# Accelerating a moving object with force

1. May 15, 2012

Example dyno graph: http://i1105.photobucket.com/albums/h355/magnethead494/turbo_busa_dyno_hahn_stg1.jpg

I'm familiar with F = MA/G => F*G/M = A for english units (G = 32.17).

This works for accelerating from a standstill (IE, approx'ing to top of first gear),

160 ft-lb * 1.596 primary * 2.615 first gear * 4.0625 chain drive = 2,712 ft-lb at the axle
2,712 * 12 inches / 13.25 inch rolling radius = 2,456 pound force
2,456 pound-force / 700 pound-weight = 3.50 G-forces = 112.87 feet per second squared average acceleration

I know that seems like a high acceleration, but for a 300HP 700 pound machine, that's a 2.3 pound-per-HP ratio.

And the formula

(Vf)2 = (V0)2 + 2ad

for subsequent gears.... But I don't know x.

Applying that formula to previous calculation,

(75.68 feet/sec)2 = (0MPH)2 + 2 * 112.87 fps2 * d

5,727.4624 feet2 per second2 = 225.74 fps2 * d

5,727.4624 feet2 per second2 / 225.74 feet per second2 = 25.372 feet to the top of first gear (AKA 25.372 feet to 51.6 miles per hour, at average 3.5G's)

But how can I calculate the acceleration for subsequent gears?
I'm going to state my 3-part hypothesis..am I right or wrong?

Part 1: Using F=Ma to get acceleration per gear

Expanded:

top of gear:
primary * chain = 6.48375
rolling radius multiplier = 0.88888
Engine Torque at 10,500 RPM: 160 ft-lb
Composite value = 160 * 6.48375 * 0.8888888 = 922.13333
1st: 922.1333 * 2.615 / 700 * 32.17 = 110.82 feet per second squared (3.44G)
2nd: 922.1333 * 1.937 / 700 * 32.17 = 82.08 feet per second squared (2.55G)
3rd: 922.1333 * 1.526 / 700 * 32.17 = 64.67 feet per second squared (2.01G)
4th: 922.1333 * 1.285 / 700 * 32.17 = 54.45 feet per second squared (1.69G)
5th: 922.1333 * 1.136 / 700 * 32.17 = 48.14 feet per second squared (1.50G)
6th: 922.1333 * 1.043 / 700 * 32.17 = 44.20 feet per second squared (1.37G)

Part 2: Getting the distance per gear

1st: 75.682 = 02 + 2 * 110.82 * d => 5,727.46/221.64 = 25.84ft
2nd: 102.182 = 75.682 + 2 * 82.08 * d => 10,440.75 - 5,727.46 = 164.16 * d => 4,713.29 / 165.16 = 28.53ft
3rd: 129.712 = 102.182 + 2 * 64.67 * d => 16,824.68 - 10,440.75 = 129.34 * d => 6,383.93 / 129.34 = 49.35ft
4th: 154.032 = 129.712 + 2 * 54.45 * d => 23,725.24 - 16,824.68 = 108.90 * d => 6,900.56 / 108.90 = 63.36ft
5th: 174.242 = 154.032 + 2 * 48.14 * d => 30,359.58 - 23,725.24 = 96.28 * d => 6,634.04 / 96.28 = 68.90ft
6th: 189.77 = 174.242 + 2 * 44.20 * d => 36,012.65 - 30,359.58 = 88.40 * d => 5,653.07 / 88.40 = 63.95ft

25.84 + 28.5 + 49.35 + 63.36 + 68.90 + 63.95 = 299.9 feet (should equal something near 660ft)

Something is clearly wrong with the calculation...but what is it? The G forces seem realistic, the MPH calculations (spreadsheet and google to FPS calcs) seem correct, what else could be wrong?

Part 3: getting the time per gear

d = t/2 * (Vf+V0)

1st: 75.682 = 02 + (2 * 110.82 * t/2 * (75.68 + 0)) => 5,727.46 = 8,386.86 * t => 0.683 seconds
2nd: 102.182 = 75.682 + (2 * 82.08 * t/2 * (102.18 + 75.68)) => 10,440.75 = 5,727.46 + 14,598.75t => 0.323 seconds
3rd: 129.712 = 102.182 + (2 * 64.67 * t/2 * (129.71 + 102.18)) => 16,824.68 = 10,440.75 + 14,996.33t => 0.426 seconds
4th: 154.032 = 129.712 + (2 * 54.45 * t/2 * (154.03 + 129.71)) => 23,725.24 = 16,824.68 + 15,449.64t => 0.447 seconds
5th: 174.242 = 154.032 + (2 * 48.14 * t/2 * (174.24 + 154.03)) => 30,359.58 = 23,725.24 + 15,802.92t => 0.420 seconds
6th: 189.77 = 174.242 + (2 * 44.20 * t/2 * (189.77 + 174.24)) => 36,012.65 = 30,359.58 + 16,089.24t => 0.351 seconds

0.683 + 0.323 + 0.426 + 0.447 + 0.420 + 0.351 = 2.65 seconds (should equal something near 4.95 seconds)

Again, something is clearly wrong. I have a feeling it's in the main equation...

(Vf)2 = (V0)2 + 2ad

A = G * F/m

d = t/2 * (Vf+V0)

If someone could point my mistakes, it would be appreciated.

2. May 15, 2012

### haruspex

You've neglected air resistance. That would be pretty significant long before you reach top speed.

3. May 15, 2012

One would think so, but it's surprisingly not as big an issue as one would think. The body panels and ground effects are done up to where, once you have the power on tap, air resistance is a very small loss, maybe 4% on MPH.

Either way, that's nothing to do with the question I asked. I asked where my calculations were off by 200% or so.

Last edited: May 15, 2012
4. May 15, 2012

### haruspex

Then you're assuming constant power (300HP) and no drag.
In that case we can skip the details and go straight to the energy.
A 700 pound (318kg) mass at 189.77 fps (57.86 mps, 129Mph) has 532.5kJ.
A power of 300HP (225000 W, or thereabouts) delivers that in 2.37 seconds.
If that is not reflected in reality then you must be ignoring a significant loss somewhere.