Yeah, I've not found it either. Tried a few times. So as a lunchtime task, I've tried to recreate this from first principles:
dc = distance of the average unoccupied cycle parking rack from shop door;
dp = distance of the average unoccupied car parking space from shop door;
W = average walking speed in m/s;
C = average cycling speed in m/s;
P = average driving speed in m/s;
hc = cycling distance between home and shop door;
hp = driving distance between home and shop door.
So time taken to drive+walk to shop = Td = ( (hp - dp) / P ) + ( dp / W )
and time taken to cycle+walk to shop = Tc = ( (hc - dc) / C ) + ( dc / W )
So I want to find the point where the balance tips which is Td = Tc which means:
( (hp - dp) / P ) + ( dp / W ) = ( (hc - dc) / C ) + ( dc / W )
Let's plug some assumptions in to get a rule of thumb!
Average walking speed = W = 3mph = 1.34m/s (it's crossing a car park, so I'm expecting some pauses and detours - I'm using metres and seconds for inputs so that the final outputs should be in metres or seconds);
average cycling speed in town (my track data) = C = 10mph = 4.47m/s;
average driving speed on urban roads = P =
18.6mph = 8.31m/s (2dp);
cycling and driving distances roughly equal because this is the UK and cycleways are as likely to have detours as shortcuts = hc = hp = h.
So:
( (hp - dp) / P ) + ( dp / W ) = ( (hc - dc) / C ) + ( dc / W )
Plug in the assumed numbers:
( (h - dp) / 8.31 ) + ( dp / 1.34 ) = ( (h - dc) / 4.47 ) + ( dc / 1.34 )
Split up the terms:
h/8.31 - dp/8.31 + dp/1.34 = h/4.47 - dc/4.47 + dc/1.34
Subtract h/8.31 from both sides and group factors:
(1/1.34 - 1/8.31)dp = (1/4.47 - 1/8.31)h + (1/1.34 - 1/4.47)dc
Work out the arithmetic:
0.626dp = 0.103h + 0.523dc
Divide both sides by 0.626:
dp = 0.16h + 0.84dc
dp =~= h/6 + 5dc/6
dp - 5dc/6 =~= h/6
So if the cycle parking is very close to the door (so we can ignore the dc term) then it would be quicker to cycle if the average unoccupied car parking space is further from the door than a sixth of the distance from home. Hmmm.
This seems to work in town. Distance from a friend's home to a town centre shop is 1060m. Cycle parking is pretty close to the door. The average unoccupied car parking space would have to be within 176m of the door, but I think it's 220m away. Cycling is definitely quicker, in reality as well as theory.
Also, hitting 18.6mph average driving speed in those streets is very unlikely, so the differential is probably larger. They're not A roads, but I've not found a better estimate of average urban driving speeds - anyone got one?
But the rule of thumb formula doesn't work for me: a shop near me is 5080m away, so I'd need the average car parking space to be 843m away from the shop door. But I know it's only 60m (the shop has its own car park across its service road) and that cycling is roughly similar on time to the door. So am I sacrificing time for the health benefits of cycling or have I got something wrong above?
