Outfield Fence Calculation

[SB23]

is there a formula to determine the distance needed to clear a fence ?

Example - the distance a ball needs to be hit to clear a 4' fence which is 200' away - Is there a multiplier as the fence gets taller ?

 

[CoogansBluff]

It would depend somewhat on trajectory. Technically, you can get a ball over by hitting it 200 feet and a few inches.

 

[SoAZDad]

need to measure a couple more variables, trajectory and velocity, then it can be calculated.

 

[Greenmonsters]

The Pythagorean theorem is an equation that relates the three sides of a right triangle (i.e., a triangle with one 90 degree angle)

a squared + b squared = c squared

(200 x 200) + (4 x 4) = c squared

40000 + 16 = c squared

200.039 ft = c

For a 20-foot high 200 ft fence, c= 200.99 ft

Based on these 2 examples, it is reasonable to assume that the distance the ball must travel to clear the fence is the distance to the fence plus a smidgen

 

[SB23]

The Pythagorean theorem is an equation that relates the three sides of a right triangle (i.e., a triangle with one 90 degree angle)

a squared + b squared = c squared

(200 x 200) + (4 x 4) = c squared

40000 + 16 = c squared

200.039 ft = c

For a 20-foot high 200 ft fence, c= 200.99 ft

Based on these 2 examples, it is reasonable to assume that the distance the ball must travel to clear the fence is the distance to the fence plus a smidgen

That smidgeon is the # I am looking for

 

[BobInMadison]

The Pythagorean theorem is an equation that relates the three sides of a right triangle (i.e., a triangle with one 90 degree angle)

a squared + b squared = c squared

(200 x 200) + (4 x 4) = c squared

40000 + 16 = c squared

200.039 ft = c

For a 20-foot high 200 ft fence, c= 200.99 ft

Based on these 2 examples, it is reasonable to assume that the distance the ball must travel to clear the fence is the distance to the fence plus a smidgen

Leave it to someone called Greenmonsters to discuss how to hit a ball over a high fence.

 

[SoAZDad]

That smidgeon is the # I am looking for

try this: Baseball HR Simulator -- Java Applet

 

[guero_gordo]

If you have a Mac, you can open Terminal and use this. 200' and 20' fence, (assuming the ball launches from the point of the plate).

echo "scale=4; sqrt( (200)^2 + (20)^2 )" | bc

If you want to assume the ball was contacted 2' off the ground and 7"in front of the plate:

echo "scale=4; sqrt( (200-2)^2 + (20-2)^2 )" | bc

As mentioned above, this is only part of the story, since balls have to travel a lot higher and further to clear a 20' fence even though the top is only 1' further away.

 

[Sparky Guy]

That smidgeon is the # I am looking for

The smidgeon you're looking for can be found by doing the following calculation.

Cost of HOTTEST bat around + Cost of hitting lessons + Hitting cage time + Bucket of Balls * Pitch speed * Bat Speed - Head wind / Fence height = Smidgeon

After extensive studies it's been found the force of the sound waves from the cheering parents has a negligible effect on the Smidgeon factor. So it was not included into the calculation.

 

[Greenmonsters]

Leave it to someone called Greenmonsters to discuss how to hit a ball over a high fence.

Fenway - plate to fence along left field foul line (measured, not reported) = 304 ft, the Greenmonstah is 37 ft tall, c= 306.24 ft