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TASER Electronic Control Devices Review Of Safety Literature

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This gives a safety margin of 74:1 = 140 mA ÷ 1.9 mA for the X26 and M26.<br />

Beigelmeier Model: 74:1 <strong>Safety</strong> Margin<br />

Beigelmeier’s model holds that long-term AC (5 seconds or more) requires 1/20 th<br />

of the current to fibrillate as does a single 20 ms (millisecond) pulse (which single<br />

pulse would have to be into the T-wave). 17 The single pulse VF threshold can be<br />

predicted from the Jones and Geddes study which found a ratio of 52:1 for a 10<br />

ms pulse (the widest pulse they investigated). This would give a single 10 ms<br />

pulse a VF threshold of 1.4 A = 52 • 27 mA.<br />

Dividing this by Beigelmeier’s ratio of 20 gives a predicted AC threshold of<br />

70 mA. Doubling this for humans gives a human threshold of 140 mA and a<br />

safety margin of:<br />

74:1 for the <strong>TASER</strong> X26 and <strong>TASER</strong> M26<br />

(which matches the results from the Knickerbocker model)<br />

Dalziel Model: 84:1 <strong>Safety</strong> Margin<br />

In the 1960s Dalziel performed a meta-analysis of VF thresholds for mammals. 38<br />

This is shown in Figure 8. This includes the animal fibrillation data from Ferris<br />

dating back to the 1930s. 39 Conservatively, Dalziel included swine which are very<br />

easy to fibrillate due to subtle cardiac conduction system differences. 40<br />

The average current required to cause fibrillation is given by the formula:<br />

I = 3.68 W + 28.5 mA<br />

Where W is the body weight in kg. For lbs (pounds) the formula is:<br />

I = 1.67 W + 28.5 mA<br />

For a typical excited delirium arrest-related death of 200 lbs (91 kg) 41 the average<br />

current required to fibrillate would be:<br />

I = 362 mA<br />

which gives an average safety margin of 191:1. However, Dalziel recognized that<br />

sensitivities to electrical currents vary and he thus performed a statistical analysis.<br />

He plotted the lower limit of sensitivity as the middle line in the figure. Note<br />

that this middle line gives a fibrillating current of over 170 mA for a 200 lb resisting<br />

human subject. This gives:<br />

<strong>Safety</strong> margin<br />

= 170 mA ÷ 1.9 mA<br />

= 89:1<br />

30

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