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AntPlusStrideSdm

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For a footpod cumulativeStrides is provided in strides and instantaneousCadence is provided in strides/min. Does this mean we must double the provided values to get the "footfall" values (step count and steps/min), or is this already correct?

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R      

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blackramlabs.com

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Afaik: yes and no.
Why not simply try it out with a reference product?      
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Thanks for the input old_man_biking. My main concern is that, as it is not clearly documented, it has not been considered and may not be consistent across new versions of the Android API.      

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In terms of fitness data, a 'stride' almost always refers to two steps. Additionally, the SDM profiles states "Accumulated strides. This value is incremented once for every two footfalls."      

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Dynastream Developer

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ShaneG - 11 August 2014 09:53 AM
In terms of fitness data, a 'stride' almost always refers to two steps. Additionally, the SDM profiles states "Accumulated strides. This value is incremented once for every two footfalls."


"Almost always" is not useful for developers though. I have assumed that in this case the API will always be for strides = 2 steps, as per the profile documentation, but I would prefer to follow official info on how the API will work.

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R      

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How do we deal with the optional capabilities? Specifically how to identify if 0 cadence or speed is the actual value or just that the device does not support that data type?

From the protocol doc I would expect to request the capabilities, but this does not appear possible. The events (with a 0 value) are occurring when the device does not support that capability.

If this information is not available, my workaround will be to assume the capability is not supported until the first non-zero value. Then use the 0's as "not moving". For the accumulated values a 0 always indicates no support for that capability.

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R      

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Hi R,

We will be adding support for request/decode of the capabilities page in a future release to deal with the ambiguity. In the meantime your current proposal is a perfectly valid solution.

Cheers,
Harrison