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**Note:** This page is a list of quick reference tools to help with knowledge of crash investigations! Your use of Crash Math is **"FREE"**.

The information within Crash Math is in all attempts, accurate, and the results of any formula should be verified. In other words considered correct, but unless you have checked and/or done the math yourself and qualified the information provided yourself in that nothing on this page is implied as expert testimony, nor implied as evidentiary, for litigation, or any other form of court use, and by no means a official report.

__**PLEASE ENJOY AND CHECK THE ANSWERS - DO THE MATH! Recommend a Scientific Calculator or a good cell phone App.__## Order of Operations and Mentions:

The order of operations is a rule that tells the correct sequence of steps for evaluating a math expression.

We can remember the order using P.E.M.D.A.S.: **P**arentheses, **E**xponents, **M**ultiplication and **D**ivision (from left to right), **A**ddition and **S**ubtraction (from left to right).

☝ More info - Click to see More in or Examples

###### Example

**Note:**30Df ±1 could look like (30Df ±1) or 30 * D * f ±1

**30 * 120' * 0.8**± 1

**-1**

**√**2879

###### ☝ Click image to enlarge

Consider this, It is not necessary to know any derivation by heart, but it is necessary to know or have a very good understanding of what we are doing. When we apply our work to a profession environment (I.e. Court), remember we must do our work correctly (In other words; Check, check, and check again, then get someone else to check, check, and check again!), show the math, and clearly display our method used in getting our results in our explanation.

Consider this, We need to and should take note of a conversion factor that we will use in nearly every formula derivation.

That is the number 1.466. This number is found by dividing the number of feet in a mile = 5280 feet' in a mile by the total number of seconds in an hour 3600 sec. seconds in an hour. The result gives us 1.4666666, which we will to 1.466.

Note: We do not round any further and must us this result as is and 1.466 is used to convert (Feet per Second)fps. into (Miles Per Hour)Mph. or vice versa. As an example: 60 Mph. times 1.466 equals 88 fps. or 88 fps. divided by 1.466 equals 60 Mph..

We see Velocity (V), which is in fps., is replaced by (S x 1. 466), which changes to fps. to Mph. and we have not altered the equations.

Consider this, we will see 1.6 which = 1.6 seconds(sec.) 1.6 secs. is a representation of a average reaction time that we as humans have when we have no other external or internal impeding stimulus effecting us.

If we don't remember any other thing, it is imperative that we remember this mention. If you don't understand it, contact someone who does and get some assistant and better understanding.
**K _{e} = .05(M)V^{2}** is the formula for energy in motion

Consider this, We we see the numerical value 30 mentioned in expressions, 30 is a derivative. What are Derivatives? They are the result of performing a differentiation process upon a function or an expression. Derivative notation is the way we express derivatives mathematically.

## Conversions:

## Legend (Key):

**″** = Symbol for unit of measure Inches.
**′** = Symbol for unit of measure Feet.
**K _{e} = .05(M)V^{2}**

**K**inetic

**E**nergy formula for energy of motion.

**D**= Distance in feet.

**S**= Speed in miles per hour (Mph.).

**V**= Velocity in feet per second (fps.).

**f**= Drag factor (Coefficient of Friction).

☚ More info.

**f**=

_{adj}__Adj__ust Drag factor (Coefficient of Friction) adjusted.

**t**= Time (In Seconds ≡ Sec.).

**R**= Radius (Feet ').

**C**= Chord (Feet ').

**g**= Acceleration of gravity or 32.2

feet per second squared (32.2 fps.)

**R**= Radius (Adjusted) (Feet ').

_{adj}**H**= Height (Feet ').

**S**= Combined Speed (Mph.).

_{c}**W**= Weight (lbs.) = How much gravity is pulling down on an object.

**n**= % Breaking Efficiency.

☚ More info.

**M** = Mass = How much matter is contained within an object.

of an object divided by the acceleration of gravity or W/g.
**30** = A constant that arises in the process of the derivation.
**15** = ½ of the constant of 30 that arose in the process of the derivation.
**1.466 ** = A conversion factor: 5280 '/3600 sec. in a hour.
**1.6 sec.** = A normal average reaction time in seconds.
**μ** = Coefficient of Friction on Level Surface
__ Note: Here for a level surface μ = f__.

**e**= Road Grade elevation 2% would be converted to .02

**Se**= Super elevation 2% would be converted to .02

**V**= Original Velocity (fps.).

_{o}## Expressions:

** GRADE ADJUSTMENT COEFFICIENT OF FRICTION(μ) AND ROAD GRADE(m)**☚ more info!

✎ The

__level__surface coefficient of friction value (

**μ**) for the road is known and we know the crash occurred on a uphill or downhill grade (

**m**).

So, We have a mini van that weighs(**W**) 4000 lbs, traveling on a level surface, and as it travels it Kinetic Energy (**F**) is 3000 lbs. We need to find μ for our Coefficient of Friction (f).
**μ = F ÷ W** Will work!

☝ More info!

###### Example

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We know that a vehicle skidding downhill will have a reduced(-) less coefficient of friction and a vehicle skidding uphill will have a greater(+) more coefficient of friction.

Consider this, A mini van crash takes place where the road surface is Asphalt and we have a 2% downhill grade.

Using our Asphalt (Well Worn) has a Coefficient of Friction range of .45 - .75. Giving the benefit to the operator (Note: Use the higher decimal) We determined the Coefficient of Friction (μ) is .75 with a Downhill grade of 2%.
**f = µ ± m** Will Work!

**KNOWNS:**

**µ**= .75

**f**= -.02

**RESULT:**

f = µ ± m

**f**= .75 ±

**-**.02 ← Down hill Grade we use negative decimal

**f**= 0.73

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**Special Note:**Finding f = μ ± m like other formulas listed are derived from a more complex algebraic formulas. Being familiar with these other formulas could help you visualize the mathematical progression.

** ADJUSTING COEFFICIENT OF FRICTION(f) BASED ON BREAKING EFFICIENCY(n) **☚ more info!

✎ We determined only two of the the four wheels breaks were working or at 50%, we adjust the Coefficient of Friction (f).

Consider this, So the above equations referred to the grade being level and the breaking system operating at 100%.

What if the, We determined only the front two breaks were working with vehicle involved in our collision and the vehicle was Front Wheel Drive?

We need to get the (**n**) breaking efficiency for our crash vehicle.

**Note:** If there is __one brake (or more__) that was __not working__ on a vehicle, the coefficient of friction __must__ be adjusted to represent the lack of (or diminished) braking efficiency on each wheel.

☝ More info.

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We said the vehicle is FRONT WHEEL DRIVE (FWD) and according to Breaking Efficiency diagram above the LEFT FRONT(LF) tire provides 35% = .35 of the vehicles breaking and the RIGHT FRONT (RF) tire provides 35% = .35 of our vehicle's breaking, and the two rear tires (LR, RR) each provide 15% = .15, but the two rear breaks were Not Working(NW), so they are providing no breaking or 0% = .0.

So, we add 35% = .35 for the LEFT FRONT(LF) tire, 35% = .35 for the RIGHT FRONT(RF) tire, 0% = .0 for the LEFT REAR(LR) tire, and 0% = .0 for the RIGHT REAR(RR) tire.

(**n**) = Total Breaking Percentage or our vehicles Breaking Efficiency is __70% = .70__

**Special Note:**Finding D = Skid Distance (D) like other formulas listed are derived from a more complex algebraic formulas. Being familiar with these other formulas could help you visualize the mathematical progression.

** COEFFICIENT OF FRICTION(f) OF LEVEL SURFACE (μ), PERCENTAGE OF BREAKING(n), AND ROAD GRADE(m)**☚ more info!

✎ When the Coefficient Of Friction(f) needs adjusting knowing (f) = Level Coefficient of Friction (

**μ**) for the percentage of braking (

**n**), and the percentage of the road grade (

**m**).

Our Front Wheel Drive (FWD) vehicle is skidding on a surface (Asphalt) with diminished breaking efficiency due to only the front two breaks working. We also determined the rear breaks are Not Working(NW) at all, and we have a Uphill road grade of 2%.

We know that we have to adjust the Coefficient of Friction (**f**) when one or more breaks are not working, and we have to adjust the Coefficient of Friction (**f**) as well for the road grade (**m**).

Using our "f = Coefficients" chart the Coefficient of Friction (**μ**) for (Asphalt) ** level** (Well Worn and Dry) we get a range of .60 - .75 . Giving the benefit to the operator we use the higher of the range (

**.75**), then using our "n = Breaking Efficiency" chart we have the determined the braking efficiency (

**n**) to have been at 70% or

**.7**, and our Uphill(+) road grade elevation(

**m**) to be 2% or

**.02**

**f = μ(n) ± m**will work.

**KNOWNS:**

**µ**= .75

**n**= .7

**m**= .02

**RESULT:**

**f**= µ(n) ± m

**f**= .75(.7) ± .02

**f**= 0.525 ± .02

**f = 0.545**

☝ More info.

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**Special Note:**Finding f = μn ± m like other formulas listed are derived from a more complex algebraic formulas. Being familiar with these other formulas could help you visualize the mathematical progression.

** MINIMUM SPEED (S) FOR A VEHICLE FROM SKID MARK DISTANCE (D)**☚ more info!

✎ A measured distance (D') of the skid mark in feet', the drag factor (f) is constant and known, and the road grade and super-elevation are appearing to be

__level__.

☛ **Minimum Speed from Skid Marks formula** - When using this formula we can derive the minimum speed of a vehicle from skid marks, again providing us with the __vehicle's minimum speed__.

Unlike __critical speed__ which provides us with the exact speed of the vehicle, minimum speed is just that...The vehicles __minimum__ speed where the skid marks begin.

Simply put, the vehicle could not possibly be going any slower than the calculated minimum speed, but the vehicle's actual speed might be faster.

So...If we have a vehicle that ends up in a collision with a another motor vehicle or a rigid object, and we have measured skid marks, and we assume the drag factor being constant with the road's grade and/or super elevation are __level__.
**S = √30 (D)f** will work.

###### Example

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D_{Avg.} = S_{1} + S_{2} / Number of Skid Marks measured.

D_{Avg.} = 120' + 120' / 2

D_{Avg.} ≡ D = 120'

We've determined that the Asphalt has a Drag Factor of (f = .80). We've measured the tire skid marks and the Distance(D) of the skid marks is D = 120 feet'. Since all four tires were locked and skidding we've determined the vehicle breaking system is working with 100% breaking efficiency, also our road grade and super-elevations are "0".

S = √30 Df

S = √30 * 120' * 0.8

S = √3600' * 0.8

S = √2880'

S = 53.66 Mph ☚ The vehicles minimum speed where skid marks begin.

Consider this, What if the same vehicle skids between different surfaces such as from asphalt, onto concrete, and then onto grass?
The drag factors of *all surfaces* must be measured and included, and then used in the Combined Speed formula.
**S _{C} = √S^{2}_{Asphalt} + S^{2}_{Concrete} + S^{2}_{Grass}**

Consider this, What if in the same vehicle had the left side tires skidding on asphalt and the right side tires skidding on a different surface, lets say the noted grass?

We then utilize a speed formula derived by changing the constant = (30) to ½ that of the constant = (15) and then adding the two different drag factors, one for the Asphalt f_{Asphalt} road for the left tires and the drag factor of the Grass f_{Grass} surface for the right tires.
**S = √15 * (f _{Asphalt} + f_{Grass}) * D**

**Special Note:**The minimum speed formula from skid marks like other formulas listed are derived from a more complex algebraic formula. Being familiar with the derivation of the minimum speed formula could help you visualize the mathematical progression.

** SPEED FROM SKID MARKS WITH DIFFERENT COEFFICIENT OF FRICTIONS(f)**☚ more info!

✎ Use when ½ vehicle's tires skidding on one surface and other ½ tires are skidding on a different surface, the road grade and super-elevation appearing to be

__level__.

Consider this, What if in the same vehicle had the left side tires skidding on asphalt and the right side tires skidding on a different surface, lets say the noted grass?

We then utilize a speed formula derived by changing the constant = (30) to ½ that of the constant = (15) and then adding the two different drag factors, one for the Asphalt f_{Asphalt} road for the left tires and the drag factor of the Grass f_{Grass} surface for the right tires.
**S = √15 * (f _{Asphalt} + f_{Grass}) * D**

**Special Note:**Finding D = Skid Distance (D) like other formulas listed are derived from a more complex algebraic formulas. Being familiar with these other formulas could help you visualize the mathematical progression.

**FINDING f ≡ DRAG FACTOR FOR A VEHICLE USING SKID MARK(S)**☚ more info!

✎ The vehicle Speed (S) is constant, we have our measured Distance (D) of vehicle skid marks with the road grade and super-elevation appearing to be

__level__.

**What is a drag factor (f) in crash investigation?**

Drag Factor is the deceleration coefficient for an entire vehicle.

The Coefficient of Friction is the deceleration coefficient for a sliding tire.

In other Words the coefficient of friction is the amount of friction existing between two surfaces.

**Note:**The Drag Factor and Coefficient of Friction are the same, if, and only if, all four tires of a motor vehicle are locked and sliding on a __level__ surface.

Thus **μ** = f = Coefficient of friction when surfaces are level.

So...Lets use our incident above, the coefficients of friction are the tire's to the surface of the asphalt.

*A low value of coefficient of friction means we have less force that is required for the skids to occur. A higher value means more force is required.* Which means the coefficient of friction (f) is different for concrete and grass.

f

_{Concrete}or f

_{Grass}

☚ More info.

So...We have a vehicle that ends up in a collision with a another motor vehicle, we have four measured skid marks in feet', the vehicle's speed is constant, and the road's grade and/or super elevation appears __level__.
**f = S ^{2} / 30(D)** will work.

###### Example

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So, We need to determine the Coefficient of Friction or the Drag Factor f.

We assume the speed of the vehicle is constant at 50 Mph. We've measured the measured the (Left Front Tire)S_{LF} skid mark @ 120', the (Left Rear Tire)S_{LR} skid mark @ 120', the (Right Front Tire)S_{RF} skid mark @ 120', and the (Right Rear Tire)S_{RR} skid mark @ 120'.

We get the average distance of the skid marks in feet D_{Avg}' by getting the sum of all 4 skid mark distances S_{LF}' + S_{LR}' + S_{RF}' + S_{RR}' and then dividing the result by the number of skid marks (4) and we get an average measured Distance (D) of 120 feet'. Since all four tires were locked and skidding we have determined a 100% breaking efficiency and our road grade and super-elevation are "0".

D_{Avg.'} = S_{LF} + S_{LR} + S_{RF} + S_{RR} / Number of Skid Marks measured.

D_{Avg'.}D_{Avg.'} = D = 120'

f = S

^{2}/ 30(D')

f = 50

^{2}Mph / 30(120')

f = 2500' / 3600'

f = 0.694

Note" we don't have a unit of measure for (f).

**Special Note:**Finding the f = Factor from skid marks like other formulas listed are derived from a more complex algebraic formulas. Being familiar with other formulas like the derivation of the minimum speed formula could help you visualize the mathematical progression.

** SKID DISTANCE (D) KNOWING SPEED(S) & COEFFICIENT OF FRICTION(f)**☚ more info!

✎ The vehicle Speed (S) is constant, we know the coefficient of friction (f) of the surface the skid mark would be on, and the road grade and super-elevation appearing to be

__level__.

**Special Note:**Finding D = Skid Distance (D) like other formulas listed are derived from a more complex algebraic formulas. Being familiar with these other formulas could help you visualize the mathematical progression.

. Last updated on (UTC-5/EST-Eastern Standard Time): May, 05-22-2022 08:01:57 AM.