directly. The more directly it faces the wind, the greater the percentage of the wind force that is felt by the ship. However, the greater the bracing angle, the greater the percentage of that felt force that is driving the ship sidewise rather than forward. But a watership isn't forced directly downwind like a free balloon because the lateral resistance is proportional to the density of water (much higher than air) and the resistance is increased by the keel, centerboard, etc. The bracing angle chosen compromises between increasing driving force and increasing leeway. For a wind off the beam, it would be 45o.
For an airship, which doesn't have sails, only the component of the wind in the down-course direction is helpful, the cross-course component must be fought.
Note that if wind speed exceeds the desired ground speed, drag increases even when the wind is from a favorable direction. Drag is the result of the relative difference in speed between the ship and the air, and it doesn't matter which is moving faster. Of course, it's likely that if you are in an area of strong wind of favorable direction, you will happily allow your airship to behave like a free balloon, and let its ground speed equal the wind speed and its airspeed (and drag) drop to zero. But if there was a reason you couldn't do this-perhaps you're escorting a surface ship-then you must pay the piper.
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Now, let's look at the consequences of the numbers in the table. Drag is proportional to the square of the airspeed, and the power to overcome the drag is proportional to the cube of the airspeed. The energy required for the journey is proportional to the square of the airspeed (and the distance to be traveled).
So, if the airspeed is 50% of the ground speed, then the drag is 25% of what you'd experience with the same ground speed in still air, and the power requirement just 12.5%.
Plainly, if you don't have to go very far out of the way to take advantage of a favorable wind, you should do so.