# Fastest biking route

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04-23-2012 06:22 AM
New Contributor
Hello

I'am trying to calculate the fastest route for biking between two points, the fastest route should take into account that it is faster to bike downhill than uphill. I know that it is possible to do this on raster data with Spatial Analyst Path Distance tool but does anybody know if it is possible to it with a Network Analyst tool with a network?

/Mathias
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10 Replies
New Contributor
A Newtonian mechanics approach could be used to account for the speed of a bicycle moving down a slope.

Speed of a bike when only considering the degree of the slope and the initial speed of bike is: v�?� =sin�? a�??t +a�?�t +v0

Where v�?� is the speed at the time t, �? the degree of the hill, a�?? the gravitational acceleration on the Earth, a�?� the acceleration from the bicyclist and v0 the speed at the time t = 0.

The interesting part here is that the speed of the bicycle is dependent of the speed at a previous position on the slope. Therefore will a bicyclist after riding down a hill have a larger speed at the bottom than at the top of the hill, this complicates the estimation of the travel time since the resistance from the wind is dependent of the speed of the bike. This means that it will be necessary to calculate the influence from the hill based on many different initial speeds to get an accurate value. One solution could be just to use an average initial value as speed of the bike.

In either case it is necessary to develop an algorithm  based on newton mechanics which then could be used with a DTM model to precalculate hills costs for the different edges in a network which then would be used as all other types of network costs applied to the edges.

An algorithm was not develop in the project we were writing on but it should be possible with an outset in v�?� =sin�? a�??t +a�?�t +v0 and an average speed of a bicyclist in flat terrain to make a estimate for the resistance from the terrain. This will not be 100% perfect but should give a useable approximation.