Let's say we have some mass of propellant m. We can release ALL the energy of this mass as according to mc^2, in some kind of perfect fusion or antimatter reaction. The burned mass is m_b and the unburned mass, into which we dump all the energy from the burned mass as kinetic energy, is m_u.
So
m_u + m_b = m
Relativistic kinetic energy is
KE = (γ-1) m_u c^2
We put all the burned mass's energy into this, giving
KE = m_b c^2.
We have the expression for relativistic momentum,
p = γ m_u v
where v is the exhaust velocity of the fuel.
The Lorentz factor γ is
γ = (1 - (v/c)^2)^-(1/2).
We want to find the maximum momentum for a given amount of fuel.
The above reduces to
p/m = c sqrt(1 - (m_u/m)^2)
So in other words, you get the most momentum out when you convert all of your fuel to pure energy.
Monday, August 15, 2016
Tuesday, June 07, 2016
Velomobiles are fast
So I learned about Velomobiles today.
They're pretty cool! Here's a table of roughly how fast you would go compared to on a normal bike:
They're pretty cool! Here's a table of roughly how fast you would go compared to on a normal bike:
Description | Power | Speed (velomobile) | Speed (normal bike) |
High effort | 400W | 107 km/h | 28 km/h |
Cruising (jogging equivalent) | 200W | 85 km/h | 22 km/h |
Almost no effort (walking equivalent) | 60W | 57 km/h | 15 km/h |
I got these numbers by starting from the Quest velomobile top speed of 107km/h with a high-performance rider and projecting down, and the normal bike speeds from the wikipedia page, projecting up (they match my normal riding patterns pretty well).
You can do your own projections using the following equation (change the initial speed and power ratios as appropriate):
15 km/h * ((400W / 60W)^(1 / 3)) = 28 kilometers per hour