- Going-Gravity-Free-0.mp3
- Going-Gravity-Free-0.mp4
- Going-Gravity-Free-I.mp3
- Going-Gravity-Free-I.mp4
- Going-Gravity-Free-II.mp3
- Going-Gravity-Free-II.mp4
- Going-Gravity-Free-Reggae.mp3
- Going-Gravity-Free-Reggae.mp4
- Going-Gravity-Free-electric-intro.mp3
[Intro]
Ninety-eight miles per hour
(Freedom will be ours)
In the head, feeling light
(Taking flight)
[Verse 1]
That wind
(Is blowing up a storm)
Blow out
(No, not about the norm)
[Bridge]
That wind
(Cuts me like a knife)
For sure
(The sheerest of my life)
[Chorus]
Intensity of the velocity
(Getting through to me)
Accuracy of the veracity
(Going gravity free)
Come with me…?
(Going gravity free)
[Verse 2]
That wind
(Can lift me off the ground)
Way less
(Weightless I have found)
[Bridge]
That wind
(Cuts me like a knife)
For sure
(The sheerest of my life)
[Chorus]
Intensity of the velocity
(Getting through to me)
Accuracy of the veracity
(Going gravity free)
Come with me…?
(Going gravity free)
[Bridge]
That wind
(Cuts me like a knife)
For sure
(The sheerest of my life)
[Chorus]
Intensity of the velocity
(Getting through to me)
Accuracy of the veracity
(Going gravity free)
[Outro]
Come with me…?
(Going gravity free)
A SCIENCE NOTE
To make the average-sized person feel “gravity-free” or effectively lift them off the ground, the wind speed must create enough upward force to counteract their weight (force due to gravity). Here’s how we approach the calculation:
1. Force due to Gravity (Weight)
The weight of an average person is about 700 N700 \, \text{N} (assuming a mass of 70 kg and gravity of 9.8 m/s29.8 \, \text{m/s}^2).
2. Lift Force from Wind
The upward force from the wind depends on its speed, the person’s body surface area exposed to the wind, and the drag coefficient. The force is given by:
F=12ρv2CdAF = \frac{1}{2} \rho v^2 C_d A
Where:
- FF: Force (N)
- ρ\rho: Air density (≈1.225 kg/m3\approx 1.225 \, \text{kg/m}^3 at sea level)
- vv: Wind speed (m/s\text{m/s})
- CdC_d: Drag coefficient (typically 1.0–1.3 for a standing person)
- AA: Cross-sectional area exposed to wind (≈0.5 m2\approx 0.5 \, \text{m}^2 for an average person)
3. Equating Forces
For the person to feel “gravity-free,” the upward force FF must equal their weight WW. Substituting values:
700=12×1.225×v2×1.2×0.5700 = \frac{1}{2} \times 1.225 \times v^2 \times 1.2 \times 0.5
Solving for vv:
v2=7000.3675≈1904v^2 = \frac{700}{0.3675} \approx 1904 v≈43.6 m/s ≈157 km/h ≈97.6 mphv \approx 43.6 \, \text{m/s} \, \approx 157 \, \text{km/h} \, \approx 97.6 \, \text{mph}
4. Conclusion
A wind speed of approximately 43.6 m/s (97.6 mph) is required to lift an average person off the ground and make them feel “gravity-free.” This is roughly the wind speed experienced in a strong Category 2 hurricane.