The good thing about it is I have real data to bounce off of the speed calculator as a data checker. Distances of 30 feet to 300 feet, takeoff angles of 11, 16, 22 and 28 degrees, landing angles of 8, 12 and 14 degrees and speeds from 33 to 82 mph.

Again, thanks for the input. I guess my next step is to take a class or two and figure these things out for myself so I don't have to keep bugging you folks.

Joe.

]]>**R** = bv

where R is the resistive force, a vector quantity directed opposite the direction of motion; v is the velocity of the object; and b is a constant (you know, one of those "constants" that varies in every situation) that represents the properties of the medium and shape/size of the object.

R = (1/2) D*p*Av²

where D is the "drag coefficient", *p* (should be a rho) is the density of air, A is the cross-sectional area of the object, and v is again the velocity. This formula is for objects moving at high speeds, such as planes, cars, and meteors.

I am putting together a spreadsheet that will calcualte speeds for this type of jump and I am almost there. The difficlut part is factoring in the wind resistance. Plenty of ballistics formulas out there but none with air resistance.

Again, thank you for your persistance, much appreciated.

Thanks - Joe

]]>You know the ramp's slope measures 130', but along the ground it is only 127', so all slope measurements should be ratiod accordingly.

So the 75' 4'' (75.333') becomes 75.333 x (127/130) = 73.59' (73' 7'')

Add that to the 235' and the total horizontal jump = 235 + 73.59 = 308.59'

]]>I'll try to make it a little more clear by giving the facts of the scenario. This is real world - last Saturday, a motorcycle jumper jumped a claimed 310' 4" for the record. This was measured by running a steel tape along the ground from the leading edge of the takeoff ramp to the leading edge of the landing ramp which measured 235' for the gap meaurement. A mark was made on the landing ramp where the jumper had landed (on the 130' hypotenuse) and this was measured with a steel tape from the top of the landing ramp (28' tall oppostite side - corresponds to measurement taken along the ground from the takeoff to the landing ramps) with a distance of 75' 4" down the slope. Since the actual distance of the jump should be measured along the adjacent only (from the leading edge of the takeoff to the corresponding point on the ground where he landed on the ramp), my argument is that the method used to measure the distance is incorrect. I need to prove this with a formula that accurately determines the true distance travelled with the data gathered in the method that it was recorded. Trig classes from over 20 years ago tells me they are wrong but I don't have the smarts to translate that into a formula. He still has the record (previous record of 277.5' measured the same way) but I would like to see accurate measurements.

Thanks again,

Joe

(wish I had a nifty saying)

]]>So, you have solved the right triangle using Pythagoras Theorem **a²+b²=c²**, but you now want another measurement, correct?

BTW A sketch would be handy - you can upload it with your message (press Post reply and look for the upload slots selection)

But I think I can understand what you want ...

First let's solve the big triangle:

a²+b²=c² =⇒ a = √(130² - 28² ) = √(130² - 28² ) = 127

Is that what you got?

Now, if you measure down the hypotenuse 75' you would be 127-75 = 52' up the slope.

Now you have a new (smaller) triangle that has an adjacent of 52', and the same angle in the corner, so you could just ratio off the lengths.

So, the smaller triangles hyptenuse would be (52/127) * 130 = 53.2

And the smaller triangles opposite would be (52/127) * 28 = 11.5

Or then again, I may have got this all mixed up

]]>Thanks.

]]>I managed to figure out the overall adjacent length with the known opposite and hypotenuse lengths but can't get that last part.

Any help at all would be most appreciated. Thanks in advance.

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