 # Horizontal Flight

### How long goes it take to accelerate?

In order to keep your velocity in a horizontal direction your thrust has to be pointed forward to cancel the drag force due to air friction and upwards to cancel the downwards pull of gravity. In order to change velocity you need to experience an acceleration; in order to accelerate there needs to be a net force acting on you. Expressing the net horizontal force acting on a character who is accelerating in a horizontal direction gives us the following (ugh...) differential equation:

m a = Fthrust cos ( b ) - Ffriction
m ( dv / dt ) = Fthrust cos ( b ) - cfriction v2

b is defined as the angle between the direction of Fthrust and the direction of travel. It can be shown that the value of cos ( b ) is found to be:

Fthrust sin ( b ) = m g
b
= sin-1 ( m g / Fthrust )

To make the differential equation easier to work with, I will also define the following, where vcritical is just some special value of velocity that satisfies the definition:

Fthrust cos ( b ) = cfriction vcritical2

which makes the differential equation look like:

m ( dv / dt ) = cfriction vcritical2 - cfriction v2
m ( dv / dt ) = cfriction ( vcritical2 - v2 )

The solution to this differential equation is found by rearranging the terms and: 1) integrating over velocity from zero to v, and 2) integrating over time from zero to the answer of our question, taccel. Because most readers capable of following the steps required to obtain this result can probably carry them out without my guidance, I'll state that the end result of these integrations is:

taccel = m / ( 2 cfriction vcritical ) * ln { ( vcritical + v ) / ( vcritical - v ) }

At this point we should think about the value of vcritical. Remember that when solving for the maximum value of Fthrust we made a diagram of the forces acting on a character at maximum velocity. Looking at that diagram we see that: Fthrust sin ( b ) = m g Fthrust cos ( b ) = cfriction vmax2 Assuming that we want to accelerate as fast as possible and will use our maximum thrust, we see that vcritical and vmax are equal!

Finally we have an expression for the time required to accelerate from zero velocity to any desired horizontal velocity:

taccel = m / ( 2 cfriction vmax ) * ln { ( vmax + v ) / ( vmax - v ) }

This calculation can become quicker if we make use of the convention: v = N vmax , which gives us:

taccel = m / ( 2 cfriction vmax ) * ln { ( 1 + N ) / ( 1 - N ) }

fN = �  ln { ( 1 + N ) / ( 1 - N ) }

taccel = m fN / ( cfriction vmax )

The logarithmic term does not depend on Fthrust, vmax, or any other character-specific quantity. This means we can make a table of values for this quantity, and players will only have to enter in their character's mass and maximum velocity in order to get the acceleration time to a certain percentage of their maximum velocity. For any character with flight powers:

 N 0.99 0.90 0.75 0.50 0.25 0.10 0.05 0.01 fN 2.6467 1.4722 0.97296 0.54931 0.25541 0.10034 0.05004 0.01000

### Sample Results

Assuming cfriction = ( 0.08125 kg/m), for a 250 lb ( m = 113.4 kg ) character with Sonic Flight, and thus a maximum velocity of 700 mph ( 312.9 m/s), it takes 11.8 seconds to get up to 99% of his vmax, 2.45 seconds to get to 50% of vmax, and 0.045 seconds to get to 1% of vmax . As a comparison, it would take a first level character with Flight: Wingless ( vmax = 220 mph = 98.3 m/s ) 37.6 seconds to get to 99% of vmax, 7.8 seconds to get to 50% of vmax, and 0.14 seconds to get to 1% of vmax .

### How long does it take to stop?

When characters are already at a given velocity, and wish to stop, they must direct their thrust upwards to cancel gravity and backwards, to assist the deceleration caused by air friction. Because the thrust is pointed in a different direction, the differential equation expressing the net force on the character is slightly different and will have a different solution:

m ( dv / dt ) = Fthrust cos ( b ) + cfriction v2
m ( dv / dt ) = cfriction ( vmax2 + v2 )

tdecel = m / ( cfriction vmax ) * tan-1 ( v / vmax )
tdecel = m / ( cfriction vmax ) * tan-1 ( N )

sN = tan-1 ( N )

tdecel = m sN / ( cfriction vmax )

Just like in the case of acceleration from zero velocity, we end up with a result that depends only on the mass and maximum velocity available to a character. For any character with flight powers:

 N 0.99 0.90 0.75 0.50 0.25 0.10 0.05 0.01 sN 0.7804 0.7328 0.6435 0.4636 0.2450 0.09967 0.04996 0.01000

### More Sample Results

Assuming cfriction = ( 0.08125 kg/m), for a 250 lb ( m = 113.4 kg ) character with Sonic Flight, and thus a maximum velocity of 700 mph ( 312.9 m/s), it takes 3.48 seconds to come to a full stop from 99% of his vmax, 2.07 seconds to stop from 50% of vmax, and 0.045 seconds to stop from 1% of vmax . As a comparison, it would take a first level character with Flight: Wingless ( vmax = 220 mph = 98.3 m/s ) 11.07 seconds to stop from 99% of vmax, 6.58 seconds to stop from 50% of vmax, and 0.14 seconds to stop from 1% of vmax .

Compare these results to those obtained above for acceleration: It takes less time for a character to decelerate than to accelerate because air friction always helps slow you down. For low velocities air friction has only a small effect, and it takes about the same time to accelerate as it does to decelerate, which means that fN and sN will be about equal.