*If h is the difference in height between the two sides of the gap, θ is the angle of the exit’s slope, V is the velocity, and g is the standard acceleration of free fall at 9.8 m/s ^{2}, the required velocity would equal to square root of 9.8m/s^{2} *6m^{2 }divided by 2(h-6tanθ)*cos^{2}*

*θ…*

We have now gotten so much conflicting feedback on this equation, that we have come to beg for a crowd consensus. Please excuse poor Illustrator skills.

A car is travelling up a slope and is about to jump a gap.

Yes, no?

Update: everything is funnies to you, isn’t it?

SandyF

I’m a librarian who Is now referring you to the post from the rocket scientist for an answer. As a reader, I can definitely say that I can live with “the car flew over the gap” without the brain breaking math.

Scott

Is the engine in the front or back of the car? Weight distribution will impact how it lands.

JdH

This is an explanation for the math of an Evel Knievel motorcycle jump.

https://youtu.be/FZ0td-SCzPo

Luciana Barroso

A few have already caught the two errors I think are there: in the numerator it should be a 36 (instead of 6) and in the denominator the subtraction needs to be changed to addition. This assumes that the height of the jump-off ramp is higher than that of the landing-ramp and that ‘h’ is the magnitude of that difference in height (and so a positive number). I tried writing up my solution and attaching the picture – hopefully it is readable and understandable.

All of the above models the car as a particle: it has no dimension and is not a deformable body. It also neglects the effects of air drag. Neglecting air drag probably reasonable: will depend on the aerodynamics of the car and likely minor. And I think you can reasonably neglect the deformability of the car (unless the car is very long, I don’t expect any significant bending or twisting of the car to be happening). However, neglecting the dimensions of the car can lead to significant differences in the result, even when treating it as a rigid body. One of the issues is the location of the center of mass of the car – is it close to the geometric center? If not, you’ll see the car rotate about that geometric center – if a car is front-heavy, that is why the front/nose of the car will start to dip relative to the center of the car. This is also all impacted by the wheels – even in mid-flight, as whether they keep spinning or not will impact angular momentum. This now becomes a much more complex problem…

Sivi

+1 I’m really bad at Maths. Your discription makes logical sense for the bits that ican understand- the diagram though is blowing my mind

Evans

I have taught this stuff many times in calculus classes.

It seems that you intend L = 6 m, which you did not mention. If that is so, your formula is almost correct. To fix it, you should change the “minus 6” in the denominator to “+ 6 m”. Then you have a correct formula for the flight of a tiny particle that does not encounter air resistance as it flies exactly from the vertical edge of one wedge to the vertical edge of the other. You might think of applying this to the bottom point on the front wheels. If I were actually going to do fly this gap, in car or motorcycle, I would use the formula, then increase the speed by fifty percent, or maybe even double it, to be safe. Then I would try it with a crash dummy before I tried it with my own body.

Luciana Barroso

Hmmm… I have L=6 defined near the top as one of the givens… I don’t see the negative you mention in the denominator (though it was on the original posted solution, so maybe your reply was to the original post and not to mine?) . It does almost look like there is a negative in the numerator… That is a scanner problem.

And I teach this topic to entering students on a regular basis at the University level. Dynamics is my area of specialization both in teaching as well as research. So this was a fun little review problem for me. 🙂

Jennifer

You need to factor in the weight of the car plus the weight of what/who it is carrying.

Most cars are aerodynamic shaped – decreased air resistance.

NASA has a branch in Texas, their computers could spit out the equation in a blink😃

Anita

The answer is 42. 😁

MargaretTheOther

Yes!!!! “Eddie’s in the space time continuum” will account for anything. (Oh I needed that this morning!!) (Eddie who?)

Billye

LOL! GO DOUGLAS ADAMS! How wheres my towel?

Ryan J Pettis

If we are ignoring wind resistance, centripetal rotation, and the rest of that stuff, and treating this like a simplified physics question then the easiest solution would be as follows:

Let @ represent the angle of the incline, H be the height difference from the end of the incline to the edge of the “landing point”, L be the horizontal distance between the incline and the “landing point”, and V be the speed of the vehicle on the incline, thus Vx is the horizontal speed of the vehicle. By the properties of triangles, Vx = V*cos@. Since the vehicle must travel the distance L within a given amount of time T, based on Vx, we can say Vx*T=L, thus T=L/Vx or T=L/(V*cos@).

Given this, we also know that the vehicle must land at the “landing point” after T seconds, and not below it. Therefore, let Vy be the vertical velocity of the vehicle. By the simple equation for distance (D), we know that D=Dstart + V*T+A*(T^2)/2, where A is the acceleration. In this case, A is gravity with a value of 9.8meters/(second^2), D is H, and Dstart is 0 (zero, though it makes little difference if D is 0 and Dstart is H, or if D is H2 and Dstart is H1, as given in the question), and V is actually Vy, or V*sin@ (by the properties of triangles).

Hence, H=Vy*T-9.8*(T^2)/2, equivalent to H=Vy*T-4.9*(T^2), for simplicity. Substituting V*sin@ for Vy and L/(V*cos@) for T, we get H=V*sin@*L/(V*cos@)-4.9*(L^2)/((V*cos@)^2). Simplifying, this becomes H=L*tan@-4.9*(L^2)/((V^2)*((cos@)^2).

By moving the V components to one side and the rest of the components to the other side gives: 4.9*(L^2)/((V^2)*((cos@)^2))=L*tan@-H. Solving to get V by itself on one side of the equality gives us this result:

V^2 = numerator [4.9*(L^2)] / denominator [L*(sin@)*(cos@) – H*((cos@)^2)]

or: V=sqrt( (9.8*(L^2)) / (2*(L*tan@ – H)*((cos@)^2)) (your answer, * -1)

So in summary, if L=sqrt(6) or your equation is saying “9.8m/s2 * (6m)^2” in the numerator, then our answers of effectively equal, accounting for a factor of -1 in the denominator, most likely due to variable placements in the vertical distance calculation. I’ll also note, it doesn’t matter if the “landing point” is also a slope, or whether it is above or below the end of the initial incline, so long as you are ignoring the centripetal rotation of the vehicle.

I would be more than happy to assist with this further if you would like to discuss it.

All the best.

Norbert

All correct, but your definition of H is inverted from the one that Ilona has given, so H = -h.

Amanda

Precisely the answer I got.

Secador

Me, too… what HE said.

Sherri

I think we neglected momentum …maybe…um…

Sherri

Gah! It’s already in there. Sorry.

Dean

Since h is a drop, the denominator will include the term an addition rather than a subtraction. Also, the numerator will have a 36 rather than a 6 – but you can easily take that out of the square root, to make it 6m * sqrt ( 9.8m/s^2 / [2cos@^2(6tan@ m + h m)] )

My main concern is that the units aren’t carried through in the text. if you start with 36m^2, you should carry on putting the units in for every instance that a length is included.

Heh. That was fun. More! 😀

M

Yes.

From my genius family member:

I’m unsure what the correct answer is, but that equation would leave you with velocity measured in meters^(1/2) per second, so it can’t be correct.

Breann

I’m imagining the car to be an Alfa 4C Spider in red (or maybe black). I have no idea if that’s correct, but that’s the way the “movie” of the jump is playing in my head. Don’t know if the math works, but the jump looks cool! 😊

Gordon

I think it is. We wanted something fast and Italian.

Breann

Yay! I imagined correctly even if I couldn’t math correctly! 😁

Brooke

Short answer: It’s easier to think of the cos as being outside the square root, even if it’s not mathematically simplified.

1 (4.9 * L * L)

v= __________ * sqrt { _________________ }

cos(theta) (h + L * tan (theta) )

I don’t know if this would help, but it’d be easier to reverse the question. How fast is it reasonable for the car to be going at the end of a ramp? How high and far of a jump are you looking at? Then calculate your angle. Or how high is the jump and what kind of angle are you looking at? Then calculate your distance. Since you’re making up the scenario, you can decide what numbers you want and then write the scene with information going backwards.

But then again, I don’t know if you’re looking for the correct FORMULA to write into the book and not just the numbers…

However, there are a couple of assumptions being made with this formula. First that the car ceases all upward movement once it leaves the ramp, and second that the car is a point object (as described by at least one other commentator). The answer above was made using the basic equations of motions.

If you’re looking to see if the car has any upward movement after it leaves the ramp, that’s a whole different can of worms. You would need a couple different sets of equations – the first to find out when upward velocity hits zero, and the second for when gravity takes affect. This would be extremely time consuming given the limited information. Again, it would be a lot easier to go about it backwards from what you’re looking for right now.

Can’t wait to read it and find out how you tackle this. 🙂

Brooke

Arg! The formula looked so good in the comment section!

It should be 1 / cos (theta)

and the square root of

(4.9 * L * L)

/

(h + L * tan (theta) )

Teej

Yeah I was considering that the issues are

-how much time will it take the car to travel the horizontal distance btw ramps?

-how much will the car rise after it leaves the first ramp?

-how much will the car drop before it gets to the 2nd ramp?

And in the real world you would need to answer those questions for the back tires of the car, assuming that the front end of the car does not rotate to be lower than the rear axles. This sounds like a problem for a Stunt Coordinator! Know anybody in Hollywood?

AndrewC

I’m pretty sure my head just exploded.

Meg

Mine too, but the car landed safely just before the math did me in, LOL.

Emily

Mine too, I’m glad that I’m not the only mathematical dunce in the BDH!

Curlyelena

Wow! Cudos for all who tried! Including Ilona and Gordon!

My contribution, aside from admiration, is worthless since I definitely don’t know, but I feel your pain… and will happily wait for it to be solved by someone else 🙂

Elodie

+1 😂😂

Cheryl Chapnick

Your fan base is so smart!! Wow!

Ms. Kim

ditto!

booge

The Mythbusters did this with a bus…

https://www.youtube.com/watch?v=QQE8fYWZMYQ

Jukebox

It worked for Sandy and Keanu with a bus… kinda?

Wenonah Lyon

This reminds me of grade five. word problems. I hated grade five word problems. I’m a grown up. I don’t have to do them anymore. If it’s OK to ignore biology by having 5000 year old evil geniuses, it’s OK to abuse physics?

Heidi

My thoughts exactly

Reenie

Wrong series! Physics abuse happens in the Innkeeper series!

Moira

Practical engineer here. Let’s go build something and test it out!

Brooke

Hahaha! Fun!

Lara

The answer is a flying car from Harry Potter. No math needed then.

Billye

LOL! That’s a good one. Here’s another. Alessandro car is tricked out and custom built, with extra weight where it needs to be to ensure its well balance. The entire structure and suspension system is top quality. This is something the AL’s have already made clear in the series with the comments on how a lot of the vehicles used by the Houses are, despite their seeming normal off the lot looks, are custom made with armor with all of the support structure and suspension required. Rogan’s SUV is a very good example. Alessandro knows he and his car can make the jump because he’s done it before. Either because he is asecret James Bond fan and a daredevil at heart, or because the house insisted on special training which happened to include this… Now how well the car fares on the other side …. Yes, all of this additional weight and everything will affect the numbers but remember, the motor will also have to beefed up to compensate for the extra eight. Even if

We all see jumps like this in the movies that we know, or should know is not really possible in real life with out some EXTRA help. Dukes of Hazzard come to mind has well has a couple of cops shows. But we all are willing to suspend disbelieve for the sake of the story and the action.

Has a final, a kudo to the AL’s for wanting to make sure their formula is correct.

AL QUOTE:

Update: everything is funnies to you, isn’t it?

Life is too short to stress out over the big things, that’s way I always eat desert first. 😛

Secador

Or Chitty-Chitty Bang-Bang.

Jo O

Only just read this and as I gave up Physics and Maths after my O levels, I channeled my inner Tom Petty and started singing – I got as far as Learning to fly but I ain’t got wings when my 5 year old grandson asked how could I fly without wings then answered his own question with: Jetpacks!

Forget the equations go for jetpacks 😀

Debie

All I can say is why would you want to?

Cessie

Hmm the physics break my brain! The only reason I even passed physics at o levels was (1) teacher and (2) being an asian, fail is a 4 letter bad word.

Cessie

I meant to emphasis cute teacher

George Bailey

Dang it , Jim, I’m a chemist, not a physicist.

And this type of thing is why !!!! 😉

Can’t wait to read how it comes out !

Lizz D.

My husband LOVES math and physics… I’ve sent this post his way… we’ll see if he has an answer different than those above. 😉

To all those above, including Ilona and Gordon… congratulations for making my head explode 20 minutes into work. 😉 You guys are awesome… I hate math. xD

Verslint

owwwwwwwww, my brain hurrrrttts

Becky C

It’s your universe. Forget the maths and provide Alessandro with a secondary, here-to-fore unknown, talent.

Wendy S

I can totally get behind this!

Reenie

Well, they have said that lots of primes have secret secondary talents, so this wouldn’t be too far fetched. Kind of a cop out though and the AL don’t usually go for those.

Kathryn

Legend says that When the Star Trek pilot screenplay was reviewed there were a multitude of pages which gave a precise description for a shuttle to get from the Enterprise to the surface of the planet. Filming this scene would have taken up over half the pilot to show. Roddenberry put a pen through the scene and just wrote “the crew transported down to the planet surface”😂

Tyger

Awesome story / urban legend!

Kris Ten-Eyck

The math – it hurts. It HURTS!

Kathryn

As someone who used to watch “The Dukes of Hazzard”, the car makes it across just fine.

Brooke

Nice!

Becky

I can’t help with the math, but….. Mythbusters did try to recreate the Dukes of Hazard car jump in one of their episodes. Special episode 9 – Mega Movie Myths. Wikipedia said the episode came out in 2006. Hopefully that might help with some of the practicalities of what you’re trying to do.

Kate F

https://www.discovery.com/tv-shows/mythbusters/videos/jumpin-a-gorge

free data recovery software download

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Robert Groover

The equation is wrong. Here’s how to think about it. (Simplifying assumptions are at the end.)

1) The car’s speed on the ramp is V. Therefore its vertical velocity Vv is Vsin@, and its horizontal velocity Vh is Vcos@.

2) The hang time t is just the horizontal gap L over the horizontal velocity Vh. Here L = 6m (abt 20 feet).

3) The vertical distance travelled by the car in the air is t*Vv + 4.9*t^2. (Because 9.8 m/s/s is the acceleration of gravity.)

4) So for example, assuming:

a 6 degree angle (which corresponds to a 10.4% grade, or about as steep as you’d ever see on a paved road);

speed of 22 m/s (abt 50 mph);

horizontal gap of 6m (abt 20 feet); then

the vertical rise across the gap is .27m.

For a 12 degree angle (like a severe jeep track) and 30 mph, the rise is about .25m.

For a 6 degree angle at 100 mph, the car can span 137 feet before it comes down to its takeoff height.

Hope this helps – Excel on request.

Rob

Amanda

This would be rather terrifying as, most cars do not have a center of mass in the center of the car. To either have the car tip forward (watching the ground come rushing at you) if the engine is in the front, or to not see the ground at all (if it is in the back). Also is the car front or rear wheel drive? That changes the effective gap by the length of the car.

Ranger

So in future these jumps need to be made in very-well-balanced electric cars. Just don’t split the battery open.

ShellyP

Hi Ilona,

I am a fan who also happens to be a high school physics teacher. I teach regular and AP Physics. As I was reading your blog last night, I really wanted to help. Part of the problem is what you are wanting to accomplish physics-wise in your scene. I found this short tutorial on youtube:

https://www.youtube.com/watch?v=rNvaEnZT0u4

It is only about four minutes long but is exactly about a ramp jumping to another ramp. I think this will help you isolate what you really want to know. Just plug in your numbers 🙂

Hopefully this helps.

Margot D.

To be completely honest, I totally skimmed/skipped this part.

I trust you 100% .

Elena

I found other results in the posts that match mine but with numbers. I left everything with letter si it is as understandable as possible. Hope it helps.

Bat

I will willingly suspend all disbelief on the car making the jump as long as you don’t throw anymore Thetas at us…. promise 😐

Rifraff

I expect this is too late to be of use, but I asked my cousin and he was able to answer right off the top of his head as follows (the blighter!):

It is almost correct. The quick way to see that it isn’t correct is that if you put in h=0 (both are at the same height), the equation turns out to be the square root of a negative number – which isn’t very useful. But someone just made a mistake with the minus sign – it should be plus not minus.

The correct equation is the same except that the bottom line of the fraction should be 2(h + Ltanθ)cos2(superscript not working for some reason?)θ. And there is a typo in the top line of the fraction as it should be gL2(superscript) or 9.8*62(superscript).

Incidentally, to be completely accurate, this is far more a physics questions than a maths question.

Danielle

I don’t do math but I am wondering if you were quoting Squealy Dan from Letterkenny there at the end ?

Richard Hainsworth

Don’t the fancy supercars have aerodynamic shaping to pull the body close to the ground at high speeds so that the wheels get good traction?

Which means that aerodynamics cannot be ignored and your car cannot be modeled as a point mass.

That being said, if the car in flight could be adjusted to act as a wing, it would ‘fly’ much further than a point mass. But also if the front pulls down in flight (due to engine or aerodynamic stablisation), the front wheels would descend much sooner than a point mass only affected by gravity.

Nicole

My brain is so simple that I just assumed all the pretty numbers were right. Flying car. Wheeeee. Feel free to leave it as is ;). I’m also assuming you aren’t going to pancake Lina…

Alex R.

Nope. She’s gone in chapter 4. Arabella takes over.

Of course, they’re not pancaking Catalina!

I don’t think ….